Fundamentals of Materials Science and Engineering An Integrated Approach,5th Edition William D. Call

SOLUTION MANUAL

Reserve Questions and Problems Solutions Manual CHAPTER 1 INTRODUCTION

Reserve Problem 01a (question pool) Materials Science and Engineering is the study of material behavior & performance and how this is simultaneously related to structure, properties, and processing. Which of the following is the best example of a material property? (a) Density (b) Annealing (c) Forging (d) Single-crystal (e) Crystalline

Solution: Material structure refers to the arrangement of a material’s constituent atoms, ions, molecules, etc.

Material properties are typically either conceptual traits or specific numeric values that describe a material’s response to applied stimuli. A few examples of numeric property examples are density, elastic modulus, hardness, tensile strength, and electrical conductivity. A few conceptual properties include brittle, ductile, and toughness.

Material processing refers to conditions, stimuli, and situations that a material is exposed to, such as manufacturing steps such as melting & casting, annealing, forging, rolling, cutting, extrusion. It is often convenient to consider service conditions as processing parameters as well, since factors like operating temperature, ambient atmosphere, humidity, etc. may further influence a material’s structure, properties, and performance.

Material performance & behavior typically pertains to how a material responds while in service.

Example: A polycrystalline (structure) steel specimen that has been carburized (processing) should exhibit a higher surface hardness (property) compared to un-carburized steel. The surface of the carburized steel specimen will therefore be more difficult to cut or abrade (performance / behavior).

A thin steel specimen that has been quenched from a glowing red-hot temperature (processing) is expected to form martensite (structure), the hardest and most brittle form of steel (property). If the quenched specimen is subsequently loaded, it will not exhibit much strain prior to failure (performance / behavior).

In this case (b) and (c) are process options, (d) is a conceptual property, and (e) is a structural description.

Reserve Problem 02a (question pool) Materials Science and Engineering is the study of material behavior & performance and how this is simultaneously related to structure, properties, and processing. Which of the following is the best example of material processing? (a) Extrusion (b) Crystalline (c) Amorphous (d) Glassy (e) Elastic Modulus

Solution: Material structure refers to the arrangement of a material’s constituent atoms, ions, molecules, etc.

Material properties are typically either conceptual traits or specific numeric values that describe a material’s response to applied stimuli. A few examples of numeric property examples are density, elastic modulus, hardness, tensile strength, and electrical conductivity. A few conceptual properties include brittle, ductile, and toughness.

Material processing refers to conditions, stimuli, and situations that a material is exposed to, such as manufacturing steps such as melting & casting, annealing, forging, rolling, cutting, extrusion. It is often convenient to consider service conditions as processing parameters as well, since factors like operating temperature, ambient atmosphere, humidity, etc. may further influence a material’s structure, properties, and performance.

Material performance & behavior typically pertains to how a material responds while in service.

Example: A polycrystalline (structure) steel specimen that has been carburized (processing) should exhibit a higher surface hardness (property) compared to un-carburized steel. The surface of the carburized steel specimen will therefore be more difficult to cut or abrade (performance / behavior).

A thin steel specimen that has been quenched from a glowing red-hot temperature (processing) is expected to form martensite (structure), the hardest and most brittle form of steel (property). If the quenched specimen is subsequently loaded, it will not exhibit much strain prior to failure (performance / behavior).

In this case (a) is a process, (b), (c), and (d) are structural descriptions and (e) is a property.

Reserve Problem 03a (question pool) Materials Science and Engineering is the study of material behavior & performance and how this is simultaneously related to structure, properties, and processing. Which of the following is the best example of material structure? (a) Single-phase (b) Elastic Modulus (c) Sintering (d) Magnetic Permeability (e) Brittle

Solution: Material structure refers to the arrangement of a material’s constituent atoms, ions, molecules, etc.

Material properties are typically either conceptual traits or specific numeric values that describe a material’s response to applied stimuli. A few examples of numeric property examples are density, elastic modulus, hardness, tensile strength, and electrical conductivity. A few conceptual properties include brittle, ductile, and toughness.

Material processing refers to conditions, stimuli, and situations that a material is exposed to, such as manufacturing steps such as melting & casting, annealing, forging, rolling, cutting, extrusion. It is often convenient to consider service conditions as processing parameters as well, since factors like operating temperature, ambient atmosphere, humidity, etc. may further influence a material’s structure, properties, and performance.

Material performance & behavior typically pertains to how a material responds while in service.

Example: A polycrystalline (structure) steel specimen that has been carburized (processing) should exhibit a higher surface hardness (property) compared to un-carburized steel. The surface of the carburized steel specimen will therefore be more difficult to cut or abrade (performance / behavior).

A thin steel specimen that has been quenched from a glowing red-hot temperature (processing) is expected to form martensite (structure), the hardest and most brittle form of steel (property). If the quenched specimen is subsequently loaded, it will not exhibit much strain prior to failure (performance / behavior).

In this case (a) is a structural description, (b), (d), and (e) are properties, and (c) is a process.

Reserve Problem 04a (question pool) Which class of material is generally associated with the highest density values at room temperature? (a) Composites (b) Ceramics (c) Metals (d) Polymers

Solution: Metals tend to be the most dense material class because (i) the atoms are generally heavy and (ii) the heavy atoms tend to pack fairly efficiently Reserve Problem 05a (question pool) By how many orders of magnitude (powers of ten, approximately) does density vary for metals? (a) 0.13 (b) 1.3 (c) 13 (d) 130

Solution: 3

3

Fig. 1.4 indicates that the density of metals ranges from about 1 g/cm to about 30 g/cm .

3

The change from 1 to 10 g/cm is 1 order of magnitude of variation.

3

The change from 10 to 30 g/cm about 0.3 order of magnitude.

The variation in density for metals is therefore about 1.3 orders of magnitude.

Reserve Questions and Problems Solutions Manual CHAPTER 2 ATOMIC STRUCTURE AND INTERATOMIC BONDING

Reserve Question 01: Atomic mass The atomic mass of an atom may be expressed as the sum of the masses of • • •

Electrons Neutrons Protons

Choose all that apply.

Solution The atomic mass of an atom may be expressed as the sum of the masses of protons and neutrons within the nucleus.

Reserve Question 02: Atomic nucleus The nucleus of an atom contains • • •

Electrons Neutrons Protons

Choose all that apply.

Solution An atom consists of a very small nucleus that contains protons and neutrons (surrounded by moving electrons).

Reserve Question 03: Atomic number The atomic number of an electrically neutral atom is equal to the number of: • • •

protons electrons neutrons

Choose all that apply.

Solution The atomic number of an electrically neutral atom is the number of protons, which is equal to the number of electrons.

Reserve Problem 04 174

Hafnium has six naturally occurring isotopes: 0.16% of Hf, with an atomic weight of 173.940 amu; 176 177 5.26% of Hf, with an atomic weight of 175.941 amu; 18.60% of Hf, with an atomic weight of 178 179 176.943 amu; 27.28% of Hf, with an atomic weight of 177.944 amu; 13.62% of Hf, with an atomic 180 weight of 178.946 amu;. and 35.08% of Hf, with an atomic weight of 179.947 amu. Calculate the average atomic weight of Hf. Give your answer to three decimal places.

Solution The average atomic weight of halfnium A Hf is computed by adding fraction-of-occurrence—atomic weight products for the six isotopes—i.e., using Equation 2.2. (Remember: fraction of occurrence is equal to the percent of occurrence divided by 100.) Thus

AHf = f174 A174 + f176 A176 + f177 A177 + f178 A178 + f179 A179 + f180 A180 Hf

Hf

Hf

Hf

Hf

Hf

Hf

Hf

Hf

Hf

Hf

Hf

Including data provided in the Chapter 02, Problem statement we solve for A Hf as

AHf = (0.0016)(173.940 amu) + (0.0526)(175.941 amu) + (0.1860)(176.943 amu) +(0.2728)(177.944 amu) + (0.1362)(178.946 amu) + (0.3508)(179.947) = 178.485 amu Reserve Problem 05 79

81

Bromium has two naturally occurring isotopes: Br, with an atomic weight of 78.918 amu, and Br, with an atomic weight of 80.916 amu. If the average atomic weight for Br is 79.903 amu, calculate the fraction-of-occurrences of these two isotopes.Give your answer to three decimal places.

79

Fraction-of-occurency for Br: _____ 81

Fraction-of-occurency for Br: _____

Solution The average atomic weight of indium ( ABr ) is computed by adding fraction-of-occurrence—atomic weight products for the two isotopes—i.e., using Equation 2.2, or

ABr = f79 A79 + f81 A81 Br

Br

Br

Br

Because there are just two isotopes, the sum of the fracture-of-occurrences will be 1.000; or f79 + f81 = 1.000 Br

Br

which means that

= 1.000 − f81

f79

Br

Br

Substituting into this expression the one noted above for f79 , and incorporating the atomic weight Br

values provided in the problem statement yields

79.903 amu = f79 A79 + f81 A81 Br

(

79.903 amu = 1.000 − f81

Br

Br

(

Br

Br

)A A 81Br

79Br + f81Br A81Br

) (78.918 amu) + f (80.916 amu) 80.916 amu = 78.918 amu − (78.918 amu) ( f ) + (80.916 amu) ( f ) 79.903 amu = 1.000 − f81

81Br

Br

81Br

Solving this expression for f81

Br

yields f81

Br

81Br

= 0.493 . Furthermore, because

f79 = 1.000 − f81 Br

Br

then

f79 = 1.000 − 0.493 = 0.507 Br

Reserve Problem 06: Electron configuration 2

2

6

An element that has the electron configuration 1s 2s 2p has how many electrons?

Enter numeric values only.

Solution Both the 1s and 2s shells are filled; therefore, they contain a total of 4 electrons (2 electrons for each s orbital). Thus the total number of electrons in the atom is 4 plus the number of electrons in the 2p shell—in this case 5—for a total of 9.

Reserve Problem 07: The four electron subshells M, K, L, N Which subshells are found in each of the following shells? a) Electron subshells -M shell • s • p • d • f b) Electron subshells – K shell • s • p • d • f c) Electron subshells – L shell • s • p • d • f d) Electron subshells – N shell • s • p • d • f Solution: a) For the M shell, the value of the quantum number n is 3 (as noted in the first paragraph under the Quantum Numbers subsection). From Table 2.1, for n = 3, s, p, and d subshells are possible. b) For the K shell, the value of the quantum number n is 1. From Table 2.1, for n = 1, the s subshell is possible. c) For the L shell, the value of the quantum number n is 2. From Table 2.1, for n = 2, s and p subshells are possible. d) For the N shell, the value of the quantum number n is 4. From Table 2.1, for n = 4, s, p, d, and f subshells are possible.

Reserve Problem 08: The number of electrons in subshells M, K, L, N What is the maximum number of electrons that each of the following shells can contain? (a) M shell

____

electrons

(b) K shell

____

electrons

(c) L shell

____

electrons

(d) N shell

____

electrons

Solution: (a) For the M shell, the value of the quantum number n is 3 (as noted in the first paragraph under the Quantum Numbers subsection). From Table 2.1, for n = 3, s, p, and d subshells are possible. (b) For the K shell, the value of the quantum number n is 1. From Table 2.1, for n = 1, the s subshell is possible. (c) For the L shell, the value of the quantum number n is 2. From Table 2.1, for n = 2, s and p subshells are possible. (d) For the N shell, the value of the quantum number n is 4. From Table 2.1, for n = 4, s, p, d, and f subshells are possible.

Reserve Question 09: The electrons that occupy the outermost…

The electrons that occupy the outermost filled shell are called ____ electrons.

Solution: The electrons that occupy the outermost filled shell are called valence electrons.

Reserve Question 10: When all the electrons in an atom occupy…

When all the electrons in an atom occupy the lowest possible energy states, the atom is said to be in its: -ground state -ionized state -cold state -regular state

Solution: When all the electrons in an atom occupy the lowest possible energy states, the atom is said to be in its ground state.

Reserve Problem 11 How many p electrons at the outermost orbital do the Group VIIA elements have?

Solution Each of the elements in Group VIIA has five p electrons.

Reserve Problem 12 To what group in the periodic table would an element with atomic number 119 belong?

• • • •

Group 0 (or 18) Group IA (or 1) Group IIA (or 2) Group VIIA (or 17)

Solution From the periodic table (Figure 2.8) the element having atomic number 119 would belong to group IA. Reserve Problem 13 Ideally speaking, bonds tend to form between two particles such that they are separated by a distance where _______ net force is exerted on them, and their overall energy is _________. (a) a positive, maximized (b) a negative, maximized (c) a positive, minimized (d) a positive, minimized (e) zero, maximized (f) zero, minimized

Solution: The force between (and therefore the energy of) two interacting particles is a function of their separation distance. The separation distance associated with the equilibrium bond distance is the separation associated with zero net force on the particles and a corresponding minimum of system energy.

Reserve Problem 14: Attraction energy Calculate the energy of attraction between a cation with a valence of +2 and an anion with a valence of –2, the centers of which are separated by a distance of 3.7 nm.

Solution The attractive energy between positive and negative ions may be calculated using Equation 2.9, that is EA = −

A r

Furthermore, the value of A is given in Equation 2.10 as

A=

1 4πε 0

( Z1e )( Z2e )

For this problem we will assume that Z1 = +2 and Z2 = –2; also form the problem statement that r = –9 3.7 nm = 3.7 × 10 m. Finally, the value of the attractive energy is computed as follows:

EA = − =−

1 4πε 0 r

( Z1 e )( Z2 e )

( +2 [1.602 × 10 F/m)(3.7 × 10 m)

1 (4)(π )(8.85 × 10

−12

−9

−19

)(

C] −2 [1.602 × 10 −19 C]

)

= −2.5 × 10 −19 J Reserve Problem 15: Attraction force Calculate the force of attraction (in N) between a cation with a valence of +2 and an anion with a valence of –3, the centers of which are separated by a distance of 8.6 nm.

Solution:

To solve this problem for the force of attraction between these two ions it is necessary to use Equation 2.13, which takes on the form of Equation 2.14 when values of the constants e and ε0 are included—that is

FA =

( )( Z2 )

(2.31 × 10 −28 N ⋅ m2 ) Z1 r2

If we take ion 1 to be the cation that has a charge of +2 and ion 2 to be the anion with charge –3; –9 also, from the problem statement r = 8.6 nm = 8.6 × 10 m. Thus, using Equation 2.14, we compute the force of attraction between these two ions as follows:

FA =

( )( −3 ) = 1.9 × 10−11 N

(2.31 × 10 −28 N ⋅ m2 ) +2 (8.6 × 10 −9 m)2

Reserve Problem 16 _______ bonding is similar to ionic bonding, except there are no high-electronegativity atoms present to accept any electrons that the present atoms are willing to donate.

(a) Ionic (b) Covalent (c) Metallic (d) Hydrogen

Solution: Metallic bonding occurs when low-electronegativity atoms (metal atoms) give away some of their electrons, which are then free to move within the bulk specimen. These “free” electrons no longer “belong” to a certain atom and are therefore described to be “delocalized” since they don’t tend to spend time near any particular atom. These electrons remain free because there are no high-electronegativity atoms (non-metals) present that might otherwise incorporate these electrons into their valence shells.

Reserve Problem 17 _________ bonds are the only primary bonds that are directionally dependent.

(a) Covalent (b) Metallic (c) Ionic (d) Van der Waals

Solution: A covalent bond arises when electrons from two adjacent atoms simultaneously spend time in valence shells of the bonding atoms. As electrons only exist in certain probabilistic volumes (molecular orbitals), covalent bonds can only be established at certain locations at the periphery of an atom. In other words, covalent bonds are established along vectors where electron sharing occurs.

The specific types of orbitals that are present in the bonding atoms will also dictate specific bonding 3 angles and therefore directional constraints on the bond. Diamond, for instance, consists of sp o 3 hybridized carbon atoms that bond at 109.5 angles due to the tetrahedral orbital configuration of sp 2 o hybridization. Graphite, on the other hand, consists of sp hybridized carbon atoms that bond at 120 2 angles due to the trigonal orbital configuration of sp hybridization.

It is worth noting that a given ionic bond is not directionally dependent, in that the bond can by formed in an infinite number of symmetrically equivalent ways when considering the bond as a tangential contact of the anion and cation spheres. Other factors such as charge neutrality and likecharge proximity impose constraints on how large numbers of ions pack as a crystal, but these factors do not alter the fact that a given ionic bond is “radially independent”.

Reserve Problem 18 _________ bonds are responsible for binding atoms together within a molecule of propane, whereas ________ bonds bind separate propane molecules together in a condensed state (liquid or crystal).

(a) Covalent, Ionic (b) Ionic, Covalent (c) Covalent, Metallic (d) Metallic, Covalent (e) Covalent, Van der Waals (f) Van der Waals, Covalent (g) Ionic, Van der Waals (h) Van der Waals, Ionic

Solution: The atoms of a molecule are generally held together by covalent bonds, since electron sharing is occurring between adjacent bonding atoms in the molecule. A form of secondary bonding (Van der Waals) bonding is responsible for binding separate molecules together.

Reserve Problem 19 Materials whose constituent particles are bound by which type of bond are generally expected to have the lowest melting temperatures?

(a) Covalent (b) Metallic (c) Ionic (d) Van der Waals (e) Hydrogen

Solution: Van der Waals interactions are generally the weakest bond type. Therefore, materials whose particles are bound together by this bonding will tend to exhibit inferior mechanical and thermal properties (e.g. lower elastic modulus, lower melting temperature).

Reserve Problem 20: Ionic character Using the figure below, calculate the percent ionic character of the interatomic bonds for the following materials:

(a) CaCl2 (b) CsBr

Solution

The percent ionic character is a function of the electronegativities of the ions XA and XB according to 2 %IC = ⎡1 − exp −0.25( X A − X B ) ⎤ × 100 ⎢⎣ ⎥⎦

(a) For CaCl2, XCa = 1.0 and XCl = 3.0, such that %ICCaCl = ⎡⎢1 − exp −0.25(3.0 −1.0) ⎤⎥ × 100 = 63.2% 2 ⎣ ⎦ 2

(b) For CsBr, XCs = 0.7 and XBr = 2.8 and, therefore, 2 %ICCsBr = ⎡⎢1 − exp −0.25(2.8−0.7) ⎤⎥ × 100 = 66.8% ⎣ ⎦

Reserve Problem 21

(a) Calculate %IC of the interatomic bonds for the intermetallic compound TiAl3. (b) On the basis of this result what type of interatomic bonding would you expect to be found in TiAl3?

• • • •

van der Waals ionic metallic covalent

Solution

(a) The percent ionic character is a function of the electron negativities of the ions XA and XB according to Equation 2.16. The electronegativities for Al and Ti (Figure 2.9) are both 1.5 and.

Therefore, the percent ionic character is determined using Equation 2.16 as follows: %IC = ⎡1 − exp( −0.25)(1.5 − 1.5)2 ⎤ × 100 = 0 % ⎣ ⎦

(b) Because the percent ionic character is zero and this intermetallic compound is composed of two metals, the bonding is completely metallic.

Reserve Questions and Problems Solutions Manual CHAPTER 3 STRUCTURES OF METALS AND CERAMICS

Reserve Problem 01a (question pool) Which structure is most consistent with a polycrystalline structure? (a)

(b)

(c)

(d)

Solution (a) and (b) depict single crystals, since there is a single pattern throughout these arrangements. (c) depicts a non-crystalline arrangement, since no long-range periodicity or symmetry (d) depicts an aggregate of separate single crystals with varying orientation, which is the definition of a polycrystal.

Reserve Problem 02a (question pool) Which structure is most consistent with an amorphous structure? (a)

(b)

(c)

(d)

Solution (a) and (b) depict single crystals, since there is a single pattern throughout these arrangements. (c) depicts a non-crystalline arrangement, since no long-range periodicity or symmetry (d) depicts an aggregate of separate single crystals with varying orientation, which is the definition of a polycrystal.

Reserve Problem 03 Which of the following microstructures is expected to be most similar to a single crystal in terms of structure and properties? Assume all of the options offer the same volumes and only consider grain boundaries as a crystalline defect for this question. (a) Textured polycrystal with about 10,000 grains (b) Random polycrystal with about 1,000,000 grains (c) Random polycrystal with about 1,000,000,000 grains (d) Amorphous Solution: A single crystal can be described as a single grain with one crystallographic orientation in space.

Therefore, the option that has the fewest and most oriented grains will be most similar to the single crystal. In this case the textured polycrystal also features the lowest grain count so both characteristics are a better match compared to the random polycrystal options, which both feature larger grain counts. Amorphous configurations are, structurally, the least similar to a single crystal.

Reserve Problem 04a (question pool) Which of the following is the primitive unit cell for the lattice that is depicted?

Solution: The parallelogram (c) is correct since it is a valid unit cell and fulfills the requirement of being primitive.

A primitive unit cell features one lattice point total within the unit cell, which requires that the unit cell contains partial lattice points at its corners only. All of the other options are valid unit cells, but they are not primitive, as they feature more than one lattice point per unit cell.

Reserve Problem 05a (question pool) Which of the following is candidate unit cells is invalid?

Solution: Candidate (c) is invalid because it is not able to produce full / complete lattice points via integer translations along its lattice vectors (axes parallel to the edges of the triangle). In this configuration specifically, the top, lower left, and lower right sixths each circle would never be generated.

Reserve Problem 06a (question pool) Which of the following unit cells exhibits the highest symmetry?

Solution: Unit cell (a) features the highest symmetry in this case. Specifically, it features 6-fold symmetry because it is a hexagon and therefore encounters six equivalent settings upon one complete rotation through 360 degrees.

The other options feature two-fold symmetry as they only register an equivalent position for every 180 degrees of rotation.

Reserve Problem 07a (question pool) Which of the following partial lattices does not exhibit 6-fold symmetry? (a)

(b)

(c)

(d)

Solution: The partial lattice (a) does not feature 6-fold symmetry, but rather exhibits 2-fold as its highest symmetry. The other options are all hexagonal lattices with varying perimeters. Note that a hexagonal lattice (a 6-fold lattice) exhibits 2- and 3-fold symmetry by definition.

Reserve Problem 08 Using atomic weight, crystal structure, and atomic radius data tabulated inside the front cover, compute the theoretical densities of lead, chromium, copper, and cobalt, and then compare these values with the measured densities listed in this same table. The c/a ratio for cobalt is 1.623.

Solution Since Pb has an FCC crystal structure, n = 4, and VC = 16R3 2 (Equation 3.6). Also, R = 0.175 nm –8

(1.75 × 10 cm) and APb = 207.2 g/mol. Employment of Equation 3.8 yields

ρ= =

nAPb VC N A

(4 atoms/unit cell)(207.2 g/mol)

{⎡⎣(16)(1.75 × 10 cm) ( 2 )⎤⎦ /(unit cell) }(6.022 × 10 atoms/mol) −8

3

23

= 11.35 g/cm3 3

The value given in the table inside the front cover is 11.35 g/cm .

3

⎛ 4R ⎞ Chromium has a BCC crystal structure for which n = 2 and VC = a = ⎜ ⎟ (Equation 3.4); also, ⎝ 3⎠ 3

ACr = 52.00 g/mol and R = 0.125 nm. Therefore, employment of Equation 3.8 leads to

ρ=

(2 atoms/unit cell)(52.00 g/mol) 3 ⎧⎡ ⎫ ⎪ (4)(1.25 × 10 −8 cm) ⎤ ⎪ 23 ⎥ /(unit cell) ⎬ (6.022 × 10 atoms/mol) ⎨⎢ 3 ⎪⎩ ⎣⎢ ⎪⎭ ⎦⎥ = 7.18 g/cm3 3

The value given in the table is 7.19 g/cm .

Copper also has an FCC crystal structure and therefore

ρ=

(4 atoms/unit cell)(63.55 g/mol) 3 ⎧⎡ ⎫ −8 23 ⎨ ⎣(2)(1.28 × 10 cm)( 2 ) ⎤⎦ /(unit cell ) ⎬ (6.022 × 10 atoms/mol) ⎩ ⎭ = 8.90 g/cm3 3

The value given in the table is 8.90 g/cm .

Cobalt has an HCP crystal structure, and from the solution to Chapter 03, Problem 3.6 (Equation 3.S1),

VC = 6R2c 3

and, since c = 1.623a and a = 2R, c = (1.623)(2R); hence

VC = 6R2 (1.623)(2 R) 3 = (19.48)( 3) R3 = (19.48)( 3)(1.25 × 10−8 cm)3 = 6.59 × 10−23 cm3 /unit cell Also, there are 6 atoms/unit cell for HCP. Therefore, the theoretical density is

ρ= =

nACo VC N A

(6 atoms/unit cell)(58.93 g/mol)

(6.59 × 10

−23

cm3 /unit cell)(6.022 × 1023 atoms/mol) = 8.91 g/cm3 3

The value given in the table is 8.9 g/cm .

Reserve Problem 09a-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a square when they are connected by lines through their centers. Cubic unit cells cut along {100} planes will feature this shape. The edges of the square are edges of the cube, so they are <100> directions. Atoms are touching along edges of the cube. The cubic structure that features close-packing along cube edges is simple cubic.

Reserve Problem 09b-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a square when they are connected by lines through their centers. Cubic unit cells cut along {100} planes will feature this shape. The edges of the square are edges of the cube, so they are <100> directions. Atoms are touching along edges of the cube. The cubic structure that features close-packing along cube edges is simple cubic.

Reserve Problem 09c-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a square when they are connected by lines through their centers. Cubic unit cells cut along {100} planes will feature this shape. The edges of the square are edges of the cube, so they are <100> directions. Atoms are touching along edges of the cube. The cubic structure that features close-packing along cube edges is simple cubic.

Reserve Problem 09d-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a square when they are connected by lines through their centers. Cubic unit cells cut along {100} planes will feature this shape. The edges of the square are edges of the cube, so they are <100> directions. Atoms are touching along edges of the cube. The cubic structure that features close-packing along cube edges is simple cubic.

Reserve Problem 10a-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a square when they are connected by lines through their centers. Cubic unit cells cut along {100} planes will feature this shape. The edges of the square are edges of the cube, so they are <100> directions. We see no atoms touching along any vectors of that lie in the {100} plane. The cubic structure that features no close-packing on cube faces is BCC.

Reserve Problem 10b-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a square when they are connected by lines through their centers. Cubic unit cells cut along {100} planes will feature this shape. The edges of the square are edges of the cube, so they are <100> directions. We see no atoms touching along any vectors of that lie in the {100} plane. The cubic structure that features no close-packing on cube faces is BCC.

Reserve Problem 10c-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a square when they are connected by lines through their centers. Cubic unit cells cut along {100} planes will feature this shape. The edges of the square are edges of the cube, so they are <100> directions. We see no atoms touching along any vectors of that lie in the {100} plane. The cubic structure that features no close-packing on cube faces is BCC.

Reserve Problem 10d-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a square when they are connected by lines through their centers. Cubic unit cells cut along {100} planes will feature this shape. The edges of the square are edges of the cube, so they are <100> directions. We see no atoms touching along any vectors of that lie in the {100} plane. The cubic structure that features no close-packing on cube faces is BCC.

Reserve Problem 11a-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a square when they are connected by lines through their centers. Cubic unit cells cut along {100} planes will feature this shape. The edges of the square are edges of the cube, so they are <100> directions. We see that atoms are touching along face diagonals of that lie in the {100} plane. The cubic structure that features close-packing along face diagonals of the cube is FCC.

Reserve Problem 11b-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a square when they are connected by lines through their centers. Cubic unit cells cut along {100} planes will feature this shape. The edges of the square are edges of the cube, so they are <100> directions. We see that atoms are touching along face diagonals of that lie in the {100} plane. The cubic structure that features close-packing along face diagonals of the cube is FCC.

Reserve Problem 11c-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a square when they are connected by lines through their centers. Cubic unit cells cut along {100} planes will feature this shape. The edges of the square are edges of the cube, so they are <100> directions. We see that atoms are touching along face diagonals of that lie in the {100} plane. The cubic structure that features close-packing along face diagonals of the cube is FCC.

Reserve Problem 11d-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a square when they are connected by lines through their centers. Cubic unit cells cut along {100} planes will feature this shape. The edges of the square are edges of the cube, so they are <100> directions. We see that atoms are touching along face diagonals of that lie in the {100} plane. The cubic structure that features close-packing along face diagonals of the cube is FCC.

Reserve Problem 12a-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a rectangle when they are connected by lines through their centers. Specifically, the longer edge of the rectangle is larger than the smaller edge by a factor of √ 2 , consistent with the ratio of the rectangular plane that results when a cubic unit cell is cut along {110} plane. The short edges of the rectangle are edges of the cube, so they are <100> directions. The longer edges of the rectangle are face diagonals of the cube.

Note that atoms are touching along <100> directions and there are gaps between the atoms along {110} directions. This is consistent with SC.

Reserve Problem 12b-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a rectangle when they are connected by lines through their centers. Specifically, the longer edge of the rectangle is larger than the smaller edge by a factor of √ 2 , consistent with the ratio of the rectangular plane that results when a cubic unit cell is cut along {110} plane. The short edges of the rectangle are edges of the cube, so they are <100> directions. The longer edges of the rectangle are face diagonals of the cube.

Note that atoms are touching along <100> directions and there are gaps between the atoms along {110} directions. This is consistent with SC.

Reserve Problem 12c-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a rectangle when the corner circles are connected by lines through their centers. Note that a right triangle is formed by the indicated dimensions and the unmarked vertical distance. The Pythagorean theorem reveals that this distance is simply the lattice parameter, or cube edge length, a. Therefore, this rectangular section is consistent with a cubic unit cell that has been cut along a {110} plane.

Note that atoms are touching along <100> directions and there are gaps between the atoms along {110} directions. This is consistent with SC.

Reserve Problem 12d-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a rectangle when they are connected by lines through their centers. Specifically, the longer edge of the rectangle is larger than the smaller edge by a factor of √ 2 , consistent with the ratio of the rectangular plane that results when a cubic unit cell is cut along {110} plane. The short edges of the rectangle are edges of the cube, so they are <100> directions. The longer edges of the rectangle are face diagonals of the cube.

Note that atoms are touching along <100> directions and there are gaps between the atoms along {110} directions. This is consistent with SC.

Reserve Problem 13a-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a rectangle when the corner circles are connected by lines through their centers. Note that a right triangle is formed by the indicated dimensions and the unmarked vertical distance. The Pythagorean theorem reveals that this distance is simply the lattice parameter, or cube edge length, a. Therefore, this rectangular section is consistent with a cubic unit cell that has been cut along a {110} plane.

Note that atoms are touching along a √ 3 dimension, which is a body diagonal of a cubic unit cell.

Close-packing along the cubic body diagonal is consistent with BCC.

Reserve Problem 13b-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a rectangle when the corner circles are connected by lines through their centers. Note that a right triangle is formed by the indicated dimensions and the unmarked vertical distance. The Pythagorean theorem reveals that this distance is simply the lattice parameter, or cube edge length, a. Therefore, this rectangular section is consistent with a cubic unit cell that has been cut along a {110} plane.

Note that atoms are touching along a √ 3 dimension, which is a body diagonal of a cubic unit cell.

Close-packing along the cubic body diagonal is consistent with BCC.

Reserve Problem 13c-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a rectangle when they are connected by lines through their centers. Specifically, the longer edge of the rectangle is larger than the smaller edge by a factor of √ 2 , consistent with the ratio of the rectangular plane that results when a cubic unit cell is cut along {110} plane. The short edges of the rectangle are edges of the cube, so they are <100> directions. The longer edges of the rectangle are face diagonals of the cube.

Note that atoms are touching along a √ 3 dimension, which is a body diagonal of a cubic unit cell.

Close-packing along the cubic body diagonal is consistent with BCC.

Reserve Problem 13d-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a rectangle when the corner circles are connected by lines through their centers. Note that a right triangle is formed by the indicated dimensions and the unmarked vertical distance. The Pythagorean theorem reveals that this distance is simply the lattice parameter, or cube edge length, a. Therefore, this rectangular section is consistent with a cubic unit cell that has been cut along a {110} plane.

Note that atoms are touching along <100> directions and there are gaps between the atoms along {110} directions. This is consistent with SC.

Reserve Problem 14a-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a rectangle when the corner circles are connected by lines through their centers. Note that a right triangle is formed by the indicated dimensions and the unmarked vertical distance. The Pythagorean theorem reveals that this distance is simply the lattice parameter, or cube edge length, a. Therefore, this rectangular section is consistent with a cubic unit cell that has been cut along a {110} plane.

Note that atoms are touching along a √ 2 dimension, which is a face diagonal of a cubic unit cell.

Close-packing along the cubic face diagonal is consistent with FCC.

Reserve Problem 14b-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a rectangle when the corner circles are connected by lines through their centers. Note that a right triangle is formed by the indicated dimensions and the unmarked vertical distance. The Pythagorean theorem reveals that this distance is simply the lattice parameter, or cube edge length, a. Therefore, this rectangular section is consistent with a cubic unit cell that has been cut along a {110} plane.

Note that atoms are touching along a √ 2 dimension, which is a face diagonal of a cubic unit cell.

Close-packing along the cubic face diagonal is consistent with FCC.

Reserve Problem 14c-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a rectangle when they are connected by lines through their centers. Specifically, the longer edge of the rectangle is larger than the smaller edge by a factor of √ 2 , consistent with the ratio of the rectangular plane that results when a cubic unit cell is cut along {110} plane. The short edges of the rectangle are edges of the cube, so they are <100> directions. The longer edges of the rectangle are face diagonals of the cube.

Note that atoms are touching along a √ 2 dimension, which is a face diagonal of a cubic unit cell.

Close-packing along the cubic face diagonal is consistent with FCC.

Reserve Problem 14d-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The circular cross sections of the atoms shown form a rectangle when the corner circles are connected by lines through their centers. Note that a right triangle is formed by the indicated dimensions and the unmarked vertical distance. The Pythagorean theorem reveals that this distance is simply the lattice parameter, or cube edge length, a. Therefore, this rectangular section is consistent with a cubic unit cell that has been cut along a {110} plane.

Note that atoms are touching along a √ 2 dimension, which is a face diagonal of a cubic unit cell.

Close-packing along the cubic face diagonal is consistent with FCC.

Reserve Problem 15a-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) SC {110} (c) SC {111} (d) FCC {100} (e) FCC {110} (f) FCC {111}

Solution: The circular cross sections of the atoms shown form an equilateral triangle when the corner circles are connected by lines through their centers. Each edge of the triangle is a factor of √ 2 larger than the cube edge, a, so each edge is a face diagonal of the cube. This sectioning is therefore consistent with a cubic unit cell that has been cut along a {111} plane.

Note that no atoms are touching on the plane and the atoms are not close-packed on this plane.

The plane is therefore consistent with SC.

Reserve Problem 15b-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) SC {110} (c) SC {111} (d) FCC {100} (e) FCC {110} (f) FCC {111}

Solution: The circular cross sections of the atoms shown form an equilateral triangle when the corner circles are connected by lines through their centers. Each edge of the triangle is a factor of √ 2 larger than the cube edge, a, so each edge is a face diagonal of the cube. This sectioning is therefore consistent with a cubic unit cell that has been cut along a {111} plane.

Note that no atoms are touching on the plane and the atoms are not close-packed on this plane.

The plane is therefore consistent with SC.

Reserve Problem 15c-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) SC {110} (c) SC {111} (d) FCC {100} (e) FCC {110} (f) FCC {111}

Solution: The circular cross sections of the atoms shown form an equilateral triangle when the corner circles are connected by lines through their centers. Each edge of the triangle is a factor of √ 2 larger than the cube edge, a, so each edge is a face diagonal of the cube. This sectioning is therefore consistent with a cubic unit cell that has been cut along a {111} plane.

Note that no atoms are touching on the plane and the atoms are not close-packed on this plane.

The plane is therefore consistent with SC.

Reserve Problem 15d-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) SC {100} (b) SC {110} (c) SC {111} (d) FCC {100} (e) FCC {110} (f) FCC {111}

Solution: The circular cross sections of the atoms shown form an equilateral triangle when the corner circles are connected by lines through their centers. Each edge of the triangle is a factor of √ 2 larger than the cube edge, a, so each edge is a face diagonal of the cube. This sectioning is therefore consistent with a cubic unit cell that has been cut along a {111} plane.

Note that no atoms are touching on the plane and the atoms are not close-packed on this plane.

The plane is therefore consistent with SC.

Reserve Problem 16a-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) FCC {100} (b) FCC {110} (c) FCC {111} (d) SC {100} (e) SC {110} (f) SC {111}

Solution: The circular cross sections of the atoms shown form an equilateral triangle when the corner circles are connected by lines through their centers. Each edge of the triangle is a factor of √ 2 larger than the cube edge, a, so each edge is a face diagonal of the cube. This sectioning is therefore consistent with a cubic unit cell that has been cut along a {111} plane.

Note that the atoms are close-packed on this plane (they pack with a 2D coordination number of 6).

The plane is therefore consistent with FCC.

Reserve Problem 16b-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) FCC {100} (b) FCC {110} (c) FCC {111} (d) SC {100} (e) SC {110} (f) SC {111}

Solution: The circular cross sections of the atoms shown form an equilateral triangle when the corner circles are connected by lines through their centers. Each edge of the triangle is a factor of √ 2 larger than the cube edge, a, so each edge is a face diagonal of the cube. This sectioning is therefore consistent with a cubic unit cell that has been cut along a {111} plane.

Note that the atoms are close-packed on this plane (they pack with a 2D coordination number of 6).

The plane is therefore consistent with FCC.

Reserve Problem 16c-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) FCC {100} (b) FCC {110} (c) FCC {111} (d) SC {100} (e) SC {110} (f) SC {111}

Solution: The circular cross sections of the atoms shown form an equilateral triangle when the corner circles are connected by lines through their centers. Each edge of the triangle is a factor of √ 2 larger than the cube edge, a, so each edge is a face diagonal of the cube. This sectioning is therefore consistent with a cubic unit cell that has been cut along a {111} plane.

Note that the atoms are close-packed on this plane (they pack with a 2D coordination number of 6).

The plane is therefore consistent with FCC.

Reserve Problem 16d-1 (question pool) Which of the following options is consistent with the unit cell plane depicted below?

(a) FCC {100} (b) FCC {110} (c) FCC {111} (d) SC {100} (e) SC {110} (f) SC {111}

Solution: The circular cross sections of the atoms shown form an equilateral triangle when the corner circles are connected by lines through their centers. Each edge of the triangle is a factor of √ 2 larger than the cube edge, a, so each edge is a face diagonal of the cube. This sectioning is therefore consistent with a cubic unit cell that has been cut along a {111} plane.

Note that the atoms are close-packed on this plane (they pack with a 2D coordination number of 6).

The plane is therefore consistent with FCC. Reserve Problem 17a (question pool) Certain crystal structures are sometimes similar to other crystal structures in a variety of ways. This problem demonstrates one of the similarities that exists between two of the cubic crystal structures discussed in chapter 3. Specifically, a family of planes in one of these crystal structures is equivalent to a family of planes in the other crystal structure.

For now, we are just considering how equal-sized spheres, but you may also consider this as atoms of the same element packing together to form two crystals, each of which exhibits one of these different crystal structures.

The figure below depicts a plane of atoms of radius R in some crystal. The plane intersects the atoms through their centers. Therefore, we do not see atoms above and below this plane. We just see the circular cross-sections of the atoms that lie within the plane.

If we consider only simple cubic (SC), body-centered cubic (BCC), and face-centered cubic (FCC) crystal structures as options, and we have no more information to rely on, what are the two possible identities for the plane shown above?

Select all that apply. (a) SC {100} (b) BCC {100} (c) FCC {100} (d) SC {110} (e) BCC {110} (f) FCC {110} (g) SC {111} (h) BCC {111} (i) FCC {111}

Solution: The atoms form a square array with close-packed axes that are 90 degrees apart, consistent with SC {100} and FCC {100}.

Reserve Problem 18: Atomic radius – simple cubic For a metal that has the simple cubic crystal structure, calculate the atomic radius if the metal has a 3 density of 2.05 g/cm and an atomic weight of 77.84 g/mol.

Solution The density of a metal may be calculated using Equation 3.8—that is

ρ=

nA VC N A

For the simple cubic crystal structure, one atom is associated with each unit cell (i.e., n = 1). Furthermore, the atomic radius and unit cell length are related as a = 2R; and since the unit cell has cubic symmetry VC = a3 = (2 R)3 = 8R3

Which means that, for this problem, Equation 3.8 takes the following form:

ρ=

nA 8R3 N A

And solving for R in this expression yields 1/3

⎛ nA ⎞ R = ⎜⎜ ⎟⎟ ⎝ 8ρ N A ⎠

Substitution of values for parameters given in the problem statement leads to the following value of R: ⎡ ⎤ (1atom/unit cell)(61.12 g/mol) R=⎢ ⎥ 3 23 ⎢⎣ (8)(7.02g/cm )(6.022 × 10 atoms/mol ⎥⎦

1/3

1.22 × 10 −8 cm = 0.122nm

Reserve Problem 19: Cubic unit cell Some metal is known to have a cubic unit cell with an edge length of 0.437 nm. In addition, it has a 3 density of 4.37 g/cm and an atomic weight of 54.85 g/mol. Indicate the letter of the metal listed in the following table that has these characteristics.

Metal Crystal Structure Atomic Radius (nm) A

BCC

0.219

B

FCC

0.309

C

FCC

0.155

D

HCP

0.125

Solution First of all, since the unit cell is cubic, the HCP crystal structure is eliminated because its unit cell has hexagonal symmetry. Now all we need do is compute the atomic radius for both FCC and BCC crystal structures from the unit cell edge length provided in the problem statement, and compare these values with R values provided. For FCC this relationship is according to a rearranged form of Equation 3.1—that is

R=

a 2 2

and for BCC a rearranged form of Equation 3.4—viz.

R=

a 3 4

Let's begin by computing R for the BCC crystal structure, as follows:

R=

(0.428nm) 3 = 0.185nm 4

Since this value does not correspond to any of the values provided in the table, the crystal structure is not BCC. We will now carry out the computation for FCC using the appropriate equation cited above. Hence

R=

0.428nm 2 2

= 0.151nm

This is the value of R for Metal C, which has the FCC crystal structure.

Reserve Problem 20: Crystal Lengths III In terms of the atomic radius, R, determine the distance between the centers of adjacent atoms for the BCC crystal structure along the [110] direction.

2

4* R*

3

Solution The [110] direction within a BCC unit cell is shown below:

From this figure we want to solve for the distance PO. Upon consideration of triangle POQ it is the case that 2

2

PQ = PO + OQ

2

Inasmuch as PQ = 4R and OQ = a, we may write 2

2

(4R) = PO + a

2

Now from triangle PNO 2

2

PO = PN + NO 2

2

2

= a + a = 2a

2

Making this substitution into the previous equations gives 2

2

2

(4R) = 2a + a = 3a

2

or

a= 2

4R 3

2

Since PO = 2a then

⎛ 4R ⎞ 4R 2 PO = a 2 = ⎜ ⎟ 2= 3 ⎝ 3⎠ Reserve Question 21: Miller indices III Below is shown the atomic packing of a plane for the simple cubic crystal structure; atoms drawn to full size are represented by the circles.

From the list below select all possible sets of indices for this plane. a) b) c) d) e) f)

(001) (11-1) (111) (-1-1-1) (101) (100)

Solution This packing corresponds to a (111)-type plane for simple cubic; therefore (11-1), (111) and (-1-1-1) are all possible answers. This plane is sketched in the illustration below.

Reserve Problem 22: BCC unit cell volume If the atomic radius of a metal that has the body-centered cubic crystal structure is 0.181 nm, calculate the volume of its unit cell.

Solution For the BCC crystal structure, the atomic radius and unit cell edge length are related according to Equation 3.4—that is

a=

4R 3 3

Because the unit cell has cubic symmetry, its volume VC is equal to a , or 3

⎛ 4R ⎞ 64 R 3 VC = a = ⎜ ⎟ = 3 3 ⎝ 3⎠ 3

And, for this problem, R = 0.130 nm, then VC =

(64)(0.130nm)3 3 3

= 0.0271nm3

Reserve Problem 23: FCC unit cell volume If the atomic radius of a metal that has the face-centered cubic crystal structure is 0.123 nm, calculate the volume of its unit cell.

Solution Since this metal has an FCC crystal structure, the volume of its unit cell may be determined using Equation 3.6 as follows: VC = 16 R3 2 = (16)(0.225nm)3 ( 2) = 0.258nm3

Reserve Problem 24: Hexagonal close-packed structure For the hexagonal close-packed crystal structure: (a) How many atoms are associated with the unit cell? (b) What is the coordination number? (c) What is the atomic packing factor?

Solution The unit cell for HCP is shown below. z

a

c a2

D a3

C

E A

a1

(a) There are six atoms associated with each unit cell; one-sixth of each of the 12 top and bottom face corner atoms, one-half of each the two center face atoms and the three midplane atoms. (b) The center atom of the top face has 6 neighboring atoms within its plane. Also the 3 atoms in the midplane below the top face are also nearest neighbors; in addition there are three more nearest neighbor atoms in an equivalent midplane for the next unit cell above the one shown. Therefore, the coordination number is 12. (c) To determine the atomic packing factor it is necessary to calculate the unit cell volume, which is just the product of the base area times the cell height, c. This base area is just three times the area of the parallelepiped ACDE shown below.

C

D a = 2R

30º A

60º B

E a = 2R

a = 2R

The area of ACDE is the length of CD times the height BC. Since CD is just a or 2R and

BC = 2 R cos(30°) =

2R 3 2

Thus, the base area is just ⎛ 2R 3 ⎞ 2 Area = (3)(CD)(BC) = (3)(2 R) ⎜ ⎟ = 6R 3 ⎜ 2 ⎟ ⎝ ⎠

and since c = 1.633a = 2R(1.633), then the volume of the unit cell is

VC = (Area)(c) = (6R2 3)(c) = (6R2 3)(2R)(1.633) = (12R3 )(1.633) 3 Thus, the atomic packing factor may be calculated as follows: APF =

(6)( 34 πR3 ) total sphere volume 2π = = = 0.740 3 total unit cell volume (12 R )(1.633) 3 3 3(1.633)

Reserve Problem 25: Orthorhombic unit cell A hypothetical metal has an orthorhombic unit cell for which the a, b, and c lattice parameters are 0.472, 0.732, and 0.826 nm, respectively. (a) If there are 8 atoms per unit cell and the atomic packing factor is 0.549, then determine the atomic radius.

3

(b) If the density is 6.04 g/cm , then calculate the metal's atomic weight.

Solution (a) The atomic packing factor, APF, is defined by Equation 3.3 as follows:

APF =

volume of atoms in a unit cell VS = VC total unit cell volume

For the orthorhombic unit cell, the unit cell volume is equal to the product of the unit cell edge lengths—that is VC = abc

Furthermore, the total atom volume is equal to the volume per atom times the number of atoms per unit cell (for this problem 8)—therefore

⎛4 ⎞ VS = (8) ⎜ π R3 ⎟ ⎝3 ⎠ Substitution of the expressions for VC and VS into the APF equation and solving for the atomic radius, R, yields the following: 1/3

⎛ 3(APF)abc ⎞ R=⎜ ⎟ 32π ⎝ ⎠

Substitution of the appropriate values from the problem statement into this expression yields

⎡ (3)(0.412)(0.394nm)(0.649 nm)(0.950 nm) ⎤ R=⎢ ⎥ 32π ⎣ ⎦ = 0.144nm

1/3

(b) It is possible to solve for the atomic weight by rearrangement of Equation 3.8 as follows:

A=

ρVC N A n

=

ρabcN A n

Substitution of appropriate values of these parameters given in the problem statement (and converting values of a, b, and c) into centimeters leads to the following: A=

(6.93g/cm3 )(3.94 × 10 −8 cm)(6.49 × 10 −8 cm)(9.50 × 10 −8 cm)(6.022 × 10 23atoms/mol) 8atoms/unit cell = 127 g/mol

Reserve Problem 26: Rhodium structure 3

Rhodium has an atomic radius of 0.1345 nm, a density of 12.41 g/cm and an atomic weight of 102.91 g/mol. What is rhodium's crystal structure? a) Simple cubic b) BCC c) FCC

Solution The number of atoms associated with the rhodium unit cell is a function of the unit cell volume, the density, and the atomic weight according to a rearranged form of Equation 3.8—that is

n=

ρVC N A A

Substitution into this expression values for rand A provided in the problem statement (as well as the value for NA) yields the following: n=

(12.41g/cm3 )(6.022 × 1023 atoms/mol)(VC ) 102.91g/mol = 7.26 × 1022 (VC )atom/cm3

For the three possible crystal structures (SC, FCC, BCC), the dependence of unit cell volume on the atomic radii are according to the following equations: SC VC = a3 = (2 R)3 = 8R3 3

⎛ 4R ⎞ 64 R 3 BCC VC = a = ⎜ ⎟ = 3 3 ⎝ 3⎠ 3

FCC VC = a3 = (2 R 2)3 = 16 R3 2 Values of a for BCC and FCC crystal structures are noted in Equations 3.4 and 3.1, respectively. We now insert each of these expressions for VC into the above expression for n and incorporate the value of R provided in the problem statement (0.1345 nm = 1.345 ´ 10-8 cm). If this value of n is 1.0, the crystal structure is SC; if the value is 2.0 the crystal structure is BCC; and if the value is 4.0 it is FCC. These computations are as follows: For SC:

n = (7.26 × 1022 atom/cm3 )(8R3 ) = (7.26 × 1022 atoms/cm3 )(8)(1.345 × 10−8 cm)3 = 1.41atoms For BCC

⎛ 64 R3 ⎞ n = (7.26 × 1022 atom/cm3 ) ⎜ ⎟ ⎜3 3 ⎟ ⎝ ⎠ − 8 3 ⎡ (64)(1.345 × 10 cm) ⎤ = (7.26 × 1022 atom/cm3 ) ⎢ ⎥ = 2.18atoms 3 3 ⎢⎣ ⎥⎦

While for FCC

n = (7.26 ×1022 atom/cm3 )(16R3 3) = (7.26 ×1022 atom/cm3 )(16)(1.345 ×10−8 cm)3 ( 2) = 4.0 Therefore, the crystal structure for rhodium is FCC; for FCC only, is this computed value of n the same as the theoretical value.

Reserve Problem 27: Unit cell length 3

A hypothetical metal has a cubic unit cell, a density of 6.79 g/cm , a coordination number of 6, and an atomic weight of 78.57 g/mol. Calculate the unit cell edge length for this material.

Solution The density of a metal may be calculated using the following equation:

ρ=

nA VC NA

Since the unit cell has cubic symmetry, then VC = a3 ,

where a is the unit cell edge length. Substitution of this expression for VC into the previous equation, and solving for a leads to ⎛ nA ⎞ a = ⎜⎜ ⎟⎟ ⎝ ρNA ⎠

(

)

1/3

⎡ 1 atoms / unitcell ( 91.6g/mol ) ⎤ ⎢ ⎥ a= ⎢ 9.6g/cm3 6.022 × 1023 atoms / mol ⎥ ⎢⎣ ⎥⎦

(

)(

)

1/3

= 2.51 × 10−8 cm

Reserve Problem 28 Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol.

Compute and compare its theoretical density with the experimental value found inside the front cover of the book.

Solution This problem calls for a computation of the density of iron. According to Equation 3.8

ρ=

nAFe VC N A

For BCC, n = 2 atoms/unit cell, and ⎛ 4R ⎞ VC = ⎜ ⎜ 3 ⎟⎟ ⎝ ⎠

3

Thus,

ρ=

=

nAFe 3

⎛ 4R ⎞ ⎜⎜ ⎟⎟ N A ⎝ 3⎠

(2 atoms/unit cell)(55.85 g/mol) 3

⎡ (4) (0.124 × 10 −7 cm) / 3 ⎤ / (unit cell) (6.022 × 1023 atoms/mol) ⎣ ⎦ 3

= 7.90 g/cm 3

The value given inside the front cover is 7.87 g/cm .

Reserve Problem 29 Niobium (Nb) has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical density for Nb.

Solution According to Equation 3.8:

ρ=

nANb VC N A 3

For BCC, n = 2 atoms/unit cell. Furthermore, because VC = a , and a = 3

⎛ 4R ⎞ VC = ⎜ ⎟ ⎝ 3⎠

4R 3

(Equation 3.4), then

Thus, realizing that ANb = 92.91 g/mol, using the above version of Equation 3.8, we compute the theoretical density of Nb as follows:

ρ=

=

nANb 3

⎛ 4R ⎞ ⎜ ⎟ NA ⎝ 3⎠

(2 atoms/unit cell)(91.92 g/mol) 3

⎡ (4)(0.143 × 10 −7 cm) ⎤ ⎢ ⎥ /(unit cell)(6.022 × 1023 atoms/mol) ⎢⎣ ⎥⎦ 3 3

= 8.48 g/cm 3 The experimental density for Nb is 8.57 g/cm .

Reserve Problem 30: Rhenium has an HCP crystal structure, an atomic radius of 0.137 nm, and a c/a ratio of 1.615.

Compute the volume of the unit cell for Re. Solution In order to compute the volume of the unit cell for Re, it is necessary to use Equation 3.S1 (found in Problem 3.6), that is

VC = 6 R2c 3 The problem states that c = 1.615a, and a = 2R. Therefore

VC = (1.615) (12 3) R3 = (1.615) (12 3 )(1.37 × 10−8 cm)3 = 8.63 × 10−23 cm3 = 8.63 × 10−2 nm3 Reserve Problem 31

The unit cell for MgFe2O4 (MgO-Fe2O3) has cubic symmetry with a unit cell edge length of 0.836 nm.

3 If the density of this material is 4.52 g/cm , compute its atomic packing factor. For this computation, you will need to use ionic radii listed in Table 3.4.

Solution

This problem asks us to compute the atomic packing factor for MgFe2O4 given its density and unit cell edge length. It is first necessary to determine the number of formula units in the unit cell in order to calculate the sphere volume. Solving for n' from Equation 3.9 leads to n' =

=

=

ρVC N A

∑ AC + ∑ AA ρVC NA

AMg + 2 AFe + 4 AO

(4.52 g/cm3 ) ⎡⎣(8.36 × 10 −8 cm)3 /unit cell ⎤⎦ (6.022 × 10 23 formula units/mol) (1)(24.31 g/mol) + (2)(55.85 g/mol) + (4)(16.00 g/mol)

= 8.0 formula units/unit cell 2+

3+

2-

Thus, in each unit cell there are 8 Mg , 16 Fe , and 32 O ions. From Table 3.4, rMg2+ = 0.072 nm, rFe3+ = 0.069 nm, and rO2- = 0.140 nm. Thus, the total sphere volume in Equation 3.3 (which we denote as VS), is just

⎛4 ⎞ ⎛4 ⎞ ⎛4 ⎞ VS = (8) ⎜ π ⎟ (7.2 × 10 −9 cm)3 + (16) ⎜ π ⎟ (6.9 × 10 −9 cm)3 + (32) ⎜ π ⎟ (1.40 × 10−8 cm)3 ⎝3 ⎠ ⎝3 ⎠ ⎝3 ⎠ –22 3 = 4.02 × 10 cm Now, the unit cell volume (VC) is just

VC = a3 = (8.36 × 10-8 cm)3 = 5.84 ×10−22 cm3 Finally, the atomic packing factor (APF) from Equation 3.3 is just

APF =

VS 4.02 × 10−22 cm3 = = 0.688 VC 5.84 × 10−22 cm3

Reserve Problem 32 For each statement below, choose the bonding type that best completes each phrase.

a) The crystal structures of ____ ceramics are constrained by bond angles associated with the locations of shared electrons within the material. b) The coordination number of ____ ceramics are constrained by the relative sizes of the compound’s component species. c) The crystal structures of ____ ceramics are constrained by the relative charges of the compound’s component species. d) The crystal structures of ____ ceramics are constrained by ratio of the compound’s charged component species such that the structure maintains charge neutrality.

Solution: [a] covalent

[b] ionic [c] ionic [d] ionic

Reserve Problem 33 2+

Consider the fluorite (CaF2) crystal structure. The coordination number of Ca the coordination number of F ions is __[b]__.

ions is __[a]__, and

Solution [a] 8

[b] 4 Reserve Problem 34a (question pool) 4+

Consider the ideal barium titanate (BaTiO3) structure. What is the coordination number of the Ti 2ion in terms of surrounding O ions? [a] 1 [b] 2 [c] 3 [d] 4 [e] 5 [f] 6 [g] 7 [h] 8

Solution: 4+

According to Fig. 3.10, the central Ti six cube faces.

2-

ion is surrounded by six O ions, one residing at each of the

Reserve Problem 35 2+

Consider the ideal barium titanate (BaTiO3) structure. What is the coordination number of the Ba 4+ ion in terms of surrounding Ti ions? [a] 4 [b] 6 [c] 8 [d] 10 [e] 12

Solution: Careful consideration of Fig. 3.10, and consideration of the perovskite compound stoichiometry 4+ 2+ reveals that 12 Ti ions surround each Ba ion. A change in perspective is recommended to help 24+ visualize this, where the unit cell cube corners feature O ions in the corners, Ti ions centered 2+ along each cube edge, and Ba ions centered in the middle of the cube.

Reserve Problem 36 Which of the following options correctly depicts a {110} plane from a diamond cubic unit cell? a)

b)

c)

d)

e)

f)

g)

h)

Solution: {110} planes intersect cubic unit cells as rectangles and feature atoms at each corner so all of the options are valid in this respect. For diamond cubic specifically, there should be one atom centered along each face diagonal, and all options correctly depict this feature. Finally, along just one face diagonal, there should be two atoms within the rectangle, and each of these will be tangent to two atoms in the plane: the central atom along the face diagonal and one of the corner atoms. Thus, atoms will be tangent in a zigzag way along just one of the long edges of the rectangle.

Reserve Problem 37 Compute the PPF of {110} planes for the diamond cubic crystal structure. [a] 0.29 [b] 0.34 [c] 0.42 [d] 0.56 [e] 0.68 [f] 0.71 [g] 0.74 [h] 0.82

Solution:

DC = PPF 110

Area Occupied [4(1/4) + 2(1/2) + 2(1)]π R2 4π R2 = = 2 Total Area a(a 2) a 2 a 3 8R = 2R a = 4 3 DC PPF 110 =

4π R2 ⎛ 8R ⎞ ⎜ ⎟ ⎝ 3⎠

=

2

2

3π 16 2

= 0.417

Reserve Problem 38 Compute the planar density of atoms, in atoms per square centimeter, on a {110} plane of a defectfree diamond cubic crystal, whose atoms have a radius of [R] nanometers.

Solution:

DC PD 110 , R = 0.077 nm =

Atom Count [4(1/4) + 2(1/2) + 2(1)]π R2 4 = = 2 Total Area a(a 2) a 2 a 3 8R = 2R a = 4 3

DC , R = 0.077 nm = PD 110

4 ⎛ 8R ⎞ ⎜ ⎟ ⎝ 3⎠

=

2

2

4 ⎛ 8(0.077 × 10 −7 cm) ⎞ ⎜⎜ ⎟⎟ 3 ⎝ ⎠

= 2.24 × 1015 atoms/cm2

2

2

Reserve Problem 39 Compute the PPF of {100} planes for the diamond cubic crystal structure. [a] 0.29 [b] 0.34 [c] 0.42 [d] 0.56 [e] 0.68 [f] 0.71 [g] 0.74 [h] 0.82

Solution:

DC PPF 100 =

Area Occupied [4(1/4) + 1(1)]π R2 2π R2 = = 2 Total Area a( a) a a 3 8R = 2R a = 4 3 DC PPF 100 =

2π R2 ⎛ 8R ⎞ ⎜ ⎟ ⎝ 3⎠

2

=

3π = 0.29 32

Reserve Problem 40 Compute the planar density of atoms, in atoms per square centimeter, on a {100} plane of a defectfree diamond cubic crystal, whose atoms have a radius of [R] nanometers.

Solution:

DC PD 110 , R = 0.077 nm =

4(1/4) + 1 ⎛ 8R ⎞ ⎜ ⎟ ⎝ 3⎠

2

2

=

2

= 1.58 × 1015 atoms/cm2

⎛ 8R ⎞ ⎜ ⎟ ⎝ 3⎠

2

a 3 8R = 2R a = 4 3 DC PD 110 , R = 0.077 nm =

2 ⎛ 8(0.077 × 10 −7 cm) ⎞ ⎜⎜ ⎟⎟ 3 ⎝ ⎠

Reserve Question 41: Lattice parameters I Which crystal system(s) listed below has (have) the following relationship for the unit cell edge lengths? a=b=c

a) b) c) d) e) f) g)

Cubic Hexagonal Tetragonal Rhombohedral Orthorhombic Monoclinic Triclinic

Solution For both cubic and rhombohedral crystal systems all the unit cell edge lengths are equal.

Reserve Question 42: Lattice parameters II Which crystal system(s) listed below has (have) the following relationship for the unit cell edge lengths? a=b≠c

a) b) c) d) e) f) g)

Cubic Hexagonal Tetragonal Rhombohedral Orthorhombic Monoclinic Triclinic

Solution For both hexagonal and tetragonal crystal systems two of the unit cell edge lengths are equal to one another, but unequal to the third length.

Reserve Question 43: Lattice parameters III Which crystal system(s) listed below has (have) the following relationship for the unit cell edge lengths? a≠b≠c

a) b) c) d) e) f) g)

Cubic Hexagonal Tetragonal Rhombohedral Orthorhombic Monoclinic Triclinic

Solution For orthorhombic, monoclinic and triclinic crystal systems, all three of the unit cell edge lengths are unequal to one another.

Reserve Question 44: Unit cell geometries I Which crystal system(s) listed below has (have) the following interaxial angle relationship?

α ≠ β ≠ γ ≠ 90° a) b) c) d) e) f) g)

Cubic Hexagonal Tetragonal Rhombohedral Orthorhombic Monoclinic Triclinic

Solution Triclinic is the only crystal system for which none of the interaxial angles are equal to one another and also not equal to 90°.

Reserve Question 45: Unit cell geometries II Which crystal system(s) listed below has (have) the following interaxial angle relationship?

α = β = γ = 90° a) b) c) d) e) f) g)

Cubic Hexagonal Tetragonal Rhombohedral Orthorhombic Monoclinic Triclinic

Solution Cubic, tetragonal and orthorhombic crystal systems all have the three interaxial angles equal to 90°.

Reserve Question 46: Unit cell geometries III Which crystal system(s) listed below has (have) the following interaxial angle relationship? α = β = 90°, γ = 120°

a) b) c) d) e) f) g)

Cubic Hexagonal Tetragonal Rhombohedral Orthorhombic Monoclinic Triclinic

Solution Only the hexagonal crystal system has two of the interaxial angles equal to 90°, while the third angle is equal to 120°.

Reserve Question 47: Unit cell geometries IV Which crystal system(s) listed below has (have) the following interaxial angle relationship? α = β = γ ≠ 90°

a) b) c) d) e) f) g)

Cubic Hexagonal Tetragonal Rhombohedral Orthorhombic Monoclinic Triclinic

Solution Only the rhombohedral crystal system has all three interaxial angles equal to one other, but not equal to 90°.

Reserve Question 48: Unit cell geometries V Which crystal system(s) listed below has (have) the following interaxial angle relationship?

α = γ = 90° ≠ β

a) b) c) d) e) f) g)

Cubic Hexagonal Tetragonal Rhombohedral Orthorhombic Monoclinic Triclinic

Solution Only the monoclinic crystal system has two of the interaxial angles equal to one another and to 90°, which is different than the third interaxial angle.

Reserve Problem 49

Sketch a tetragonal unit cell, and within that cell indicate locations of the

1 2

1

1 2

and

113 424

point

indices.

Solution

A tetragonal unit in which is shown the

1 2

1

1 2

and

113 424

point indices is presented below

Reserve Problem 50 1

111

2

434

Sketch an orthorhombic unit cell, and within that cell indicate locations of the 1 0 and indices.

point

Solution 1

An orthorhombic unit cell in which is shown the location of the 1 0 point indices is presented 2

below.

An orthorhombic unit cell in which is shown location of the

111 434

point indices is presented below.

Reserve Problem 51: Crystal lengths II In terms of the atomic radius, R, determine the distance between the centers of adjacent atoms for the BCC crystal structure along the [111] direction.

2* R

Solution The [111] direction within a BCC unit cell is shown below:

Since the atoms in a BCC unit cell touch one another in this [111] direction (i.e., along the cell diagonal), the distance between the centers of these adjacent atoms is 2R. Reserve Problem 52: SC indices For the simple cubic crystal structure, in terms of the atomic radius, R, determine the distance between the centers of adjacent atoms along the [120] direction.

2* R* 5

Solution

To solve this problem it is first necessary to construct the [120] direction vector. This is possible using rearranged forms of Equations 3.10a, 3.10b, and 3.10c. We will assume that x1 = 0a, y1 = 0b, and z1= 0c (that is the vector tail resides at a unit cell corner and the origin of the reference coordinate system). For this [120] direction u = a v = 2b w = 0c The vector head coordinates are determined using the rearranged forms of Equations 3.10a-3.10c as follows: x 2 = ua + x1 = 1a + 0 a = a y2 = vb + y1 = 2b + 0 b = 2bp

z2 = wc + z1 = 0c + 0c = 0c

For the simple cubic crystal structure, it is the case that a = b = c = 2R, where R is the atomic radius.

Therefore, to construct this [120] vector, we move along the x axis a (2R units) from the origin, from this point 2b (4R) units parallel to the y axis; since the z2 coordinate is 0c (0R), it is not necessary to move parallel to the z axis—that is the vector lies in the x-y plane. This direction is shown in the simple cubic unit cell below with the vector head located at point p.

The problem now is to determine the length of the [120] vector from the origin to point p. This is possible using the following expression:

p2 = (2R)2 + (4R)2 and solving for p gives

p = 2R 5 Reserve Question 53: Cubic direction indices 01 What are the indices for the direction represented by the vector that has been drawn within a unit cell?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number. z

c

a b x

a) b) c) d)

[-102] [120] [121] 102

y

Solution We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows:

x1 = 0a

y1 = 0b

z1 = 0c

y2 = b

z2 = 0c

Whereas head coordinates are as follows: x2 = a/2

From Equations 3.10a, 3.10b, and 3.10c assuming a value of 1 for the parameter n

⎛ x −x ⎞ ⎛ a/2 − 0a ⎞ 1 u = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟= a ⎝ ⎠ 2 ⎝ a ⎠ ⎛ y −y ⎞ ⎛ b − 0b ⎞ v = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟ =1 ⎝ b ⎠ ⎝ b ⎠ ⎛ z −z ⎞ ⎛ 0c − 0c ⎞ w = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟=0 ⎝ c ⎠ ⎝ c ⎠ In order to reduce these values to the lowest set of integers, we multiply each of the indices by 2.

Therefore, this direction is [120].

Reserve Question 54: Cubic direction indices 02 What are the indices for the direction represented by the vector that has been drawn within a unit cell?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number. z

c

a b x

a) b) c) d)

[-002] [120] [101] [012]

y

Solution We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows:

x1 = 0a

y1 = 0b

z1 = 0c

y2 = b/2

z2 = c

Whereas head coordinates are as follows: x2 = 0a

From Equations 3.10a, 3.10b, and 3.10c assuming a value of 1 for the parameter n

⎛ x −x ⎞ ⎛ 0a − 0a ⎞ u = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟=0 ⎝ a ⎠ ⎝ a ⎠ ⎛ y −y ⎞ ⎛ b/2 − 0b ⎞ 1 v = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟= b ⎝ ⎠ 2 ⎝ b ⎠ ⎛ z −z ⎞ ⎛ c − 0c ⎞ w = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟ =1 ⎝ c ⎠ ⎝ c ⎠ In order to reduce these values to the lowest set of integers, we multiply each of the indices by 2.

Therefore, this direction is [012].

Reserve Question 55: Cubic direction indices 03 What are the indices for the direction represented by the vector that has been drawn within a unit cell?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number. z

c

a b x

a) b) c) d)

[111] [1-11] [11-1] [-111]

y

Solution We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows:

x1 = 0a

y1 = 0b

z1 = 0c

y2 = b

z2 = -c

Whereas head coordinates are as follows: x2 = a

From Equations 3.10a, 3.10b, and 3.10c assuming a value of 1 for the parameter n

⎛ x −x ⎞ ⎛ a − 0a ⎞ u = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟ =1 ⎝ a ⎠ ⎝ a ⎠ ⎛ y −y ⎞ ⎛ b − 0b ⎞ v = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟ =1 ⎝ b ⎠ ⎝ b ⎠ ⎛ z −z ⎞ ⎛ −c − 0c ⎞ w = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟ = −1 ⎝ c ⎠ ⎝ c ⎠ Therefore, this direction is [11-1].

Reserve Question 56: Cubic direction indices 04 What are the indices for the direction represented by the vector that has been drawn within a unit cell?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number. z

c

a b x

a) b) c) d)

[0-11] [1-11] [01-1] [-111]

y

Solution We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows:

x1 = 0a

y1 = 0b

z1 = 0c

y2 = -b

z2 = c

Whereas head coordinates are as follows: x2 = 0a

From Equations 3.10a, 3.10b, and 3.10c assuming a value of 1 for the parameter n

⎛ x −x ⎞ ⎛ 0a − 0a ⎞ u = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟=0 ⎝ a ⎠ ⎝ a ⎠ ⎛y −y ⎞ ⎛ −b − 0b ⎞ v = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟ = −1 ⎝ b ⎠ ⎝ b ⎠ ⎛ z −z ⎞ ⎛ c − 0c ⎞ w = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟ =1 ⎝ c ⎠ ⎝ c ⎠ Therefore, this direction is [0-11].

Reserve Question 57: Cubic direction indices 05 What are the indices for the direction represented by the vector that has been drawn within a unit cell?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number. z

c 1 2

a b x

a) b) c) d)

[210] [200] [-210] [2-11]

y

Solution We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows:

x1 = a

y1 = -b/2

z1 = 0c

y2 = 0b

z2 = 0c

Whereas head coordinates are as follows: x2 = 0a

From Equations 3.10a, 3.10b, and 3.10c assuming a value of 1 for the parameter n

⎛ x −x ⎞ ⎛ 0a − a ⎞ u = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟ = −1 ⎝ a ⎠ ⎝ a ⎠ ⎛ y −y ⎞ ⎡ 0b − (−b/2) ⎤ 1 v = n ⎜ 2 1 ⎟ = (1) ⎢ ⎥=2 b ⎣ ⎦ ⎝ b ⎠ ⎛ z −z ⎞ ⎛ 0c − 0c ⎞ w = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟=0 ⎝ c ⎠ ⎝ c ⎠ In order to reduce these values to the lowest set of integers, we multiply each of the indices by 2.

Therefore, this direction is [-210].

Reserve Question 58: Cubic direction indices 06 What are the indices for the direction represented by the vector that has been drawn within a unit cell?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number. z

c y a b x

a) b) c) d)

[0-21] [001] [201] [021]

Solution We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows:

x1 = 0a

y1 = 0b

z1 = 0c

y2 = 2b

z2 = c

Whereas head coordinates are as follows: x2 = 0a

From Equations 3.10a, 3.10b, and 3.10c assuming a value of 1 for the parameter n

⎛ x −x ⎞ ⎛ 0a − 0a ⎞ u = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟=0 ⎝ a ⎠ ⎝ a ⎠ ⎛y −y ⎞ ⎛ 2b − 0b ⎞ v = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟=2 ⎝ b ⎠ ⎝ b ⎠ ⎛ z −z ⎞ ⎛ c − 0c ⎞ w = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟ =1 ⎝ c ⎠ ⎝ c ⎠ Therefore, this direction is [021].

Reserve Question 59: Cubic direction indices 07 What are the indices for the direction represented by the vector that has been drawn within a unit cell?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number. z

c

a b x

a) b) c) d)

[-0-2-1] [-1-1-1] [-0-1-1] [-0-0-2]

y

Solution We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows:

x1 = a

y1 = b

z1 = c

y2 = 0b

z2 = 0c

Whereas head coordinates are as follows: x2 = 0a

From Equations 3.10a, 3.10b, and 3.10c assuming a value of 1 for the parameter n ⎛x −x ⎞ ⎛ 0a − a ⎞ u = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟ = −1 a ⎝ a ⎠ ⎝ ⎠

⎛y −y ⎞ ⎛ 0b − b ⎞ v = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟ = −1 ⎝ b ⎠ ⎝ b ⎠ ⎛ z −z ⎞ ⎛ 0c − c ⎞ w = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟ = −1 ⎝ c ⎠ ⎝ c ⎠ Therefore, this direction is [-1-1-1].

Reserve Question 60: Cubic direction indices 08 What are the indices for the direction represented by the vector that has been drawn within a unit cell?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number. z

1, 1 2 2

c

a

y

b x

Solution We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows:

x1 = a/2

y1 = b/2

z1 = c

Whereas head coordinates are as follows: x2 = a

y2 = b

z2 = c

From Equations 3.10a, 3.10b, and 3.10c assuming a value of 1 for the parameter n

⎛ x −x ⎞ ⎛ a − a/2 ⎞ 1 u = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟= ⎝ a ⎠ 2 ⎝ a ⎠ ⎛y −y ⎞ ⎛ b − b/2 ⎞ 1 v = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟= ⎝ b ⎠ 2 ⎝ b ⎠ ⎛ z −z ⎞ ⎛ c −c ⎞ w = n ⎜ 2 1 ⎟ = (1) ⎜ ⎟=0 ⎝ c ⎠ ⎝ c ⎠ In order to reduce these values to the lowest set of integers, we multiply each of the indices by 2.

Therefore, this direction is [110].

Reserve Question 61: Cubic direction indices 09 What are the indices for the direction represented by the vector that has been drawn within a unit cell?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number. z

1 3

1 2

c

a

y

b x

a) b) c) d)

[-321] [-420] [-431] [-430]

Solution We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows:

x1 = a

y1 = b/2

z1 = c

Whereas head coordinates are as follows: x2 = a/3

y2 = b

z2 = c

From Equations 3.10a, 3.10b, and 3.10c assuming a value of 3 for the parameter n

⎛ x −x ⎞ ⎛ a/3 − a ⎞ u = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟ = −2 ⎝ a ⎠ ⎝ a ⎠ ⎛ y −y ⎞ ⎛ b − b/2 ⎞ 3 v = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟= ⎝ b ⎠ 2 ⎝ b ⎠ ⎛ z −z ⎞ ⎛ c −c ⎞ w = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟=0 ⎝ c ⎠ ⎝ c ⎠ In order to reduce these values to the lowest set of integers, we multiply each of the indices by 2.

Therefore, this direction is [-430].

Reserve Question 62: Cubic direction indices 10 What are the indices for the direction represented by the vector that has been drawn within a unit cell?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number. z

1, 1 2 2 2 3

c

a

y

b x

a) b) c) d)

[1-5-3] [1-4-2] [1-6-3] [1-6-4]

Solution We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows:

x1 = a/2

y1 = b

z1 = c/2

Whereas head coordinates are as follows: x2 = 2a/3

y2 = 0b

z2 = 0c

From Equations 3.10a, 3.10b, and 3.10c assuming a value of 3 for the parameter n

⎛ x −x ⎞ ⎛ 2a/3 − a/2 ⎞ 1 u = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟= a ⎝ ⎠ 2 ⎝ a ⎠ ⎛ y −y ⎞ ⎛ 0b − b ⎞ v = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟ = −3 ⎝ b ⎠ ⎝ b ⎠ ⎛ z −z ⎞ ⎛ 0c − c /2 ⎞ 3 w = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟=− 2 ⎝ c ⎠ ⎝ c ⎠ In order to reduce these values to the lowest set of integers, we multiply each of the indices by 2.

Therefore, this direction is [1-6-3].

Reserve Question 63: Cubic direction indices 11 What are the indices for the direction represented by the vector that has been drawn within a unit cell?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number. z

c

1, 1 2 2

2, 2 3 3

b x

a) b) c) d)

[-1-43] [1-42] [0-43] [0-41]

a

y

Solution We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows:

x1 = 2a/3

y1 = 2b/3

z1 = 0c

y2 = 0b

z2 = c/2

Whereas head coordinates are as follows: x2 = a/2

From Equations 3.10a, 3.10b, and 3.10c assuming a value of 3 for the parameter n ⎛x −x ⎞ ⎛ a/2 − 2a/3 ⎞ 1 u = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟=− 2 a a ⎝ ⎠ ⎝ ⎠

⎛y −y ⎞ ⎛ 0b − 2b/3 ⎞ v = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟ = −2 b ⎝ ⎠ ⎝ b ⎠ ⎛ z −z ⎞ ⎛ c /2 − 0c ⎞ 3 w = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟= ⎝ c ⎠ 2 ⎝ c ⎠ In order to reduce these values to the lowest set of integers, we multiply each of the indices by 2.

Therefore, this direction is [-1-43].

Reserve Problem 64

Select the orthorhombic unit cell illustrating a [12-1] direction. Note: all angles are 90°. a)

b)

c)

d)

Solution

This problem calls for us to draw a [121] direction within an orthorhombic unit cell (a ≠ b ≠ c, α = β = γ = 90°). This direction may be plotted using the procedure outlined in Example Problem 3.8, with

which rearranged forms of Equations 3.10a-3.10c are used. Let us position the tail of the direction vector at the origin of our coordinate axes; this means that tail vector indices for this [121] direction are u=1 v=2 w = −1 Because the tail of the direction vector is positioned at the origin, its coordinates are as follows: x1 = 0a y1 = 0b z1 = 0c Head coordinates are determined using the rearranged forms of Equations 3.10a-3.10c, as follows: x 2 = ua + x1 = (1)a + 0 a = a

y2 = vb + y1 = 2b + 0b = 2b z2 = wc + z1 = (−1)c + 0c = −c Therefore, coordinates for the vector head are a, 2b, and −c. To locate the vector head, we start at the origin, point O, and move along the +x-axis a units (from point O to point A), then parallel to the +y-axis 2b units (from point A to point B). Finally, we proceed parallel to the z-axis -c units (from point B to point C). The [121] direction is the vector from the origin (point O) to point C as shown.

Reserve Problem 65 Select a monoclinic unit cell illustrating a [0-11] direction.

a)

b)

c)

Solution

This problem asks that a [011] direction be drawn within a monoclinic unit cell (a ≠ b ≠ c, and α = β = 90º ≠ γ). One such unit cell with its origin at point O is sketched below. This direction may be plotted using the procedure outlined in Example Problem 3.8, with which rearranged forms of Equations 3.10a-3.10c are used. Let us position the tail of the direction vector at the origin of our coordinate axes; this means that vector head indices for this [011] direction are u=0 v = −1 w=1

Because the tail of the direction vector is positioned at the origin, its coordinates are as follows: x1 = 0a y1 = 0b z1 = 0c Head coordinates are determined using the rearranged forms of Equations 3.10a-3.10c, as follows:

x2 = ua + x1 = (0)a + 0a = 0a y2 = vb + y1 = (−1)b + 0b = −b z2 = wc + z1 = (1)c + 0c = c Therefore, coordinates for the vector head are 0a, −b, and c. To locate the vector head, we start at the origin, point O. Since the x index is zero, it is not necessary to move from point O along the x-axis.

We now move from point O b units along the minus y-axis (from point O to point R). Since the final coordinate is c, we move from point R parallel to the z-axis, c units (to point P). Thus, the [011] direction corresponds to the vector passing from the origin to point P, as indicated in the figure.

Reserve Problem 66 What are the indices for the directions indicated by the two vectors in the sketch below?

Solution We are asked for the indices of the two directions sketched in the figure. Unit cell edge lengths are a = 0.4 nm, b = 0.5 nm, and c = 0.3 nm. In solving this problem we will use the symbols a, b, and c rather than the magnitudes of these parameters

The tail of the Direction 1 vector passes through the origin, therefore, its tail coordinates are x1 =0a y1 =0b z1 =0c And the vector head coordinates are as follows: x2 = 0a y2 = b/2 z2 = c To determine the directional indices we employ Equations 3.10a, 3.10b, and 3.10c. Because there is a 2 in the denominator for y2 we will assume n = 2. Hence ⎛x −x ⎞ ⎛ 0a − 0a ⎞ u = n ⎜ 2 1 ⎟ = (2) ⎜ ⎟=0 ⎝ a ⎠ ⎝ a ⎠ ⎛ y − y1 ⎞ ⎛ b /2 − 0 b ⎞ v = n⎜ 2 ⎟ = (2) ⎜ ⎟ =1 b b ⎝ ⎠ ⎝ ⎠

⎛ z −z ⎞ ⎛ c − 0c ⎞ w = n ⎜ 2 1 ⎟ = (2) ⎜ ⎟=2 c ⎝ c ⎠ ⎝ ⎠

Therefore, Direction 1 is a [012] direction.

We use the same procedure to determine the indices for Direction 2. Again, because the vector tail passes through the origin of the coordinate system, the values of x1, y1, and z1 are the same as for Direction 1. Furthermore, vector head coordinates are as follows: x2 = a/2 y2 = b/2 z2 = −c

We again choose a value of 2 for n because of the 2 in the denominator of the x2 and y2coordinates.

Therefore,

⎛ x −x ⎞ ⎛ a/2 − 0a ⎞ u = n ⎜ 2 1 ⎟ = (2) ⎜ ⎟ =1 a ⎝ ⎠ ⎝ a ⎠ ⎛y −y ⎞ ⎛ b/2 − 0b ⎞ v = n ⎜ 2 1 ⎟ = (2) ⎜ ⎟ =1 b ⎝ ⎠ ⎝ b ⎠ ⎛ z −z ⎞ ⎛ −(c ) − 0c ⎞ w = n ⎜ 2 1 ⎟ = (2) ⎜ ⎟ = −2 c ⎝ ⎠ ⎝ c ⎠ Therefore, Direction 2 is a [112] direction.

Reserve Problem 67 Determine the indices for the directions shown in the following cubic unit cell:

Solution For Direction A

We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows: x1 = 2a/3

y1 = 0b

z1 = c

y2 = b/2

z2 = c

Whereas head coordinates are as follows: x2 = 0a

From Equations 3.10a, 3.10b, and 3.10c assuming a value of 3 for the parameter n

⎛ x −x ⎞ ⎛ 0a − 2a/3 ⎞ u = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟ = −2 a ⎝ ⎠ ⎝ a ⎠ ⎛ y −y ⎞ ⎛ b/2 − 0b ⎞ 3 v = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟= b ⎝ ⎠ 2 ⎝ b ⎠ ⎛ z −z ⎞ ⎛ c −c ⎞ w = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟=0 ⎝ c ⎠ ⎝ c ⎠ Because the v index is 3/2, it is necessary to multiply each index by the factor 2 to eliminate the fraction. Therefore, Direction A is [430] .

For Direction B

We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows: x1 = a/3

y1 = b

z1 = 0c

y2 = 0b

z2 = 2c/3

Whereas head coordinates are as follows: x2 = a

From Equations 3.10a, 3.10b, and 3.10c assuming a value of 3 for the parameter n

⎛ x −x ⎞ ⎛ a − a/3 ⎞ u = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟=2 ⎝ a ⎠ ⎝ a ⎠ ⎛y −y ⎞ ⎛ 0b − b ⎞ v = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟ = −3 ⎝ b ⎠ ⎝ b ⎠ ⎛ z −z ⎞ ⎛ 2c /3 − 0c ⎞ w = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟=2 c ⎝ ⎠ ⎝ c ⎠ Therefore, Direction B is [232] .

For Direction C

We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows: x1 = a/3

y1 = b

z1 = c

y2 = 0b

z2 = 0c

Whereas head coordinates are as follows: x2 = 2a/3

From Equations 3.10a, 3.10b, and 3.10c assuming a value of 3 for the parameter n

⎛ x −x ⎞ ⎛ 2a/3 − a/3 ⎞ u = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟ =1 a ⎝ ⎠ ⎝ a ⎠ ⎛y −y ⎞ ⎛ 0b − b ⎞ v = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟ = −3 ⎝ b ⎠ ⎝ b ⎠ ⎛ z −z ⎞ ⎛ 0c − c ⎞ w = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟ = −3 ⎝ c ⎠ ⎝ c ⎠

Therefore, Direction C is [133] .

For Direction D

We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows: x1 = a/3

y1 = 0b

z1 = c

y2 = b/2

z2 = 0c

Whereas head coordinates are as follows: x2 = a/2

From Equations 3.10a, 3.10b, and 3.10c assuming a value of 3 for the parameter n

⎛ x −x ⎞ ⎛ a/2 − a/3 ⎞ 1 u = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟= a ⎝ ⎠ 2 ⎝ a ⎠ ⎛ y −y ⎞ ⎛ b/2 − 0b ⎞ 3 v = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟= b ⎝ ⎠ 2 ⎝ b ⎠ ⎛ z −z ⎞ ⎛ 0c − c ⎞ w = n ⎜ 2 1 ⎟ = (3) ⎜ ⎟ = −3 ⎝ c ⎠ ⎝ c ⎠ In order to eliminate the factor 2 in the denominators of u and v it is necessary to multiply each index by 2. Therefore, Direction D is [136] .

Reserve Problem 68

(a) What are the direction indices for a vector that passes from point 1 0 tetragonal unit cell? (b) Repeat part (a) for a rhombohedral unit cell.

1 1 1 to point 1 in a 3 2 2

Solution

(a) Point indices for the vector tail, 1 0

q =1

1 , means that 3 r=0

s=

1 3

or that, using Equations 3.9a-3.9c, lattice position coordinates to the three axes are determine as follows:

Px = qa = (1)a = a = x1 Py = rb = (0)b = 0b = y1

⎛1⎞ c Pz = sc = ⎜ ⎟ c = = z1 3 ⎝3⎠ Similarly for the vector head:

q=

1 2

r=1

s=

1 2

And we determine lattice position coordinates for using Equations 3.9a-3.9c, in a similar manner:

⎛1⎞ a Px = qa = ⎜ ⎟ a = = x2 2 ⎝2⎠ Py = rb = (1)b = b = y2

⎛1⎞ c Pz = sc = ⎜ ⎟ c = = z2 2 ⎝2⎠ And, finally, determination of the u, v, and w direction indices is possible using Equations 3.10a3.10c; because there is a 3 in the denominator and a 2 in another denominator of two of the lattice position coordinates, let us assume a value of 6 for the parameter n in these equations. Therefore,

⎛ x −x ⎞ ⎛ a/2 − a ⎞ u = n ⎜ 2 1 ⎟ = (6) ⎜ ⎟ = −3 ⎝ a ⎠ ⎝ a ⎠ ⎛y −y ⎞ ⎛ b − 0b ⎞ v = n ⎜ 2 1 ⎟ = (6) ⎜ ⎟=6 ⎝ b ⎠ ⎝ b ⎠ ⎛ z −z ⎞ ⎛ c /2 − c /3 ⎞ w = n ⎜ 2 1 ⎟ = (6) ⎜ ⎟ =1 c ⎝ ⎠ ⎝ c ⎠ Therefore, this vector points in a [361] direction.

(b) For a rhombohedral unit cell, the direction indices will also be [361] . Lattice position coordinates for both vector head and tail will be the same as for cubic. Likewise, incorporating these lattice position coordinates into Equations 3.10a-3.10c also yield u = −3 v=6 w=1

Reserve Problem 69 For tetragonal crystals, cite the indices of directions that are equivalent to each of the following directions:

a) [001] b) [110] c) [010] Solution For tetragonal crystals, a = b ≠ c and α = β = γ = 90°; therefore, projections along the x and y axes are equivalent, which are not equivalent to projections along the z axis.

(a) Therefore, for the [001] direction, there is only one equivalent direction: [001] . (b) For the [110] direction, equivalent directions are as follows: [110] , [110] , and [110] . (b) Also, for the [010] direction, equivalent directions are the following: [010] , [100] , and [100] .

Reserve Problem 70 Convert the [100] and [111] directions into the four-index Miller–Bravais scheme for hexagonal unit cells.

Solution For [100]

U=1 V=0 W=0

From Equations 3.11a-3.11d

u=

1 1 2 (2U − V ) = [(2)(1) − 0] = 3 3 3

v=

1 1 1 (2V − U ) = [(2)(0) − 1] = − 3 3 3

⎛2 1⎞ 1 t = − (U + V ) = − ⎜ − ⎟ = − 3 ⎝3 3⎠ w=W=0 It is necessary to multiply these numbers by 3 in order to reduce them to the lowest set of integers.

Thus, the direction is represented as [uvtw] = [2110] .

For [111], U = 1, V = 1, and W = 1; therefore,

1 1 u = [(2)(1) − 1] = 3 3 v=

1 1 [(2)(1) − 1] = 3 3

t=

⎛1 1⎞ 2 −⎜ + ⎟ = − 3 ⎝3 3⎠ w= 1

If we again multiply these numbers by 3, then u = 1, v = 1, t = –2, and w = 3. Thus, the direction is represented as Thus, the direction is represented as [uvtw] = [1123] .

Reserve Problem 71 Determine indices for the directions shown in the following hexagonal unit cells.

a)

b)

c)

d)

Solution (a)

One way solve this problem is to begin by determining the U, V, and W indices for the vector referenced to the three-axis scheme; this is possible using Equations 3.12a through 3.12c. Because the tail of the vector passes through the origin, a1′ = a2′ = 0a , and z ′ = 0c . Furthermore, coordinates for the vector head are as follows:

a1′′ = a

a2′′ =

a 2

z ′′ =

c 2

Now, solving for U, V, and W using Equations 3.12a, 3.12b, and 3.12c assuming a value of 2 for the parameter n ⎛ a′′ − a ′ ⎞ ⎛ a − 0a ⎞ =2 U = n ⎜ 1 1 ⎟ = (2) ⎜ ⎜ a ⎟⎟ ⎜ a ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ a′′ − a ′ ⎞ ⎛ a/2 − 0a ⎞ V = n ⎜ 2 2 ⎟ = (2) ⎜ ⎟⎟ = 1 ⎜ ⎜ a ⎟ a ⎝ ⎠ ⎝ ⎠ ⎛ c /2 − 0c ⎞ ⎛ z ′′ − z ′ ⎞ W = n⎜ ⎟⎟ = 1 ⎟ = (2) ⎜⎜ c ⎝ c ⎠ ⎝ ⎠

Now, it becomes necessary to convert these indices into an index set referenced to the four-axis scheme. This requires the use of Equations 3.11a through 3.11d, as follows:

1 1 u = (2U − V ) = ⎡⎣(2)(2) − (1) ⎤⎦ = 1 3 3 1 1 v = (2V − U ) = ⎡⎣(2)(1) − (2) ⎤⎦ = 0 3 3

(

)

t = −(u + v ) = − 1 + 0 = −1 w=W=1 No reduction is necessary inasmuch as all of these indices are integers; therefore, this direction in the four-index scheme is [1011]. (b)

1 1 u = ( 2u' − v) = ⎡⎣(2)(1) − 2 ⎤⎦ = 0 3 3

We solve this problem by first determining the U, V, and W indices for the vector referenced to the three-axis scheme; this is possible using Equations 3.12a through 3.12c. Because the tail of the vector passes through the origin, a1′ = a2′ = 0a , and z ′ = 0c . Furthermore, coordinates for the vector head are as follows:

a1′′ =

a 2

a2′′ = a

z ′′ = 0 c

Now, solving for U, V, and W using Equations 3.12a, 3.12b, and 3.12c assuming a value of 2 for the parameter n ⎛ a′′ − a ′ ⎞ ⎛ a/2 − 0a ⎞ U = n ⎜ 1 1 ⎟ = (2) ⎜ ⎟⎟ = 1 ⎜ ⎜ a ⎟ a ⎝ ⎠ ⎝ ⎠

⎛ a′′ − a ′ ⎞ ⎛ a − 0a ⎞ V = n ⎜ 2 2 ⎟ = (2) ⎜ =2 ⎜ a ⎟⎟ ⎜ a ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 0c − 0c ⎞ ⎛ z ′′ − z ′ ⎞ W = n⎜ ⎟⎟ = 0 ⎟ = (2) ⎜⎜ c ⎝ c ⎠ ⎝ ⎠

Now, it becomes necessary to convert these indices into an index set referenced to the four-axis scheme. This requires the use of Equations 3.11a through 3.11d, as follows:

1 1 u = (2U − V ) = ⎡⎣(2)(1) − (2) ⎤⎦ = 0 3 3 1 1 v = (2V − U ) = ⎡⎣(2)(4) − (1) ⎤⎦ = 1 3 3

(

)

t = −(u + v ) = − 0 + 1 = −1 w=W=0

No reduction is necessary inasmuch as all of these indices are integers; therefore, this direction in the four-index scheme is [1011]. (c)

We solve this problem by first determining the U, V, and W indices for the vector referenced to the threeaxis scheme; this is possible using Equations 3.12a through 3.12c. Because the tail of the vector passes through the origin, a1′ = a2′ = 0a , and z ′ = 0 c . Furthermore, coordinates for the vector head are as follows:

a1′′ = −a

a2′′ = −a

z ′′ =

c 2

Now, solving for U, V, and W using Equations 3.12a, 3.12b, and 3.12c assuming a value of 2 for the parameter n ⎛ a′′ − a ′ ⎞ ⎛ −a − 0 a ⎞ U = n ⎜ 1 1 ⎟ = (2) ⎜ = −2 ⎜ a ⎟⎟ ⎜ a ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ a′′ − a ′ ⎞ ⎛ −a − 0 a ⎞ V = n ⎜ 2 2 ⎟ = (2) ⎜ = −2 ⎜ a ⎟⎟ ⎜ a ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ c /2 − 0c ⎞ ⎛ z ′′ − z ′ ⎞ W = n⎜ ⎟⎟ = 1 ⎟ = (2) ⎜⎜ c ⎝ c ⎠ ⎝ ⎠

Now, it becomes necessary to convert these indices into an index set referenced to the four-axis scheme. This requires the use of Equations 3.11a through 3.11d, as follows:

1 1 2 u = (2U − V ) = ⎡⎣(2)( −2) − ( −2) ⎤⎦ = − 3 3 3 1 1 2 v = (2V − U ) = ⎡⎣(2)( −2) − ( −2) ⎤⎦ = − 3 3 3

⎛ 2 2 ⎞ 4 t = − (u + v ) = − ⎜ − − ⎟ = ⎝ 3 3 ⎠ 3 w=W=1 Now, in order to get the lowest set of integers, it is necessary to multiply all indices by the factor 3, with the result that this direction is a [2243] direction. (d)

We solve this problem by first determining the U, V, and W indices for the vector referenced to the three-axis scheme; this is possible using Equations 3.12a through 3.12c. Because the tail of the vector passes through the origin, a1′ = a2′ = 0a , and z ′ = 0 c . Furthermore, coordinates for the vector head are as follows: a1′′ = 0 a

a2′′ = − a

z ′′ = 0 c

Now, solving for U, V, and W using Equations 3.12a, 3.12b, and 3.12c assuming a value of 1 for the parameter n

⎛ a′′ − a ′ ⎞ ⎛ 0a − 0a ⎞ =0 U = n ⎜ 1 1 ⎟ = (1) ⎜ ⎜ a ⎟⎟ ⎜ a ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ a′′ − a ′ ⎞ ⎛ −a − 0 a ⎞ = −1 V = n ⎜ 2 2 ⎟ = (1) ⎜ ⎜ a ⎟⎟ ⎜ a ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 0c − 0c ⎞ ⎛ z ′′ − z ′ ⎞ W = n⎜ ⎟⎟ = 0 ⎟ = (1) ⎜⎜ c ⎝ c ⎠ ⎝ ⎠

Now, it becomes necessary to convert these indices into an index set referenced to the four-axis scheme. This requires the use of Equations 3.11a through 3.11d, as follows:

1 1 1 u = (2U − V ) = ⎡⎣(2)(0) − ( −1) ⎤⎦ = 3 3 3 1 1 2 v = (2V − U ) = ⎡⎣(2)( −1) − (0) ⎤⎦ = − 3 3 3

⎛1 2 ⎞ 1 t = − (u + v ) = − ⎜ − ⎟ = ⎝3 3 ⎠ 3 w=W=0 Now, in order to get the lowest set of integers, it is necessary to multiply all indices by the factor 3, with the result that this is a [1210] direction.

Reserve Problem 72 1. Select the orthorhombic unit cell with a (210) plane identified:

a)

b)

c)

2. Select the monoclinic unit cell with a (002) plane identified. a)

b)

c)

Solution 1. We are asked to draw a (210) plane within an orthorhombic unit cell. For the orthorhombic crystal system, relationships among the lattice parameters are as follows:

a≠b≠c

α = β = γ = 90° Thus, the three coordinate axes are parallel to one another. In order to construct the (210) plane it is necessary to determine intersections with the coordinate axes. The A, B, and C intercepts are computed using rearranged forms of Equations 3.14a, 3.14b, and 3.14c with h = 2, k = 1, and l = 0.

Thus, intercept values are as follows (assuming n = 1):

A=

na (1)a a = = h 2 2

B=

nb (1)b = =b k 1

C=

nc (1)c = = ∞c l 0

Thus, intercepts with the x, y, and z axes are respectively, a/2, b, and ∞c; this plane parallels the z axis inasmuch as its intercept is infinity. The plane that satisfies these requirements has been drawn within the orthorhombic unit cell below.

2. In this part of the problem we are asked to draw a (002) plane within a monoclinic unit cell. For the monoclinic crystal system, relationships among the lattice parameters are as follows: a≠b≠c

α = γ = 90° ≠ β In order to construct the (002) plane it is necessary to determine intersections with the coordinate axes. The A, B, and C intercepts are computed using rearranged forms of Equations 3.14a, 3.14b, and 3.14c with h = 0, k = 0, and l = 2. Thus, intercept values are as follows (assuming n = 1):

A=

na (1)a = = ∞a h 0

B=

nb (1)b = = ∞b k 0

C=

nc (1)c c = = l 2 2

Thus, intercepts with the x, y, and z axes are respectively, ∞a, ∞b, and c/2; this plane parallels both the x and y axes inasmuch their intercepts are infinity. The plane that satisfies these requirements has been drawn within the monoclinic unit cell below.

Reserve Problem 73 What are the indices for the two planes drawn in the sketch below?

Solution In order to solve for the h, k, and l indices for these two crystallographic planes it is necessary to use Equations 3.14a, 3.14b, and 3.14c.

Plane 1, is parallel to both the the x and z axes and intercepts the y axis b/2—that is A = ∞a

B = b/2

C = ∞c

Thus, from Equations 3.14a through 3.14c (assuming n = 1) we have the following:

h=

na (1)a = =0 A ∞a

k=

nb (1)b = =2 B b/2

l=

nc (1)c = =0 C ∞c

Therefore, Plane 1 is a (020) plane.

Plane 2 the intercepts the x, y, and z axes at a/2 (0.2 nm), −b/2 (−0.2 nm), and c (0.2 nm)—that is B = −b/2

A = a/2

C=c

And when we incorporate these values into Equations 3.14a, 3.14b, and 3.14c, computations of the h, k, and l indices are as follows (assuming that n = 1):

h= k=

nb (1)b = = −2 B −b/2

l= Therefore, Plane 2 is a (221) plane.

na (1)a = =2 A a/2

nc (1)c = =1 C c

Reserve Problem 75 Determine the Miller indices for the planes shown in the following unit cell:

Solution For plane A, since the plane passes through the origin of the coordinate system as shown, we will move the origin of the coordinate system one unit cell distance to the right along the y axis. Using this new origin, intersections with with the x, y, and z axes are, respectively, as follows: 2a/3, −b, and c/2; these are the values of A, B, and C, respectively. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.14a, 3.14b, and 3.14c as follows:

h=

na (1)a 3 = = A 2a/3 2

k=

nb (1)b = = −1 B −b

l=

nc (1)c = =2 C c /2

We can reduce these indices to the lowest set of integers by dividing each by the factor 2. Therefore, plane A is a (324) plane.

For plane B, its intersections with with the x, y, and z axes are a/2, b/2, and c, which are the values of A, B, and C, respectively. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.14a, 3.14b, and 3.14c as follows:

Therefore, plane B is a (221) plane.

h=

na (1)a = =2 A a/2

k=

nb (1)b = =2 B b/2

l=

nc (1)c = =1 C c

Reserve Problem 76 Find the indices of the direction that results from the intersection of each of the following pairs of planes within a cubic crystal:

a) The (100) and (010) planes b) The (111) and (11-1) planes c) The (10-1) and (001) planes

Solution (a) In the figure below is shown (100) and (010) planes, and, as indicated, their intersection results

in a [001], or equivalently, a [001] direction.

(b) In the figure below is shown (111) and (111) planes, and, as indicated, their intersection results in a [110] , or equivalently, a [110] direction.

(c) In the figure below is shown (101) and (001) planes, and, as indicated, their intersection results in a [010], or equivalently, a [010] direction.

Reserve Problem 77

Consider the reduced-sphere unit cell shown in the figure, having an origin of the coordinate system positioned at the atom labeled O. For the following sets of planes, determine which are equivalent:

a) (00-1), (010), and (-100) b) (1-10), (10-1) and (-1-10) c) (-1-1-1), (-11-1), (-111) and (1-11)

Solution (a) The unit cell in Problem 3.22 is body-centered tetragonal. Of the three planes given in the

problem statement the (100) and (010) are equivalent—that is, have the same atomic packing.

The atomic packing for these two planes as well as the (001) are shown in the figure below.

(b) Of the four planes cited in the problem statement, (110) and (110) are equivalent to one another—have the same atomic packing. The atomic arrangement of these planes is shown in the left drawing below. Furthermore, the (101) and (011) are equivalent to each other (but not to the other pair of planes); their atomic arrangement is represented in the other drawing. Note: the 0.424 nm dimension in the left-most drawing comes from the relationship ⎡ (0.30 nm )2 + (0.30 nm ) 2 ⎤ ⎣ ⎦

1/2

. Likewise, the 0.500 nm dimension found in the right-most

drawing comes from ⎡ (0.30 nm )2 + (0.40 nm ) 2 ⎤ ⎣

⎦

1/2

.

(c) All of the (11 1) , (11 1) , (111) , and (111) planes are equivalent, that is, have the same atomic packing as illustrated in the following figure:

Reserve Problem 78 The accompanying figure shows three different crystallographic planes for a unit cell of a hypothetical metal. The circles represent atoms:

a) To what crystal system does the unit cell belong? b) What would this crystal structure be called?

3 c) If the density of this metal is 8.95 g/cm , determine the atomic weight.

Solution The unit cells constructed below show the three crystallographic planes that were provided in the problem statement.

(a) This unit cell belongs to the orthorhombic crystal system since a = 0.30 nm, b = 0.40 nm, c = 0.35 nm, and α = β = γ = 90°. (b) This crystal structure would be called body-centered orthorhombic since the unit cell has orthorhombic symmetry, and an atom is located at each of the corners, as well as at the cell center. (c) In order to compute its atomic weight, we employ Equation 3.8, with n = 2; thus A=

=

ρVC N A n

(8.95 g/cm3 ) (3.0)(4.0)(3.5) (× 10-24 cm3 /unit cell)(6.022 × 10

23

atoms/mol)

2 atoms/unit cell

= 113.2 g/mol Reserve Problem 79 Convert the (a) (010) and (b) (101) planes into the four-index Miller–Bravais scheme for hexagonal unit cells.

Solution For (010), h = 0, k = 1, and l = 0, and, from Equation 3.15, the value of i is equal to i = − (h + k ) = − (0 + 1) = − 1

Therefore, the (010) plane becomes (0110) .

Now for the (101) plane, h = 1, k = 0, and l = 1, and computation of i using Equation 3.15 leads to i = − (h + k ) = − [1 + 0] = − 1

such that (101) becomes (10 11) .

Reserve Problem 80 Determine the indices for the planes shown in the following hexagonal unit cells:

Solution (a)

For this plane, intersections with a1, a2, and z axes are ∞a, ∞a, and c/2 (the plane parallels both a1 and a2 axes). Therefore, these three values are equal to A, B, and C, respectively. If we assume that

the value of n is 1, the values of h, k, and l are determined using Equations 3.14a, 3.14b, and 3.14c as follows:

h=

na (1)a = =0 A ∞a

k=

na (1)a = =0 B ∞a

l=

nc (1)c = =2 C c /2

Now, from Equation 3.15, the value of i is i = − (h + k ) = − [0 + 0] = 0

Hence, this is a (0002) plane. (b)

This plane passes through the origin of the coordinate axis system; therefore, we translate this plane one unit distance along the x axis, per the sketch shown below:

For this plane, intersections with a1, a2, and z axes are a, ∞a, and ∞c (the plane parallels both a2 and z axes). Therefore, these three values are equal to A, B, and C, respectively. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.14a, 3.14b, and 3.14c as follows:

h=

na (1)a = =1 A a

k=

na (1)a = =0 B ∞a

l=

nc (1)c = =0 C ∞c

Now, from Equation 3.15, the value of i is i = − (h + k ) = − [1 + 0] = − 1

Hence, this is a (10 10) plane. (c)

For this plane, intersections with a1, a2, and z axes are −a, a, and c. Therefore, these three values are equal to A, B, and C, respectively. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.14a, 3.14b, and 3.14c as follows:

h=

na (1)a = = −1 A −a

k=

na (1)a = =1 B a

l=

nc (1)c = =1 C c

Now, from Equation 3.15, the value of i is i = − (h + k ) = − [−1 + 1] = 0

Hence, this is a (1101) plane. (d)

For this plane, intersections with a1, a2, and z axes are −a/2, a, and c/2. Therefore, these three values are equal to A, B, and C, respectively. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.14a, 3.14b, and 3.14c as follows:

h=

na (1)a = = −2 A −a/2

k=

na (1)a = =1 B a

l=

nc (1)c = =2 C c /2

Now, from Equation 3.15, the value of i is i = − (h + k ) = − [−2 + 1] = 1

Hence, this is a (2112) plane.

Reserve Problem 81 Identify the (a) (1-101) and (b) (11-20) planes in hexagonal unit cell.

(a) 1-

2-

3-

4-

(b) 1-

2-

3-

4-

Solution

For (1 101) the values of h, k, i, and l are, respectively, 1, –1, 0, and 1. Now, for h, k, and l, we solve for values of intersections with the a1, a2, and z axes (i.e., A, B, and C) using rearranged forms of Equations 3.13a, 3.13b, and 3.13c (assuming a value of 1 for the parameter n) as follows:

A= B=

na (1)a = =a h 1

na (1)a = = −a k −1

C=

nc (1)c = =c l 1

Hence, this plane intersects the a1 axis at a, the a2 axis at –a, and the z axis at c. The plane having these intersections is shown in the figure below.

For (1120) the values of h, k, i, and l are, respectively, 1, 1, –2, and 0. Now, for h, k, and l, we solve for values of intersections with the a1, a2, and z axes (i.e., A, B, and C) using rearranged forms of Equations 3.13a, 3.13b, and 3.13c (assuming a value of 1 for the parameter n) as follows:

A=

na (1)a = =a h 1

B=

na (1)a = =a k 1

C=

nc (1)c = = ∞c l 0

Hence, this plane intersects the a1 axis at a, the a2 axis at a, and parallels the z axis. The plane having these intersections is shown in the figure below.

Reserve Question 82: Cubic plane indices 01 What are the Miller indices for the plane shown below?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number.

a) b) c) d)

(-100) (-110) (011) (010)

Solution Since the plane passes through the origin of the coordinate system as shown, we will move the origin of the coordinate system one unit cell distance horizontally along the y axis in the –y direction.

Referenced to this new origin, intercepts with the x, y, and z axes are, respectively, ∞a, b, and ∞c (this plane parallels both the x and z axes). If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.14a, 3.14b, and 3.14c, incorporating the following values for A, B, and C:

A = ∞a B = b C = ∞c Thus,

h=

na (1)a = =0 A ∞a

k=

nb (1)b = =1 B b

l=

nc (1)c = =0 C ∞c

Therefore, this plane is a (010) [or, equivalently a (0-10)] plane.

Reserve Question 83: Cubic plane indices 02 What are the Miller indices for the plane shown below?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number.

(a) (110) (b) (-110) (c) (010) (d) (111) Solution Since the plane does not pass through the origin of the coordinate system no new origin need be chosen. Intercepts with the x, y, and z axes are, respectively, a, b, and ∞c (since this plane parallels the z axis). If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.14a, 3.14b, and 3.14c, incorporating the following values for A, B, and C:

A = a B = b C = ∞c Thus,

Therefore, this plane is a (110) plane.

h=

na (1)a = =1 A a

k=

nb (1)b = =1 B b

l=

nc (1)c = =0 C ∞c

Reserve Question 84: Cubic plane indices 03 What are the Miller indices for the plane shown below?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number.

a) b) c) d)

(110) (-110) (010) (111)

Solution Since the plane does not pass through the origin of the coordinate system no new origin need be chosen. Intercepts with the x, y, and z axes are, respectively, a, b, and c. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.14a, 3.14b, and 3.14c, incorporating the following values for A, B, and C:

A=aB=bC=c Thus,

Therefore, this plane is a (111) plane.

h=

na (1)a = =1 A a

k=

nb (1)b = =1 B b

l=

nc (1)c = =1 C c

Reserve Question 85: Cubic plane indices 05 What are the Miller indices for the plane shown below?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number.

a) b) c) d)

(1-20) (1-10) (0-02) (0-10)

Solution Since the plane does not pass through the origin of the coordinate system no new origin need be chosen. Intercepts with the x, y, and z axes are, respectively, a, -b, and ∞c (since this plane parallels the z axis).

h=

Therefore, this plane is a (1-10) plane.

na (1)a = =1 A a

k=

nb (1)b = = −1 B −b

l=

nc (1)c = =0 C ∞c

Reserve Question 86: Cubic plane indices 06 What are the Miller indices for the plane shown below?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number.

a) b) c) d)

(102) (1-13) (112) (103)

Solution Since the plane does not pass through the origin of the coordinate system no new origin need be chosen. However, solving this problem is simplified if we move the origin one unit cell distance along the positive z axis. Intercepts relative to this origin with the x, y, and z axes are, respectively, a, ∞b (since this plane parallels the y axis), and c/3. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.14a, 3.14b, and 3.14c, incorporating the following values for A, B, and C:

A = a B = ∞b C = c/3 Thus,

Therefore, this plane is a (103) plane.

h=

na (1)a = =1 A a

k=

nb (1)b = =0 B ∞b

l=

nc (1)c = =3 C c /3

Reserve Question 87: Cubic plane indices 07 What are the Miller indices for the plane shown below?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number.

a) b) c) d)

(022) (013) (012) (103)

Solution Since the plane does not pass through the origin of the coordinate system no new origin need be chosen. This plane, as drawn, does not intercept the y axis. However, if we extend the plane in the +y direction one unit cell distance it does intersection the y axis at 2b. Therefore, intercepts with the x, y, and z axes are, respectively, ∞a (since this plane parallels the x axis), 2b, and c. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.14a, 3.14b, and 3.14c, incorporating the following values for A, B, and C:

A = ∞a B = 2b C = c Thus,

h=

na (1)a = =0 A ∞a

k=

nb (1)b 1 = = B 2b 2

l=

nc (1)c = =1 C c

In order to reduce these values to the lowest set of integers, we multiply each of the indices by 2.

Therefore, this plane is a (012) plane.

Reserve Question 88: Cubic Plane indices 08 What are the Miller indices for the plane shown below?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number.

a) b) c) d)

(40-3) (41-3) (41-2) (40-2)

Solution Since the plane does not pass through the origin of the coordinate system no new origin need be chosen. This plane, as drawn, does not intercept the z axis. However, if we extend the plane in the -x direction one unit cell distance it does intersection the z axis at –2c/3. Therefore, intercepts with the x, y, and z axes are, respectively, a/2, ∞b (since this plane parallels the y axis), and –2c/3. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.14a, 3.14b, and 3.14c, incorporating the following values for A, B, and C:

A = a/2 B = ∞b C = –2c/3 Thus,

h=

na (1)a = =2 A a/2

k=

nb (1)b = =0 B ∞b

l=

nc (1)c 3 = =− C −2c /3 2

In order to reduce these values to the lowest set of integers, we multiply each of the indices by 2.

Therefore, this plane is a (40-3) plane.

Reserve Question 89: Cubic plane indices 09 What are the Miller indices for the plane shown below?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number.

a) b) c) d)

(1-13) (-1-13) (-1-23) (1-23)

Solution Since the plane does not pass through the origin of the coordinate system no new origin need be chosen. The intercepts with the x, y, and z axes are, respectively, a, -b/2, and c/3. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.14a, 3.14b, and 3.14c, incorporating the following values for A, B, and C:

A = a B = -b/2 C = c/3 Thus,

h= k=

nb (1)b = = −2 B −b/2

l= Therefore, this plane is a (1-23) plane.

na (1)a = =1 A a

nc (1)c = =3 C c /3

Reserve Question 90: Cubic plane indices 10 What are the Miller indices for the plane shown below?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number.

a) (2-23) b) (1-23) c) (0-12) d) (2-32) Solution Since the plane passes through the origin of the coordinate system as shown, we will move the origin of the coordinate system one unit cell distance horizontally along the y axis in the +y direction.

Referenced to this new origin, intercepts with the x, y, and z axes are, respectively, a, -b, and 2c/3. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.14a, 3.14b, and 3.14c, incorporating the following values for A, B, and C: A = a B = -b C = 2c/3 Thus,

h=

na (1)a = =1 A a

k=

nb (1)b = = −1 B −b

l=

nc (1)c 3 = = C 2c /3 2

In order to reduce these values to the lowest set of integers, we multiply each of the indices by 2.

Therefore, this plane is a (2-23).

Reserve Question 91: Cubic plane indices 11 What are the Miller indices for the plane shown below?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number.

a) b) c) d)

(42-2) (31-3) (43-3) (24-3)

Solution Since the plane does not pass through the origin of the coordinate system no new origin need be chosen. The intercepts with the x, y, and z axes are, respectively, a/2, 2b/3, and –2c/3. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.14a, 3.14b, and 3.14c, incorporating the following values for A, B, and C:

A = a/2 B = 2b/3 C = –2c/3 Thus,

h=

na (1)a = =2 A a/2

k=

nb (1)b 3 = = B 2b/3 2

l=

nc (1)c 3 = =− C −2c /3 2

In order to reduce these values to the lowest set of integers, we multiply each of the indices by 2.

Therefore, this plane is a (43-3) plane.

Reserve Question 92: Cubic plane indices 12 What are the Miller indices for the plane shown below?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number.

a) b) c) d)

(13-3) (24-3) (14-1) (15-4)

Solution Since the plane does not pass through the origin of the coordinate system no new origin need be chosen. This plane, as drawn, does not intercept the z axis. However, if we extend the plane in the back into the plane of the screen, it intersections the z axis at –2c/3. Therefore, intercepts with the x, y, and z axes are, respectively, a, b/2, and –2c/3. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.14a, 3.14b, and 3.14c, incorporating the following values for A, B, and C:

A = a B = b/2 C = –2c/3 Thus,

h=

na (1)a = =1 A a

k=

nb (1)b = =2 B b/2

l=

nc (1)c 3 = =− C −2c /3 2

In order to reduce these values to the lowest set of integers, we multiply each of the indices by 2.

Therefore, this plane is a (24-3) plane.

Reserve Question 93: Hexagonal plane indices What are the Miller-Bravais indices for the plane shown below?

A negative index is indicated with a minus sign ("-") in front of (rather than over) the index number.

a) b) c) d)

(000-1) (0010) (0101) (0100)

Solution

Since the plane passes through the origin of the coordinate system as shown, we will move the origin of the coordinate system one unit cell distance vertically along the z axis in the +z direction.

Referenced to this new origin, intercepts with a1, a2, and z axes are, respectively, ∞a, ∞a, and -c (the plane parallels both a1 and a2 axes). Therefore, these three values are equal to A, B, and C, respectively. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.14a, 3.14b, and 3.14c as follows:

h=

na (1)a = =0 A ∞a

k=

na (1)a = =0 B ∞a

l=

nc (1)c = = −1 C −c

Now, from Equation 3.15, the value of i is

i = −(h + k ) = −(0 + 0) = 0 Hence, this is a (000-1) plane.

Reserve Question 94: Miller indices I Below is shown the atomic packing of a plane for the simple cubic crystal structure; atoms drawn to full size are represented by the circles.

From the list below select all possible sets of indices for this plane. a) b) c) d) e) f)

(011) (1-10) (10-1) (-110) (0-11) (110)

Solution The packing corresponds to a (110)-type plane for simple cubic; therefore (011), (1-10), (110), (-110), (0-11) and (10-1) are all possible answers. This plane is sketched in the illustration below.

Reserve Problem 95 (a) Derive linear density expressions for FCC [100] and [111] directions in terms of the atomic radius R

(b) Compute and compare linear density values for these same two directions for silver.

Solution (a) In the figure below is shown a [100] direction within an FCC unit cell.

For this [100] direction there is one atom at each of the two unit cell corners, and, thus, there is the equivalent of 1 atom that is centered on the direction vector. The length of this direction vector is just the unit cell edge length, 2 R 2 (Equation 3.1). Therefore, the expression for the linear density of this plane is

LD100 =

number of atoms centered on [100] direction vector length of [100] direction vector =

1 atom 2R 2

=

1 2R 2

An FCC unit cell within which is drawn a [111] direction is shown below.

For this [111] direction, the vector shown passes through only the centers of the single atom at each of its ends, and, thus, there is the equivalence of 1 atom that is centered on the direction vector. The length of this direction vector is denoted by z in this figure, which is equal to z=

x 2 + y2

where x is the length of the bottom face diagonal, which is equal to 4R. Furthermore, y is the unit cell edge length, which is equal to 2 R 2 (Equation 3.1). Thus, using the above equation, the length z may be calculated as follows: z=

( 4 R ) 2 + (2 R 2 ) 2 =

24 R 2 = 2 R 6

Therefore, the expression for the linear density of this direction is

LD111 =

number of atoms centered on [111] direction vector length of [111] direction vector =

1 atom 2R 6

=

1 2R 6

(b) From the table inside the front cover, the atomic radius for silver is 0.144 nm. Therefore, the linear density for the [100] direction is LD100 (Ag) =

1 2R 2

=

1 (2)(0.144 nm) 2

= 2.46 nm −1 = 2.46 × 109 m −1

While for the [111] direction LD111 (Ag) =

1 2R 6

=

1 (2)(0.144 nm) 6

= 1.42 nm −1 = 1.42 × 109 m −1

Reserve Problem 96 Figure 3.39 shows the first four peaks of the x-ray diffraction pattern for copper, which has an FCC crystal structure; monochromatic x-radiation having a wavelength of 0.1542 nm was used.

(a) Index (i.e., give h, k, and l indices) for each of these peaks. (b) Determine the interplanar spacing for each of the peaks. (c) For each peak, determine the atomic radius for Cu and compare these with the value presented in Table 3.1.

Solution (a) Since Cu has an FCC crystal structure, only those peaks for which h, k, and l are all either odd or even will appear. Therefore, the first peak results by diffraction from (111) planes.

(b) For each peak, in order to calculate the interplanar spacing we must employ Equation 3.20. For the first peak which occurs at 43.8° d111 =

nλ (1)(0.1542 nm) = = 0.2067 nm 2 sin θ ⎛ 43.8° ⎞ (2) ⎜ sin 2 ⎟⎠ ⎝

(c) Employment of Equations 3.21 and 3.1 is necessary for the computation of R for Cu as

R=

a

=

(dhkl ) (h)2 + (k )2 + (l )2

2 2

=

2 2

(0.2067 nm) (1)2 + (1)2 + (1)2 2 2

= 0.1266 nm Similar computations are made for the other peaks, which results are tabulated below: 2θ 50.8 74.4 90.4

Peak Index 200 220 311

dhkl(nm) 0.1797 0.1275 0.1087

R (nm) 0.1271 0.1275 0.1274

Reserve Problem 97 Below are listed diffraction angles for the first three peaks (first-order) of the x-ray diffraction pattern for some metal. Monochromatic x-radiation having a wavelength of 0.1254 nm was used. (a) Determine whether this metal’s crystal structure is FCC, BCC, or neither FCC or BCC. (b) If the crystal structure is either BCC or FCC, identify which of the metals in Table 3.1 gives this diffraction pattern.

Peak Number

Diffraction Angle (2θ)

1

31.2°

2

44.6°

3

55.4°

Solution (a) The steps in solving this part of the problem are as follows:

1. For each of these peaks compute the value of dhkl using Equation 3.20 in the form

dhkl =

nλ 2 sin θ

(P.1)

taking n = 1 since this is a first-order reflection, and λ = 0.1254 nm (as given in the problem statement).

2. Using the value of dhkl for each peak, determine the value of a from Equation 3.21 for both BCC and FCC crystal structures—that is

a = dhkl h2 + k 2 + l 2

(P.2)

For BCC the planar indices for the first three peaks are (110), (200), and (211), which yield the 2 2 2 respective h + k + l values of 2, 4, and 6. On the other hand, for FCC planar indices for the first 2 2 2 three peaks are (111), (200), and (220), which yield the respective h + k + l values of 3, 4, and 8.

3. If the three values of a are the same (or nearly the same) for either BCC or FCC then the crystal structure is which of BCC or FCC has the same a value.

4. If none of the set of a values for both FCC and BCC are the same (or nearly the same) then the crystal structure is neither BCC or FCC. Step 1 Using Equation P.1, the three values of dhkl are computed as follows: d1 =

d2 =

d3 =

nλ (1)(0.1254 nm) = = 0.2332 nm 2sin θ1 ⎛ 31.2° ⎞ (2) sin ⎜ ⎟ ⎝ 2 ⎠

nλ (1)(0.1254 nm) = = 0.1652 nm 2sin θ2 ⎛ 44.6° ⎞ (2) sin ⎜ ⎟ ⎝ 2 ⎠ nλ (1)(0.1254 nm) = = 0.1349 nm 2sin θ3 ⎛ 55.4° ⎞ (2) sin ⎜ ⎟ ⎝ 2 ⎠

Step 2 Using Equation P.2 let us first compute values of a for BCC. For the (110) set of planes

a1 (BCC) = d1 h2 + k 2 + l 2 = (0.2332 nm) 12 + 12 + 02 = 0.3298 nm For the (200) set of planes:

a2 (BCC) = d2 h2 + k 2 + l 2 = (0.1652 nm) 22 + 02 + 02 = 0.3304 nm And for the (211) set of planes:

a3 (BCC) = d3 h2 + k 2 + l 2 = (0.1349 nm) 22 + 12 + 12 = 0.3304 nm Inasmuch as these three values of a are nearly the same, the crystal structure is BCC. (b) Since we know the value of a for this BCC crystal structure, it is possible to calculate the value of the atomic radius R using a rearranged form of Equation 3.4; and once we know the value of R, we just find that metal in Table 3.1 that has this atomic radius. Thus, we calculate the value of R (using Equation 3.4) as follows (using a value of 0.3302 for a):

R=

a 3 (0.3302 nm)( 3 ) = = 0.1430 nm 4 4

From Table 3.1 the only BCC metal that has this atomic radius is tantalum.

Reserve Question 98: Noncrystalline: anisotropic The properties of noncrystalline materials are anisotropic.

a) True b) False Solution False. Noncrystalline materials do not have defined crystal structures, and, as a result, have no definite crystallographic directions; hence, the properties of noncrystalline materials are isotropic as opposed to being anisotropic.

Reserve Questions and Problems Solutions Manual CHAPTER 4 POLYMER STRUCTRES

Reserve Question 01: Double/triple bonds Hydrocarbon molecules that contain double and/or triple bonds are called a) unsaturated. b) saturated.

Solution Hydrocarbon molecules that contain double and/or triple bonds are called unsaturated.

Reserve Question 02: Hydrocarbon bonding Which type(s) of bonding is (are) found within hydrocarbon molecules? a) b) c) d) e)

Ionic bonding Hydrogen boning Covalent bonding Van der Waals bonding Metallic bonding

Solution Hydrocarbon molecules are covalently bonded.

Reserve Question 03: Hydrocarbon groups Select the correct name of each hydrocarbon group that is shown below.

A.

B. C.

D.

E.

Group A ____ Group B ____ Group C ____ Group D ____ Group E ____ Solution Group A is an aldehyde; Group B is an ether; Group C is an alcohol; Group D is an aromatic hydrocarbon; and Group E is an acid.

Reserve Question 04: Isomers Hydrocarbon compounds that have the same composition, but different atomic arrangements are called ____.

Solution Hydrocarbon compounds that have the same composition, but different atomic arrangements are called isomers.

Reserve Question 05: Crystallinity comparisons For the following two polymers: Linear polyethylene Lightly branched isotactic polypropylene

Is it possible to determine if one is more likely to crystallize than the other? a) Yes. Linear polyethylene b) Yes. Lightly branched isotactic polypropylene c) No Solution (a) The linear and syndiotactic poly(vinyl chloride) is more likely to crystallize; the phenyl sidegroup for PS is bulkier than the Cl side-group for PVC. Syndiotactic and isotactic isomers are equally likely to crystallize. (b) It is not possible to decide for these two polymers. Both heavily crosslinked and network polymers are not likely to crystallize. (c) The linear polyethylene is more likely to crystallize. The repeat unit structure for PP is chemically more complicated than is the repeat unit structure for PE. Furthermore, branched structures are less likely to crystallize than are linear structures. (d) The alternating poly(styrene-ethylene) copolymer is more likely to crystallize. Alternating copolymers crystallize more easily than do random copolymers.

Reserve Question 06: Repeat unit structures From the pull-down menus, select the correct name for each repeat unit structure that is shown below A.

B.

C.

D.

E.

F.

Repeat unit A ____ Repeat unit B ____ Repeat unit C ____ Repeat unit D ____ Repeat unit E ____ Repeat unit F ____ Solution Repeat unit A is polytetrafluoroethylene (PTFE); Repeat unit B is poly(methyl methacrylate) (PMMA); Repeat unit C is poly(vinyl chloride) (PVC); Repeat unit D is polystyrene (PS); Repeat unit E is polyethylene (PE); and Repeat unit F is polypropylene (PP).

Reserve Problem 07: Substitution of group Polyethylene may be fluorinated by inducing the random substitution of fluorine atoms for hydrogen.

For this polymer, determine the following: (a) The concentration of F (in wt %) that must be added if this substitution occurs for 18.6% of all of the original hydrogen atoms. (b) The concentration of F (in wt %) that must be added to completely fluorinate the material, i.e. to produce polytetrafluoroethylene (PTFE).

Atomic weights for several elements are included in the following table: Carbon 12.01 g/mol Chlorine 35.45 g/mol Fluorine 19.00 g/mol Hydrogen 1.008 g/mol Oxygen 16.00 g/mol Solution (a) For fluorinated polyethylene, we are asked to determine the weight percent of fluorine added for 8% F substitution of all original hydrogen atoms. Consider 50 carbon atoms; for polyethylene, there are 100 possible side-bonding sites. Ninety-two are occupied by hydrogen, and eight are occupied by F. Thus, the mass of these 50 carbon atoms, mC, is just mC = 50(AC) = (50)(12.01 g/mol) = 600.5 g Likewise, for hydrogen and fluorine, mH = 92(AH) = (92)(1.008 g/mol) = 92.74 g mF = 8(AF) = (8)(19.00 g/mol) = 152.00 g Thus, the concentration of fluorine, CF, is determined using Equation 5.6b as follows: CF =

=

mF × 100 mC +mH +mF

152.00g × 100 = 18.0 wt% 600.5 g + 92.74 g + 152.00 g

(b) In order to completely fluorinate the polyethylene all of the hydrogen sites are replaced with fluorine. As in part (a), we will consider 50 carbon atoms with 100 side-bonding sites. Since one atom of fluorine is bonded to each of these sites, there will be 100 fluorine atoms. As above, the mass of the 50 carbon atoms is 600.5 g. Whereas the mass of fluorine, mF, is computed as follows: mF = 100(AF) = (100)(19.00 g/mol) = 1900.00 g

In order to compute the concentration of fluorine, we use Equation 5.6a, as follows: CF =

=

mF × 100 mC +mF

1900.00g × 100 = 76.0wt% 600.5g + 1900.00 g

Reserve Question 08: Linear polymers Which of the following may form linear polymers? a) b) c) d) e) f)

Rubber Epoxy Polyethylene Phenol-formaldehyde Polystyrene Nylon

Solution Polyethylene, polystyrene and nylon may form linear polymers.

Reserve Question 09: Network polymers Which of the following form network polymers? a) b) c) d) e) f)

Rubber Epoxy Polyethylene Phenol-formaldehyde Polystyrene Nylon

Solution Epoxy and phenol-formaldehyde form network polymers.

Reserve Question 10: Cis/Trans Match the geometrical isomer descriptions with their names.

Groups bonded to adjacent doubly-bonded chain atoms are positioned on the same chain side. ____ Groups bonded to adjacent doubly-bonded chain atoms are positioned on opposite chain sides. ____

Solution Cis-type repeat units have their groups positioned on the same side, while trans-type repeat units have their groups positioned on opposite sides.

Reserve Question 11: Cis/trans identification Indicate which structure has the cis configuration, and which has the trans configuration. ____

____

Solution For the cis-type repeat unit unit, the two hydrogen atoms that bond to the doubly-bonded carbon atoms are on the same side of the chain (i.e., the lower repeat unit is cis). For the trans-type repeat unit, these two hydrogen atoms reside on opposite sides of the chain (i.e., the upper repeat unit is trans).

Reserve Question 12: Isotactic/syndiotactic/atactic Match the description of each stereoisomer with the name of its configuration classification.

All R groups are on the same side.

R groups are alternate sides of the chain.

R groups are randomly positioned along the chain.

____ ____ ____

Solution With the isotactic configuration, all of the R groups are situated on the same side of the chain; with the syndiotactic configuration, the R groups alternate sides of the chain; and with the atactic configuration, the R groups are randomly positioned along the chain.

Reserve Question 13: Polymer configuration For most polymers, which configuration predominates? a) Head-to-head b) Head-to-tail

Solution For most polymers, the head-to-tail configuration predominates.

Reserve Problem 14 Five pieces of plastic, each one made of a different polymer, feature the same degree of polymerization. Which one will feature, on average, the shortest molecules when fully extended? (a) PVC (b) PE (c) PTFE (d) PP (e) PS (e) They are equal lengths.

Solution: Each polymer has the same degree of polymerization, so the average molecule of each specimen features the same number of repeat units.

The fully extended length of a polymer molecule whose bond lengths are identical along the molecule backbone is given by the equation presented in Problem 4.9 of the textbook:

Lext = dN sin

θ 2

where d is the covalent bond length, N is the number of covalent bond lengths with length d, and θ is 3 the angle between two adjacent bond segments. For the C-C bond, d is 1.54 angstroms, and the sp o tetrahedral bonding angle of 109.5 .

For polymers of the type listed in this problem, each repeat unit consists of two total bond segments, since the outer bonds that are depicted are partial (shared amongst the neighboring repeat units).

Therefore, the number of bond segments with length d equals twice the number of repeat units (equals twice the degree of polymerization).

Lext = 2dn sin

θ 2

Since each polymer specimen in this scenario features the same degree of polymerization, the extended length will be the same for each polymer.

Reserve Problem 15a (question pool) Which of the following schematics is consistent with a polymer architecture featuring PE grafted onto a PTFE-PVC block-copolymer primary chain? Each circle is considered to be a repeat unit.

a)

b)

c)

d)

e)

Solution: Colored circles are implied to represent different repeat units. We are looking for the molecule whose backbone is a block copolymer (same-color circles appear in clusters along the backbone) and whose branches feature a different, and single color of circle. Grafting refers to the case where the branches feature a different repeat unit or a different repeat unit sequencing compared to the backbone species.

Reserve Problem 16a (question pool) Which of the following schematics is consistent with a polymer architecture featuring PE grafted onto a PP primary chain? Each circle is considered to be a repeat unit.

a)

b)

c)

d)

e)

Solution: Colored circles are implied to represent different repeat units. We are looking for the molecule whose backbone is a single repeat unit (same-color circles along the backbone) and whose branches feature a different, and single color of circle. Grafting refers to the case where the branches feature a different repeat unit or a different repeat unit sequencing compared to the backbone species.

Reserve Problem 17a (question pool) Which of the following schematics is consistent with a polymer architecture featuring PE grafted onto a PP primary chain? Each circle is considered to be a repeat unit.

a)

b)

c)

d)

e)

Solution: Colored circles are implied to represent different repeat units. We are looking for the molecule whose backbone is a single repeat unit (same-color circles along the backbone) and whose branches feature a different, and single color of circle. Grafting refers to the case where the branches feature a different repeat unit or a different repeat unit sequencing compared to the backbone species.

Reserve Problem 18a (question pool) Which of the following schematics is consistent with a polymer architecture featuring PE grafted onto a PP-PVC alternating-copolymer primary chain? Each circle is considered to be a repeat unit.

a)

b)

c)

d)

e)

Solution: Colored circles are implied to represent different repeat units. We are looking for the molecule whose backbone is an alternating copolymer (differently colored circles alternating along the backbone) and whose branches feature a different, and single color of circle. Grafting refers to the case where the branches feature a different repeat unit or a different repeat unit sequencing compared to the backbone species.

Reserve Problem 19a (question pool) Which of the following schematics is consistent with a polymer architecture featuring PP grafted onto a PP-PVC alternating-copolymer primary chain? Each circle is considered to be a repeat unit.

a)

b)

c)

d)

e)

Solution: Colored circles are implied to represent different repeat units. We are looking for the molecule whose backbone is an alternating copolymer (differently colored circles alternating along the backbone) and whose branches feature a single color of circle that also appears along the backbone.

Grafting refers to the case where the branches feature a different repeat unit or a different repeat unit sequencing compared to the backbone species.

Reserve Problem 20a (question pool) Which of the following schematics is consistent with a polymer architecture featuring PP-PVC grafted onto a PP primary chain? Each circle is considered to be a repeat unit.

a)

b)

c)

d)

e)

Solution: Colored circles are implied to represent different repeat units. We are looking for the molecule whose backbone is a single repeat unit (same-color circles along the backbone) and whose branches feature two different repeat units (colors) that alternate. Grafting refers to the case where the branches feature a different repeat unit or a different repeat unit sequencing compared to the backbone species.

Reserve Question 21: Polymer crystallinity Is it possible to produce a polymer that is 100% crystalline?

a) True b) False Solution False. It is not possible to produce a polymer that is 100% crystalline. The maximum crystallinity that can be obtained is about 95%, with the remaining material being amorphous.

Reserve Questions and Problems Solutions Manual CHAPTER 5 IMPERFECTIONS IN SOLIDS

Reserve Problem 01a (question pool) Consider the schematic nanostructure depicted below.

Which of the following statements is FALSE regarding this schematic structure?

Do not extrapolate the field of view. Consider only what you are shown.

(a) Each of the phases features a similar concentration of vacancies. (b) The microstructure features exactly two components and two different phases. (c) None of the phases present features interstitial impurities. (d) Only one phase boundary is depicted. (e) Two grain boundaries are depicted.

Solution: (a) is true because each phase occupies a similar area and each features two vacancies in the field of view. (b) is true. Each color represents a different atom (component). Red-rich and orange-rich areas represent two different phases. (c) is true since no interstitials are depicted. (d) is true since the horizontal boundary between the orange-rich and blue-rich areas is a phase boundary. Note that each phase is depicted to feature a different type of 2D packing. (e) is false since only one vertical grain boundary is present in the upper, orange-rich, phase.

Reserve Problem 02a (question pool) Consider the schematic nanostructure depicted below.

Which of the following statements is FALSE regarding this schematic structure?

Do not extrapolate the field of view. Consider only what you are shown.

(a) Each of the phases features a similar concentration of vacancies. (b) The microstructure features exactly two components and two different phases. (c) One of the phases present features interstitial impurities. (d) Only one phase boundary is depicted. (e) One grain boundary is depicted.

Solution: (a) is true because each phase occupies a similar area and each features two vacancies in the field of view. (b) is false since there are three distinct atom types (red, green, and blue) and these do not form ordered compounds, but instead form two different solid solution phases. (c) is true since only the green-rich phase features smaller red atoms embedded in interstitial sites. (d) is true since the horizontal boundary between the green-rich and blue-rich areas is a phase boundary. Note that each phase is depicted to feature a different type of 2D packing. (e) is true since only one vertical grain boundary is present in the upper, blue-rich, phase.

Reserve Problem 03a (question pool) Consider the schematic nanostructure depicted below.

Which of the following statements is FALSE regarding this schematic structure?

Do not extrapolate the field of view. Consider only what you are shown.

(a) Each of the phases features a similar concentration of vacancies. (b) The microstructure features exactly three components and two different phases. (c) Both of the phases present features interstitial impurities. (d) Only one phase boundary is depicted. (e) One grain boundary is depicted.

Solution: (a) is true because each phase occupies a similar area and each features two vacancies in the field of view. (b) is true since there are three distinct atom types (silver, green, and black) and these do not form ordered compounds, but instead form two different solid solution phases. (c) is false since only the green-rich phase features smaller black atoms embedded in interstitial sites. (d) is true since the horizontal boundary between the green-rich and blue-rich areas is a phase boundary. Note that each phase is depicted to feature a different type of 2D packing. (e) is true since only one vertical grain boundary is present in the upper, blue-rich, phase.

Reserve Problem 04: Energy from temperature The number of vacancies in some hypothetical metal increases by a factor of 5 when the temperature is increased from 1040 K to 1150 K. Calculate the energy (in kJ/mol) for vacancy formation assuming that the density of the metal remains the same over this temperature range.

Solution If the factor by which the number of vacancies increases as the temperature is increased from T2 to T1 (i.e., T1 > T2) is denoted by f, then f may be determined by the following expression, which incorporates Equation 4.1: ⎛ Q ⎞ N exp ⎜ − v ⎟ N ⎝ kT1 ⎠ = exp ⎡ − Qv ⎛ 1 − 1 ⎞ ⎤ f = v1 = ⎢ ⎜⎜ ⎟⎟ ⎥ N v2 ⎛ Qv ⎞ ⎢⎣ k ⎝ T1 T2 ⎠ ⎥⎦ N exp ⎜ − ⎟ ⎝ kT2 ⎠

Taking natural logarithms of both sides of this equation gives the following:

ln f = −

Qv ⎛ 1 1 ⎞ ⎜ − ⎟ k ⎜⎝ T1 T2 ⎟⎠

And solving this equation for the energy of vacancy formation, Qv, leads to Qv = −

k ln f 1 1 − T1 T2

From the problem statement we take T1 = 1170 K and T2 = 1030 K [and use the gas constant R (8.31 J/mol-K) instead of Boltzmann’s constant], then we compute the value of Qv as follows: Qv = −

(8.31J/mol-K)[ln(5)] = 115,000J/mol = 115kJ/mol 1 1 − 1170K 1030K

Reserve Problem 05: Metallic vacancies – temperature 24

–3

The number of vacancies present in some metal at 864°C is 1.1 × 10 m . Calculate the number of vacancies at 463°C given that the energy for vacancy formation is 1.25 eV/atom; assume that the density at both temperatures is the same Solution The number of vacancies per cubic meter may be calculated using Equation 4.1, that is

⎛ Q ⎞ N v = N exp ⎜ − v ⎟ ⎝ kT ⎠ Or, solving for N, the total number of atomic sites per cubic meter

N=

⎛Q ⎞ = N v exp ⎜ v ⎟ ⎛ Q ⎞ ⎝ kT ⎠ exp ⎜ − v ⎟ ⎝ kT ⎠ Nv

Because the densities are assumed to be the same at both temperatures, so also will the value of N be the same at both T’s. If we use the subscripts “1” and “2” to denote the lower and higher temperatures respectively, then it is the case that ⎛Q ⎞ ⎛Q ⎞ N v1 exp ⎜⎜ v ⎟⎟ = N v 2 exp ⎜⎜ v ⎟⎟ ⎝ kT1 ⎠ ⎝ kT2 ⎠ In this problem we are asked to determine the number of vacancies at the lower temperature Nv1.

Solving for Nv1 from the above expression leads to the following ⎛Q ⎞ exp ⎜ v ⎟ ⎝ kT2 ⎠ = N exp ⎛ Qv ⎞ ⎛ 1 − 1 ⎞ N v1 = N v 2 ⎟ ⎜ ⎟⎜ v2 k ⎠⎝⎜ T2 T1 ⎟⎠ ⎛ Qv ⎞ ⎝ exp ⎜ ⎟ ⎝ kT1 ⎠

Given the following values: 24

–3

Nv2 = 1.4 × 10 m T2 = 881°C Qv = 1.15 eV/atom –5 k = 8.62 × 10 eV/atom-K

The value of Nv1 at 628°C is determined as follows:

⎛ ⎞⎛ 1.15eV/atom 1 1 ⎞ N v1 = (1.4 × 1024 m−3 ) exp ⎜ − ⎟⎜ ⎟ −5 ⎝ 8.62 × 10 eV/atom-K ⎠ ⎝ 881 + 273K 628 + 273K ⎠ = 5.45 × 1022m −3 (5.45E+22m −3 ) Reserve Question 06: Vacancies vs. self-interstitials In metals, there are significantly more vacancies than self-interstitials.

a) True b) False Solution True. In metals, there are significantly more vacancies than self-interstitials; the reason for this is that the atom is significantly larger than the interstitial position in which it is situated, and, consequently significant lattice strains result.

Reserve Problem 07 Using the following data that relate to the formation of Schottky defects in some oxide ceramic (having the chemical formula MO), determine the following: T ( °C) 3

−3

(g/cm )

Ns (m )

750

5.50

9.21 × 10

1000

5.44

?

1250

5.37

5.0 × 10

19

22

(a) The energy for defect formation (in eV) (b) The equilibrium number of Schottky defects per cubic meter at 1000°C. (c) The identity of the oxide (i.e., what is the metal M?)

Solution This problem provides for some oxide ceramic, at temperatures of 750°C and 1250°C, values for density and the number of Schottky defects per cubic meter. The (a) portion of the problem asks that we compute the energy for defect formation. To begin, let us combine a modified form of Equation 5.2 and Equation 5.1 as

⎛ Q ⎞ N s = N exp ⎜ − s ⎟ ⎝ 2kT ⎠

⎛ NAρ ⎞ ⎛ Qs ⎞ = ⎜⎜ ⎟⎟ exp ⎜ − ⎟ ⎝ 2kT ⎠ ⎝ AM + AO ⎠ Inasmuch as this is a hypothetical oxide material, we don’t know the atomic weight of metal M, nor the value of Qs in the above equation. Therefore, let us write equations of the above form for two temperatures, T1 and T2. These are as follows:

⎛ N A ρ1 ⎞ ⎛ Qs ⎞ N s1 = ⎜⎜ ⎟⎟ exp ⎜⎜ − ⎟⎟ ⎝ 2kT1 ⎠ ⎝ AM + AO ⎠

(5.S1a)

⎛ N A ρ2 ⎞ ⎛ Qs ⎞ N s2 = ⎜⎜ ⎟⎟ exp ⎜⎜ − ⎟⎟ ⎝ 2kT2 ⎠ ⎝ AM + AO ⎠

(5.S1b)

Dividing the first of these equations by the second leads to ⎛ N A ρ1 ⎞ ⎛ Qs ⎞ ⎜ ⎟ exp ⎜ − ⎟ 2kT1 ⎠ AM + AO ⎠ N s1 ⎝ ⎝ = N s2 ⎛ N A ρ2 ⎞ ⎛ Qs ⎞ ⎜ ⎟ exp ⎜ − ⎟ ⎝ 2kT2 ⎠ ⎝ AM + AO ⎠ which, after some algebraic manipulation, reduces to the form

⎡ Q ⎛1 N s1 ρ 1 ⎞⎤ = 1 exp ⎢ − s ⎜⎜ − ⎟⎥ N s2 T2 ⎟⎠ ⎦⎥ ρ2 ⎣⎢ 2k ⎝ T1 Now, taking natural logarithms of both sides of this equation gives

⎛N ⎞ ⎛ρ ⎞ Q ⎛1 1⎞ − ln ⎜⎜ s1 ⎟⎟ = ln ⎜⎜ 1 ⎟⎟ − s ⎜⎜ ⎟ T2 ⎟⎠ 2k ⎝ T1 ⎝ ρ2 ⎠ ⎝ N s2 ⎠ and solving for Qs leads to the expression

⎡ ⎛N ⎞ ⎛ ρ ⎞⎤ −2k ⎢ln ⎜ s1 ⎟ − ln ⎜ 1 ⎟ ⎥ ⎝ ρ2 ⎠ ⎦⎥ ⎣⎢ ⎝ N s2 ⎠ Qs = 1 1 − T1 T2

(5.S2)

Let us take T1 = 750°C and T2 = 1250°C, and we may compute the value of Qs as

⎡ ⎛ 9.2 × 1019 m-3 ⎞ ⎛ 5.50 g/cm3 ⎞ ⎤ −(2)(8.62 × 10-5 eV/K) ⎢ln ⎜ − ln ⎟ ⎜⎜ ⎟⎥ 22 -3 3⎟ ⎢⎣ ⎜⎝ 5.0 × 10 m ⎟⎠ ⎝ 5.37 g/cm ⎠ ⎥⎦ Qs = 1 1 − 750 + 273 K 1250 + 273 K = 3.40 eV (b) It is now possible to solve for Ns at 1000°C using Equation 5.S2 above. This time let’s take T1 = 1000°C and T2 = 750°C. Thus, solving for Ns1, substituting values provided in the problem statement and Qs determined above yields

N s1 =

=

(9.2 × 1019 m-3 )( 5.44 g/cm3 ) 5.50 g/cm3

N s 2 ρ1

ρ2

⎡ Q ⎛1 1 ⎞⎤ exp ⎢ − s ⎜⎜ − ⎟⎥ T2 ⎟⎠ ⎦⎥ ⎣⎢ 2k ⎝ T1

⎡ ⎛ ⎞⎤ 1 1 3.40 eV − ⎥ exp ⎢− ⎜ ⎟ 750 + 273 K ⎠ ⎥ ⎢⎣ (2)(8.62 × 10 −5 eV/K) ⎝ 1000 + 273 K ⎦ 21

–3

= 4.0 × 10 m

(c) And, finally, we want to determine the identity of metal M. This is possible by computing the atomic weight of M (AM) from Equation 5.S1a. Rearrangement of this expression leads to

⎛ N A ρ1 ⎞ ⎛ Qs ⎞ ⎜⎜ ⎟⎟ = N s1exp ⎜⎜ ⎟⎟ ⎝ 2kT1 ⎠ ⎝ AM + AO ⎠ And, after further algebraic manipulation ⎡ ⎤ ⎢ ⎥ N A ρ1 ⎢ ⎥ =A +A M O ⎢ ⎛ Qs ⎞ ⎥ ⎢ N s1exp ⎜ ⎥ ⎟ ⎢⎣ ⎝ 2kT1 ⎠ ⎥⎦ And, solving this expression for AM gives ⎡ ⎤ ⎢ ⎥ N A ρ1 ⎢ ⎥ − A AM = ⎢ O ⎛ Qs ⎞ ⎥ ⎢ N s1exp ⎜ ⎥ ⎟ ⎢⎣ ⎝ 2kT1 ⎠ ⎥⎦

Now, assuming that T1 = 750°C, the value of AM is

⎧ ⎫ ⎪ ⎪ (6.022 × 1023 ions/mol)( 5.50 g/cm3 )(106 cm3 /m3 ) ⎪ ⎪ AM = ⎨ ⎬ − 16.00 g/mol ⎤⎪ 3.40 eV ⎪ (9.2 × 1019 ions/m3 ) exp ⎡ ⎢ ⎥⎪ -5 ⎪ ⎣ (2)(8.62 × 10 eV/K)(750 + 273 K) ⎦ ⎭ ⎩ = 136.7 g/mol 2+

Upon consultation of the periodic table in Figure 2.8, the divalent metal (i.e., that forms M ions) that has an atomic weight closest to 136.7 g/mol is barium. Thus, this metal oxide is BaO. Reserve Problem 08 Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MnO? Explain your answers.

(a) MgO (b) CaO (c) BeO (d) NiO Solution In order to have a high degree (or 100% solubility) of one ceramic compound in another ceramic compound, the following criteria must be met: the relative sizes of cations/anions must be similar (< ±15%); the valence (or charges) on the cations and anions of both compounds must be the same; and the crystals structures of the two compounds must the same or similar. Inasmuch as crystal structure information for the compounds in this problem, for the most part, is not provided in the text, we will base our criteria only on the relative ion sizes and charges on the anions and cations. Because all of these ceramics are oxides and the charge on all of the cations is +2, in making solubility determinations, only the relative sizes of the cations will be considered.

2+

2+

(a) For MgO, the ionic radii of the Mn and Mg (found inside the front cover of the book) are 0.067 nm and 0.072 nm, respectively. Therefore the percentage difference in ionic radii, Δr% is determined as follows:

Δr % = =

rMg2+ − rMn2+ rMg2+

× 100

0.072 nm − 0.067 nm × 100 = 6.9% 0.072 nm

which value is with the acceptable range for a high degree of solubility.

2+

2+

(b) For CaO, the ionic radii of the Mn and Ca (found inside the front cover of the book) are 0.067 nm and 0.100 nm, respectively. Therefore, the percentage difference in ionic radii, Δr% is determined as follows:

r 2+ − rMn2+ × 100 Δr % = Ca rCa2+ =

0.100 nm − 0.067 nm × 100 = 33% 0.100 nm

This Δr% value is much larger than the ±15% range, and, therefore, CaO is not expected to experience any appreciable solubility in MnO. 2+

2+

(c) For BeO, the ionic radii of the Mn and Be (found inside the front cover of the book) are 0.067 nm and 0.035 nm, respectively. Therefore, the percentage difference in ionic radii, Δr% is determined as follows:

r 2+ − rBe2+ Δr % = Mn × 100 rMn 2+

=

0.067 nm − 0.035 nm × 100 = 48% 0.067 nm

This Δr% value is much larger than the ±15% range, and, therefore, BeO is not expected to experience any appreciable solubility in MnO. 2+

2+

(d) For NiO, the ionic radii of the Mn and Ni (found inside the front cover of the book) are 0.067 nm and 0.069 nm, respectively. Therefore, the percentage difference in ionic radii, Δr% is determined as follows:

r 2+ − rNi2+ Δr % = Mn × 100 rMn2+ =

0.069 nm − 0.067 nm × 100 = 3% 0.069 nm

which value is, of course, within the acceptable range for a high degree of solubility.

Reserve Problem 09 +

2+

(a) Suppose that Li2O is added as an impurity to CaO. If the Li substitutes for Ca , what kind of + vacancies would you expect to form? How many of these vacancies are created for every Li added?

-

2-

(b) Suppose that CaCl2 is added as an impurity to CaO. If the Cl substitutes for O , what kind of vacancies would you expect to form? How many of the vacancies are created for every Cl added?

Solution +

2+

+

(a) For Li substituting for Ca in CaO, oxygen vacancies would be created. For each Li 2+ substituting for Ca , one positive charge is removed; in order to maintain charge neutrality, a single negative charge may be removed. Negative charges are eliminated by creating oxygen + vacancies, and for every two Li ions added, a single oxygen vacancy is formed.

-

2-

-

(b) For Cl substituting for O in CaO, calcium vacancies would be created. For each Cl substituting 2for an O , one negative charge is removed; in order to maintain charge neutrality, two Cl ions will lead to the formation of one calcium vacancy.

Reserve Problem 10 Atomic radius, crystal structure, electronegativity, and the most common valence are given in the following table for several elements; for those that are nonmetals, only atomic radii are indicated.

Element

Atomic Radius (nm)

Crystal Strructure

Electronegativity

Valence

Cu

0.1278

FCC

1.9

+2

C

0.071

H

0.046

O

0.060

Ag

0.1445

FCC

1.9

+1

Al

0.1431

FCC

1.5

+3

Co

0.1253

HCP

1.8

+2

Cr

0.1249

BCC

1.6

+3

Fe

0.1241

BCC

1.8

+2

Ni

0.1246

FCC

1.8

+2

Pd

0.1376

FCC

2.2

+2

Pt

0.1387

FCC

2.2

+2

Zn

0.1332

HCP

1.6

+2

Choose which of these elements you would expect to form the following with copper:

A substitutional solid solution having complete solubility • • • • • • • • • • • •

Ni O H Pt Pd Co C Zn Ag Al Cr Fe

A substitutional solid solution of incomplete solubility • • • • • • • • • • • •

Pd Al Cr Fe H Pt Ni Zn C Ag O Co

An interstitial solid solution • • • • • • • • • • • •

C Pd Pt H Cr O Zn Ag Al Co Fe Ni

Solution In this problem we are asked to cite which of the elements listed form with Cu the three possible solid solution types. For complete substitutional solubility the following criteria must be met: 1) the difference in atomic radii between Cu and the other element (ΔR%) must be less than ±15%, 2) the crystal structures must be the same, 3) the electronegativities must be similar, and 4) the valences should be the same, or nearly the same. Below are tabulated, for the various elements, these criteria.

Element Cu C H O Ag Al Co Cr Fe Ni Pd Pt Zn

ΔR%

–44 –64 –53 +13 +12 –2 –2 –3 –3 +8 +9 +4

Crystal Structure FCC

ΔElectronegativity

FCC FCC HCP BCC BCC FCC FCC FCC HCP

0 –0.4 –0.1 –0.3 –0.1 –0.1 +0.3 +0.3 –0.3

Valence 2+

1+ 3+ 2+ 3+ 2+ 2+ 2+ 2+ 2+

(a) Ni, Pd, and Pt meet all of the criteria and thus form substitutional solid solutions having complete solubility. At elevated temperatures Co and Fe experience allotropic transformations to the FCC crystal structure, and thus display complete solid solubility at these temperatures. (b) Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility. All these metals have either BCC or HCP crystal structures, and/or the difference between their atomic radii and that for Cu are greater than ±15%, and/or have a valence different than 2+. (c) C, H, and O form interstitial solid solutions. These elements have atomic radii that are significantly smaller than the atomic radius of Cu.

Reserve Problem 11: Wt% to concentration

The concentration of carbon in an iron-carbon alloy is 0.57 wt%. What is the concentration in kilograms of carbon per cubic meter of alloy? The densities of iron and carbon are 7.87 and 2.25 3 g/cm , respectively.

Solution 3

In order to compute the concentration in kg/m of C in a 0.15 wt% C-99.85 wt% Fe alloy we must use Equation 5.12a, that is ⎛ ⎞ ⎜ C1 ⎟ C1′′ = ⎜ C1 C2 ⎟ × 103 + ⎜ ⎟ ⎜ ρ1 ρ 2 ⎟ ⎝ ⎠

For this problem, the above equation takes the form ⎛ ⎞ ⎜ CC ⎟ CC′′ = ⎜ CC CFe ⎟ × 103 ⎜ ⎟ + ⎜ ρC ρFe ⎟ ⎝ ⎠ 3

As given in the problem statement, the densities for iron and carbon are 7.87 and 2.25 g/cm , respectively, and, therefore ⎛ ⎞ 0.15wt% ⎜ ⎟ ⎜ 99.85wt% ⎟ × 103 CC′′ = 0.15wt% + ⎜ ⎟ 3 3 ⎜ 2.25g/cm 7.87g/cm ⎟ ⎝ ⎠

= 11.8kg/m3 Reserve Problem 12 –7

The concentration of gallium in silicon is 5.0 × 10 at%. What is the concentration in kilograms of gallium per cubic meter?

Solution Strategy for solving: Step 1: Convert the concentration of Ga from at% to wt% using Equation 5.10a Step 2: Compute the concentration of kilograms per cubic meter using Equation 5.12a

Step 1: From Equation 5.10a, the concentration of Ga in weight percent is determined as follows: CGa =

′ AGa CGa × 100 ′ AGa + CSi ′ ASi CGa

where

′ = 5.0 × 10− at% CGa 7

′ = 100.00 at% − 5.0 × 10− at% = 99.9999995 at% CSi 7

AGa = 69.72 g/mol (from inside front cover) ASi = 28.09 g/mol (from inside front cover) Now, solve for CGa using the above equation

CGa =

(5.0 × 10 −7 at%)(69.72 g/mol) (5.0 × 10 −7 at%)(69.72 g/mol) + (99.9999995 at%)(28.09 g.mol)

× 100

= 1.24 × 10 −6 wt% Step 2: We now compute the number of kilograms per meter cubed (Equation 5.12a) with 3

ρSi = 2.33 g/cm (from inside front cover) 3 ρGa = 5.90 g/cm (from inside front cover) CSi = 100.00 wt% − 1.24 × 10 −6 wt% = 99.99999876 Therefore ⎛ ⎞ CGa ⎜ ⎟ ⎜ ⎟ × 103 C C ′′ CGa = ⎜ Ga + Si ⎟ ⎜ ρGaP ρSi ⎟ ⎝ ⎠ ⎛ ⎞ 1.24 ×10-6 wt% ⎜ ⎟ ⎜ ⎟ -6 = ⎜ 1.24 ×10 wt% 99.99999876 wt% ⎟ × 103 + ⎜ 5.90 g / cm3 ⎟ 2.33 g / cm3 ⎜ ⎟ ⎝ ⎠

= 2.89 × 10 −5 kg/m3 Reserve Problem 13

Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities 3 of metals A and B are 4.27 and 6.35 g/cm , respectively, whereas their respective atomic weights are 61.4 and 125.7 g/mol, determine whether the crystal structure for this alloy is simple cubic, facecentered cubic, or body-centered cubic. Assume a unit cell edge length of 0.395 nm.

Solution In order to solve this problem it is necessary to employ Equation 3.8; in this expression density and atomic weight will be averages for the alloy—that is

ρave =

nAave VC N A 3

Inasmuch as for each of the possible crystal structures, the unit cell is cubic, then VC = a , or

nA

ρave = 3 ave

a NA

And, in order to determine the crystal structure it is necessary to solve for n, the number of atoms per unit cell. For n =1, the crystal structure is simple cubic, whereas for n values of 2 and 4, the crystal structure will be either BCC or FCC, respectively. When we solve the above expression for n the result is as follows:

n=

ρave a3 N A Aave

Expressions for Aave and ρaveare found in Equations 5.13a and 5.14a, respectively, which, when incorporated into the above expression yields

⎛ ⎞ ⎜ 100 ⎟ ⎜ CA CB ⎟ a3 N A + ⎜ ⎟ ⎜ ρ A ρB ⎟ ⎠ n= ⎝ ⎛ ⎞ ⎜ 100 ⎟ ⎜C C ⎟ ⎜ A + B ⎟ ⎜ AA AB ⎟ ⎝ ⎠ Substitution of the concentration values (i.e., CA = 12.5 wt% and CB = 87.5 wt%) as well as values for the other parameters given in the problem statement, into the above equation gives ⎛ ⎞ 100 ⎜ ⎟ ⎜ 12.5 wt% 87.5 wt% ⎟ (3.95 × 10-8 nm)3 (6.022 × 1023 atoms/mol) + ⎜ ⎟ 3 6.35 g/cm3 ⎟ ⎜ 4.27 g/cm ⎠ n= ⎝ ⎛ ⎞ 100 ⎜ ⎟ ⎜ 12.5 wt% 87.5 wt% ⎟ + ⎜ ⎟ ⎜ 61.4 g/mol 125.7 g/mol ⎟ ⎝ ⎠

= 2.00 atoms/unit cell Therefore, on the basis of this value, the crystal structure is body-centered cubic.

Reserve Problem 14 For a BCC iron-carbon alloy that contains 0.15 wt% C, calculate the fraction of unit cells that contain carbon atoms.

Solution It is first necessary to compute the number of carbon atoms per cubic centimeter of alloy using Equation 5.24. For this problem

A1 = AC = 12.011 g/mol (from inside the front cover) C1 = CC = 0.15 wt% 3 ρ1 = ρC = 2.25 g/cm (from inside front cover) 3 ρ2 = ρFe = 7.87 g/cm (from inside front cover) Now using from Equation 5.24

NC =

N A CC CC AC AC (100 − CC ) +

ρC

=

ρFe

(6.022 × 1023atoms/mol)(0.15 wt%) (0.15 wt%)(12.011 g/mol) (12.011 g/mol) + (100 − 0.15 wt%) 2.25 g/cm3 7.87 g/cm3 20

3

= 5.90 × 10 C atoms/cm

We now perform the same calculation for iron atoms, where A1 = AFe = 55.85 g/mol (from inside the front cover) C1 = CFe = 100% − 0.15 wt% = 99.85 wt% 3 ρ1 = ρFe = 7.87 g/cm (from inside front cover) 3 ρ2 = ρC = 2.25 g/cm (from inside front cover) Therefore,

=

(6.022 × 1023atoms/mol)(99.85 wt%) (99.85 wt%)(55.85 g/mol) (55.85 g/mol) (100 − 99.85 wt%) + 7.87 g/cm3 2.25 g/cm3 22

3

= 8.44 × 10 Fe atoms/cm

From Section 3.4 there are two Fe atoms associated with each unit cell; hence, the number of unit cells in a centimeter cubed, ni, is equal to No. Fe atoms/cm3 No. atoms/unit cell 8.44 × 1022 atoms/cm3 = = 4.22 × 1022 unit cells/cm3 2 atoms/unit cell ni =

Now, the fraction of unit cells that contain carbon atoms, nf , is just the number of carbon atoms per 20 22 centimeter cubed (5.90 × 10 ) divided by the number of unit cells per centimeter cubed (4.22 × 10 ), or

nf =

5.90 × 1020 C atoms/cm3 4.22 × 1022 unit cells/cm3

= 1.40 × 10 −2 atoms/unit cell The reciprocal of nf is the number of unit cells per atom—viz.

1 1 = − 2 n f 1.40 × 10 atoms/unit cell = 71.5 unit cells/atom

That is, there is one carbon atom per 71.5 unit cells.

Reserve Problem 15 –6

For Si to which has had added 1.5 × 10 at% of arsenic calculate the number of As atoms per cubic meter.

Solution

To make this computation we need to use Equation 5.24. However, because the concentration of As is 6 given in atom percent, it is necessary to convert 1.5 × 10− at% to weight percent using Equation 5.10a.

For this computation C1 = CAs = concentration of As in wt%

′ atom percent of As = 1.5 × 10− at% C1′ = CAs 6

′ atom percent of Si = 100 at% − 1.5 × 10− at% = 99.9999985 at% C2′ = CSi 66

A1 = AAs = atomic weight of As = 74.92 g/mol (from Figure 2.8) A2 = ASi = atomic weight of Si = 28.09 g/mol (from inside front cover) Now using Equation 5.10a we compute the value of C1 (or CAs) as follows: C1 =

=

C1′ A1 × 100 C1′ A1 + C2′ A2

(1.5 × 10 −6 wt%)(74.92 g/mol) (1.5 × 10 −6 wt%)(74.92 g/mol) + (99.9999985 wt%)(28.09 g/mol) 4.00 × 10− wt% 8

× 100

Now employment of Equation 5.24 with N1= number of As atoms per meter cubed 8 C1 = concentration of As in weight percent = 4.00 × 10− wt% A1 = atomic weight of As = 74.92 g/mol (from Figure 2.8) 3 ρ1 = density of As = 5.78 g/cm (Internet) 3 ρ2 = density of silicon = 2.33 g/cm (inside front cover) leads to

N1 =

N A C1 C1 A1 A1 + (100 − C1 )

ρ1

=

ρ2

(6.022 × 1023 atoms/mol)(4.00 × 10−8 wt%) (4.00 × 10 −8 wt%)(74.92 g/mol) 5.78 g/cm

3

+

(74.92 g/mol) 3

2.33 g/cm

12

(100 − 4.00 × 10 −8 wt%)

= 7.49 × 10 As atoms/cm

3

Reserve Problem 16

Electronic devices found in integrated circuits are composed of very high purity silicon to which has been added small and very controlled concentrations of elements found in Groups IIIA and VA of the 21 periodic table. For Si that has had added 8.3 × 10 atoms per cubic meter of antimony compute (a) the weight percent and (b) the atom percent of Sb present.

Solution (a) This part of the problem is solved using Equation 5.24, in which

C1 = concentration of Sb in wt% 21 3 3 6 3 N1 = number of Sb atoms per cubic centimeter = (8.3 × 10 atoms/m )(1 m /10 cm ) 15 3 = 8.3 × 10 atoms/cm A1 = atomic weight of Sb = 121.76 g/mol (Figure 2.8) 3 ρ1 = density of Sb = 6.68 g/cm (Internet) 3 ρ2 = density of silicon = 2.33 g/cm (inside front cover) And, from Equation 5.24

C1 =

100 1+

N A ρ 2 ρ2 − N1 A1 ρ1

100

= 1+

atoms/mol)(2.33 g/cm3 ) 2.33 g/cm3 − (8.3 × 1015 atoms/cm3 )(121.76 g/mol) 6.68 g/cm3

(6.022 × 10

23

−5

= 7.20 × 10 wt%

(b) To convert from weight percent to atom percent we use Equation 5.9a where = atom percent of Sb C1 = weight percent of Sb = 7.20 × 10− wt% 5 C2 = weight percent of Si = 100 wt% − 7.20 × 10− wt% = 99.999928 A1 = atomic weight of Sb = 121.76 g/mol A2 = atomic weight of Si = 28.09 g/mol And from Equation 5.9a we obtain the following: 5

C1′ =

=

C1 A2 × 100 C1 A2 + C2 A1

(7.20 × 10 −5 wt%)(28.09 g/mol) (7.20 × 10 −5 wt%)(28.09 g/mol) + (99.999928 wt%)(121.76 g/mol)

× 100

= 1.66 × 10− at% 5

Reserve Problem 17 Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to approximately 20 wt% V at room temperature. Determine the concentration in weight percent of V that must be added to iron to yield a unit cell edge length of 0.289 nm.

Solution

To begin, it is necessary to employ Equation 3.8, and solve for the unit cell volume, VC, as VC =

nAave ρave N A

where Aave and ρave are the atomic weight and density, respectively, of the Fe-V alloy. Inasmuch as both of these materials have the BCC crystal structure, which has cubic symmetry, VC is just the cube of the unit cell length, a. That is

VC = a3 = (0.289 nm)3

= (2.89 × 10−8 cm)3 = 2.414 × 10−23 cm3 It is now necessary to construct expressions for Aave and ρave in terms of the concentration of vanadium, CV, using Equations 5.13a and 5.14a. For Aave we have Aave =

=

100 C V (100 − C V ) + AV AFe

100 CV (100 − CV ) + 50.94 g/mol 55.85 g/mol

whereas for ρave

ρave =

CV

+

ρV

=

100 (100 − CV )

ρFe

100 CV 3

6.10 g/cm

+

(100 − CV ) 7.87 g/cm3

Within the BCC unit cell there are 2 equivalent atoms, and thus, the value of n in Equation 3.8 is 2; hence, this expression may be written in terms of the concentration of V in weight percent as follows: VC = 2.414 × 10

=

=

–23

cm

3

nAave ρave N A

⎡ ⎤ 100 ⎢ ⎥ CV (100 − CV ) ⎥ (2 atoms/unit cell) ⎢ + ⎢ ⎥ ⎢ 50.94 g/mol 55.85 g/mol ⎥ ⎣ ⎦ ⎡ ⎤ 100 ⎢ ⎥ ⎢ CV (100 − CV ) ⎥ (6.022 × 1023 atoms/mol) ⎢ ⎥ + ⎢ 6.10 g/cm3 7.87 g/cm3 ⎥ ⎢⎣ ⎥⎦

And solving this expression for CV leads to CV = 12.9 wt%.

Reserve Problem 18: Linear defects Which of the following is a (are) linear defect(s)?

a) An edge dislocation b) A Frenkel defect c) A Schottky defect Solution Edge and screw dislocations are linear defects.

Reserve Problem 19: ASTM grain size I A photomicrograph was taken of a specimen at a magnification of 100×, and it was determined that the average number of grains per square inch was 200. What is this specimen’s ASTM grain size number?

Solution For a micrograph taken at a magnification of 100×, the number of grains per square inch, n, is related to the ASTM grain size number, G, according to Equation 4.17, that is

n = 2G −1 Taking logarithms of both sides of this equation and rearranging the resulting expression and solving for G gives the following:

G=

log n +1 log 2

Thus, for the situation when n = 15

G=

log15 + 1 = 4.9 log2

Reserve Problem 20 For a single crystal of some hypothetical metal that has the simple cubic crystal structure (Figure 3.3) would you expect the surface energy for a (100) plane to be greater, equal to, or less than a (110) plane?

Solution: The surface energy for a crystallographic plane will depend on its packing density [i.e., the planar density]—that is, the higher the packing density, the greater the number of nearest-neighbor atoms, and the more atomic bonds in that plane that are satisfied, and, consequently, the lower the surface energy.

It becomes necessary to determine the packing densities of the (100) and (110) planes for the simple cubic crystal structure. Below is shown a (100) plane for simple cubic.

For this plane there is one atom at each of the four corners, which is shared with four adjacent unit cells. Thus there is the equivalence of one atom associated with this (100) plane. The planar section is a square, where in the side lengths are equal to the unit cell edge length, 2R. Thus the area of this square is 4R 2 , which yields a planar density of

PD100 =

number of atoms centered on the (100) plane area of (100) plane

=

1 atom 4R

2

=

0.25 R2

We now repeat this procedure for a (110) plane, which is shown below.

Because each of these atoms is shared with four other unit cells, there is the equivalence of 1 atom associated with the plane. This planar section is a rectangle with dimensions of 2R × 2R 2 (Note:

2 2 = 2.83 ). Thus the area of this rectangle is 4 R2 2 = 5.66R2 . The planar density for the (110) plane is, therefore, PD110 =

1 atom 5.66R

2

=

0.18 R2

Thus, because the planar density for (100) is greater, it will have the lower surface energy.

Reserve Questions and Problems Solutions Manual CHAPTER 6 DIFFUSION

Reserve Question 01: Rate of Diffusion Diffusion by which mechanism occurs more rapidly in metal alloys? a) Vacancy diffusion b) Interstitial diffusion Solution In metal alloys, interstitial diffusion takes place more rapidly than vacancy diffusion because the interstitial atoms are smaller and are more mobile. Also, there are more vacant adjacent interstitial sites than there are vacancies.

Reserve Question 02: Temperature effect in diffusion As temperature decreases, the fraction of total number of atoms that are capable of diffusive motion a) increases. b) decreases.

Solution As temperature decreases, the fraction of the total number of atoms that are capable of diffusive motion decreases.

Reserve Problem 03 A gas mixture is found to contain two diatomic A and B species for which the partial pressures of both are 0.05065 MPa (0.5 atm). This mixture is to be enriched in the partial pressure of the A species by passing both gases through a thin sheet of some metal at an elevated temperature. The resulting enriched mixture is to have a partial pressure of 0.02026 MPa (0.2 atm) for gas A, and 0.01013 MPa 3 (0.1 atm) for gas B. The concentrations of A and B (CA and CB, in mol/m ) are functions of gas partial pressures ( pA and p B , in MPa) and absolute temperature according to the following expressions: 2

2

⎛ 25.0 kJ/mol ⎞ CA = 200 pA exp ⎜ − ⎟ 2 RT ⎝ ⎠ ⎛ 30.0 kJ/mol ⎞ CB = 1.0 × 10−3 pB exp ⎜ − ⎟ 2 RT ⎝ ⎠

Furthermore, the diffusion coefficients for the diffusion of these gases in the metal are functions of the absolute temperature as follows:

⎛ 15.0 kJ/mol ⎞ DA (m2 /s) = 4.0 × 10−7 exp ⎜ − ⎟ RT ⎝ ⎠ ⎛ 24.0 kJ/mol ⎞ DB (m2 /s) = 2.5 × 10−6 exp ⎜ − ⎟ RT ⎝ ⎠ Is it possible to purify the A gas in this manner? If so, specify a temperature at which the process may be carried out, and also the thickness of metal sheet that would be required. If this procedure is not possible, then state the reason(s) why.

Solution This problem calls for us to ascertain whether or not an A2-B2 gas mixture may be enriched with respect to the A partial pressure by allowing the gases to diffuse through a metal sheet at an elevated temperature. If this is possible, the temperature and sheet thickness are to be specified; if such is not possible, then we are to state the reasons why. Since this situation involves steady-state diffusion, we employ Fick’s first law, Equation 6.2. Inasmuch as the partial pressures on the high-pressure side of the sheet are the same, and the pressure of A2 on the low-pressure side is 2.0 times that of B2, and concentrations are proportional to the square root of the partial pressure, the diffusion flux of A, JA, is the square root of 2.0 times the diffusion flux of nitrogen JB—i.e.

J A = 2.0 J B Thus, equating the Fick’s law expressions incorporating the given equations for the diffusion coefficients and concentrations in terms of partial pressures leads to the following JA =

(200 )

1 × Δx ⎛ 25.0 kJ ⎞

⎛ 15.0 kJ ⎞

( 0.05065 MPa − 0.02026 MPa ) exp ⎜⎜⎝ − RT ⎟⎟⎠ (4.0 × 10 m /s) exp ⎜⎜⎝ − RT ⎟⎟⎠ −7

2

2.0 J B =

(1.0 × 103 )

2.0 × Δx

( 0.05065 MPa − 0.01013 MPa ) exp⎛⎜⎝ − 30.0RTkJ ⎞⎟⎠ (2.5 × 10 m /s) exp⎛⎜⎝ − 24.0RTkJ ⎞⎟⎠ −6

2

The Δx's cancel out, which means that the process is independent of sheet thickness. Now solving the above expression for the absolute temperature T gives T = 401 K (128°C) Thus, it is possible to carry out this procedure at 401 K or 128°C. Reserve Problem 04 If [m] atoms of helium pass through a [a] square meter plate area every [t] hours, and if this flux is constant with time, compute the flux of helium in units of atoms per square meter per second.

Solution: The definition of flux is the quantity of something (which could be in terms of particle count, mass, etc.) that passes through a specified unit of area in a specified unit of time.

In this case, we are provided an atom count that passes through the known area each hour.

Therefore, accounting for the seconds in each hour: J=

M [m] = 2 At ([a] m )([t ] hr(3600 s/ hr))

Reserve Problem 05 2

If water molecules pass through a membrane with a steady state flux of [j] mole/(m day), how long will it take, in hours, for [m] kg of water to pass through a [a] square centimeter of the membrane?

Solution: The definition of flux is the quantity of something (which could be in terms of particle count, mass, etc.) that passes through a specified unit of area in a specified unit of time.

In this case, we are provided the flux of water molecules and are asked to compute the time required for a certain mass of water molecules to transit a specific area of membrane.

Therefore, accounting for the number of hours in each day:

⎛ 1000 g ⎞ ⎛ 1 mole ⎞ [m] kg ⎜ ⎟⎜ ⎟ ⎛ 24 hr ⎞ M ⎝ 1 kg ⎠ ⎝ 18 g ⎠ t= = ⎜ ⎟ AJ ⎛ 1 m2 ⎞⎛ mol ⎞ ⎝ 1 day ⎠ 2 ([a] cm ) ⎜ [ j] ⎜ 1002 cm2 ⎟⎜ ⎟ ⎜ m2 day ⎟⎟ ⎠ ⎝ ⎠⎝

Reserve Problem 06a (question pool) The cornea is the transparent outer layer of the human eye. Because it must be transparent to light, it does not normally contain blood vessels. Therefore, it must receive its nutrients via diffusion.

Oxygen from the surrounding air diffuses to the cornea through the surface tears whereas other nutrients diffuse to the cornea from the inner parts of the eye, such as the vitreous humor and lens.

During operation, the cornea produces waste in the form of CO2 gas that must be expelled to keep the eye healthy and functioning. This is accomplished by the simultaneous diffusion of CO2 from the cornea to the surrounding atmosphere, which generally features a low CO2 concentration.

It is therefore critical that modern contact lens materials allow sufficient diffusion rates of oxygen and carbon dioxide. Without oxygen, the cornea will warp, loose transparency, and become susceptible to scarring. The body may also react by growing additional blood vessels into the eye, which can damage the cornea.

If an increased steady-state flow rate of O2 (oxygen molecules per second) to the cornea is desired, which of the following contact lens / ambient condition modifications is not likely to be useful?

Note: the flow rate is equal to product of the diffusion flux and an area of interest through which diffusion occurs. (a) Increase the contact lens thickness (b) Increase the diffusivity of oxygen gas by decreasing the contact lens porosity (c) Increase the ambient temperature (d) Increase the ambient partial pressure of oxygen gas (e) All of the suggestions (a-d) are useful for increasing the flow rate of oxygen

Solution: Note: Since the area of the contact lens (A) doesn’t change, the flow rate of oxygen molecules (J*A) increases whenever the diffusion flux of oxygen (J) increases. (a) If the contact lens thickness is increased and all other factors are preserved, the concentration gradient decrease and the oxygen diffusion flux should decrease. (b) If the lens features more voids, we expect the oxygen to permeate the membrane faster since oxygen should be able to transit voids faster than the bulk lens material.

(c) Increasing the temperature should increase the diffusivity and therefore increase the diffusion flux and flow rate. (d) Increasing the concentration of oxygen in the ambient environment will increase the concentration gradient, thereby increasing the diffusion flux of oxygen.

Reserve Problem 07: Non steady-state – Specific concentration at different T For a steel alloy it has been determined that a carburizing heat treatment of 16 h duration at 757°C will raise the carbon concentration to 0.5 wt% at a point 2.3 mm from the surface. Estimate the time necessary to achieve the same concentration at a 8 mm position for an identical steel and at a -5 2 carburizing temperature of 1130°C. Assume that D0 is 4.6 × 10 m /s and Qd is 104 kJ/mol.

Solution For this problem the solution to Fick's second law is according to Equation 6.6b, that is x2 = constant Dt

For two temperatures, T1 and T2, we may write the following expression:

x12 x2 = 2 D1t1 D2t2 Furthermore, the dependence of D depends on T is represented by Equation 6.8, which means that

⎛ Q ⎞ D1 = D0 exp ⎜⎜ − d ⎟⎟ ⎝ RT1 ⎠ ⎛ Q ⎞ D2 = D0 exp ⎜⎜ − d ⎟⎟ ⎝ RT2 ⎠ Combining these two expressions into the previous equation, leads to the following:

x12 ⎛ Q ⎞ t1 exp ⎜ − d ⎟ ⎝ RT1 ⎠

=

x22 ⎛ Q ⎞ t2 exp ⎜ − d ⎟ ⎝ RT2 ⎠

Assuming that t2 is unknown time to be determined, the above equation takes the form ⎛ Q ⎞ t1 x 22 D0 exp ⎜ − d ⎟ ⎝ RT1 ⎠ t2 = ⎛ Q ⎞ x12 D0 exp ⎜ − d ⎟ ⎝ RT2 ⎠

which reduces to the following

⎛ t x2 ⎞ ⎡ Q ⎛ 1 1 ⎞⎤ ⎜ t2 = 1 2 ⎟ exp ⎢− d ⎜⎜ − ⎟⎟⎥ ⎜ x2 ⎟ ⎣⎢ R ⎝ T1 T2 ⎠⎦⎥ ⎝ 1 ⎠ From our problem statement: t1 = 17 h T1 = 898°C (= 1171 K) -3 x1 = 3.1 mm (3.1 × 10 m) T2 = 1010°C (= 1283 K) -3 x2 = 5.4 mm (5.4 × 10 m) Qd = 118,000 J/mol -5 2 D0 = 2.5 ×10 m /s When these values are included into the above equation, the value of t2 is computed as follows: ⎡ (17h)(5.4 × 10 −3 m) 2 ⎤ ⎡ (118,000J/mol) ⎛ 1 1 ⎞⎤ − t2 = ⎢ ⎥ exp ⎢ − ⎜ ⎟⎥ −3 2 ⎢⎣ (3.1 × 10 m) ⎥⎦ ⎣ 8.31J/mol-K ⎝ 1171K 1283K ⎠ ⎦

= 17.9 h

Reserve Problem 08a (question pool) The figure below features diffusion profiles that developed within four separate plain-carbon steel specimens of equivalent geometry after they separately experienced a carburization process. The specimens were prepared such that the scenario of a one-dimensional semi-infinite solid applies as a solution to Fick's 2nd Law of Diffusion. Each part originally contained a uniform distribution of carbon, and each part was processed using the same carburization temperature. In the figure, x = 0 corresponds to the surfaces of the steel parts that were exposed to the carbon-rich atmosphere during the diffusion process.

Answer True or False for each of the following statements. (a) The carburization surface was maintained at 1.00 wt% carbon for each specimen. (b) Comparing the finished specimens at a depth of 0.75 mm, specimen A features the largest carbon concentration. (c) Comparing the finished specimens as a whole, specimen A features the lowest overall amount of carbon. (d) Specimen B experienced a longer carburization time compared to specimen C. (e) The initial concentration of carbon in each part (prior to carburization) was a little less than 0.25 wt% carbon.

Solution: (a) The carbon content at x = 0 equals 1.00 wt% carbon.

(b) The concentration profile of specimen D has the greatest value of at the specified position. (c) The concentration profile of specimen D has the highest carbon content at every position.

Therefore, it has the highest overall carbon content. (d) Since the temperature of each carburization was the same, a longer diffusion time will result in a profile with higher carbon content at each position. Therefore, specimen C experienced a longer carburization time than specimen B. (e) The concentration profiles approach a value slightly less than 0.25 wt% carbon as the x position increases.

Reserve Problem 09 The diffusion coefficient for aluminum in silicon is DAl in Si = 3x10 is about room temperature).

-16

2

cm /s at 300 K (note that 300 K

What is a reasonable value for DAl in Si at 600 K ?

Note: Rather than performing a specific calculation, you should be able to justify your answer from the options below based on the mathematical temperature dependence of the diffusion coefficient. (a) D < 3x10

-16

(b) D = 3x10 (c) D = 6x10

-16

-16

(d) D = 1.5x10 (e) D > 6x10

2

cm /s 2

cm /s

-16

-16

(f) D = 6x10

2

cm /s

-17

2

cm /s 2

cm /s 2

cm /s

Solution: We expect the diffusion coefficient to increase if the temperature of this system is increased.

Therefore, options (a), (b), (d), and (f) are eliminated.

Furthermore, we expect that since the diffusion coefficient is exponentially dependent on temperature, the diffusivity should increase by more than a factor of two if the absolute temperature is doubled.

Reserve Problem 10a (question pool) Depicted below are five different steady-state concentration profiles for the same gas across five separate and identical plastic membranes at the same temperature. Which concentration profile results in the lowest diffusion flux through the membrane?

Solution: Since we are comparing diffusion profiles for the same diffuser and host in each case, and the temperature is the same for each of the cases, only the concentration gradient needs to be compared.

Since the diffusion flux is directly proportional to the concentration gradient, the case featuring the lowest magnitude slope should exhibit the lowest magnitude diffusion flux.

Specifically, since case (a) has no apparent gradient (a uniform concentration is implied), there is no driving force for chemical diffusion.

Reserve Problem 11

One integrated circuit design calls for the diffusion of arsenic into silicon wafers; the background 20 3 concentration of As in Si is 2.5 × 10 atoms/m . The predeposition heat treatment is to be conducted 26 3 at 1000°C for 45 minutes, with a constant surface concentration of 8 × 10 As atoms/m . At a drivein treatment temperature of 1100°C, determine the diffusion time required for a junction depth of -3 2 1.2 μm. For this system, values of Qd and D0 are 4.10 eV and 2.29 × 10 m /s, respectively.

Solution

This problem asks that we compute the drive-in diffusion time for arsenic diffusion in silicon. It is first necessary to determine the value of Q0 using Equation 6.12. But before this is possible, the value of Dp at 1000°C must be computed with the aid of Equation 6.8. Thus,

⎛ Q ⎞ D p = D0 exp ⎜ − d ⎟ ⎜ kT p ⎟ ⎝ ⎠ ⎡ ⎤ 4.10 eV = (2.29 × 10 −3 m2 /s) exp ⎢ − ⎥ −5 ⎣⎢ (8.62 × 10 eV/atom-K)(1000°C + 273 K) ⎦⎥ = 1.36 × 10 −19 m2 /s Now for the computation of Q0 using Equation 6.12:

Q0 = 2Cs = (2)(8 × 1026 atoms/m3 )

Dpt p

π

(1.36 × 10−19 m2 /s)(45 min)(60 s/min)

π

= 1.73 × 1019 atoms/m2 We now desire to calculate td in Equation 6.13. Algebraic manipulation and rearrangement of this expression leads to

⎛ x2 ⎞ Q0 j ⎟= exp⎜ ⎜ 4Dd td ⎟ CB π Dd td ⎝ ⎠

At this point it is necessary to determine the value of Dd (at 1100°C). Thus, using Equation 6.8 ⎡ ⎤ 4.10 eV Dd = (2.29 × 10 −3 m2 /s) exp ⎢ − ⎥ −5 ⎣⎢ (8.62 × 10 eV/atom-K)(1100°C + 273 K) ⎦⎥ = 2.06 × 10 −18 m2 /s

And incorporation of values of all parameters except td in the above expression yields

⎡ ⎤ (1.2 × 10 −6 m)2 1.73 × 1019 atoms/m2 exp ⎢ = ⎥ ⎢⎣ (4)(2.06 × 10 −18 m2 /s)td ⎥⎦ (2.5 × 1020 atoms/m3 ) (π )(2.06 × 10 −18 m2 /s)t d which expression reduces to ⎛ 1.75 × 10 5 s ⎞ 2.72 × 10 7 s1/2 exp ⎜ ⎟= ⎜ ⎟ td td ⎝ ⎠

Solving for td is not a simple matter. One possibility is to use a graphing technique. Let us take the logarithm of both sides of the above equation, which gives

⎛ 2.72 × 107 s1/2 ⎞ 1.75 × 105 s ⎟ = ln ⎜ ⎜ ⎟ td t d ⎝ ⎠ Now if we plot the terms on both left and right hand sides of this equation versus td, the value of td at the point of intersection of the two resulting curves is correct answer. Below is such a plot:

As noted, the two curves intersect at about 13,900 s, which corresponds to td = 3.86 h.

Reserve Problem 12

Phosphorus atoms are to be diffused into a silicon wafer using both predeposition and drive-in heat 19 treatments; the background concentration of P in this silicon material is known to be 5 × 10 3 atoms/m . The predeposition treatment is to be conducted at 950°C for 45 minutes; the surface 26 3 concentration of P is to be maintained at a constant level of 1.5 × 10 atoms/m . Drive-in diffusion will be carried out at 1200°C for a period of 2.5 h. For the diffusion of P in Si, values of Qd and D0 are -4 2 3.40 eV and 1.1 × 10 m /s, respectively. (a) Calculate the value of Q0. (b) Determine the value of xj for the drive-in diffusion treatment. (c) Also for the drive-in treatment, compute the position x at which the concentration of P atoms is 24 -3 10 m .

Solution

(a) For this portion of the problem we are asked to determine the value of Q0. This is possible using Equation 6.12. However, it is first necessary to determine the value of D for the predeposition treatment [Dp at Tp = 950°C (1223 K)] using Equation 6.8. Thus

⎛ Q ⎞ D p = D0 exp ⎜ − d ⎟ ⎜ kT p ⎟ ⎝ ⎠ ⎡ ⎤ 3.40 eV = (1.1 × 10 −4 m 2 /s) exp ⎢ − ⎥ ⎢⎣ (8.62 × 10 −5 eV/atom-K)(1223 K) ⎥⎦ = 1.08 × 10 −18 m 2 /s The value of Q0 may be determined as follows:

Q0 = 2Cs = (2)(1.5 × 1026 atoms/m3 )

Dpt p

π

(1.08 × 10 −18 m2 /s)(45 min)(60 s/min)

π 18

= 9.14 × 10 atoms/m

2

(b) Computation of the junction depth requires that we use Equation 6.13. However, before this is possible it is necessary to calculate D at the temperature of the drive-in treatment [Dd at 1200°C (1473 K)]. Thus, ⎡ ⎤ 3.40 eV Dd = (1.1 × 10 −4 m 2 /s) exp ⎢ − ⎥ −5 ⎣⎢ (8.62 × 10 eV/atom-K)(1473 K) ⎦⎥ = 2.58 × 10 − 16 m 2 /s

Now from Equation 6.13 1/2

⎡ ⎛ ⎞⎤ Q0 ⎟⎥ x j = ⎢(4Dd td )ln ⎜ ⎜ C π D t ⎟⎥ ⎢ d d ⎠⎦ ⎝ B ⎣

1/2

⎧ ⎡ ⎤⎫ 9.14 ×1018 atoms/m2 ⎪ −16 2 ⎢ ⎥ ⎪⎬ = ⎨(4)(2.58 ×10 m /s)(9000 s) ln ⎢ (5 ×1019 atoms/m3 ) (π )(2.58 × 10−16 m2 /s)(9000 s) ⎥ ⎪ ⎪⎩ ⎣ ⎦⎭ = 1.21×10−5 m = 12.1 μm 24

3

(c) For a concentration of 10 P atoms/m for the drive-in treatment, we compute the value of x using Equation 6.11. However, it is first necessary to manipulate Equation 5.11 so that x is the dependent variable. Taking natural logarithms of both sides leads to

⎛ Q ⎞ x2 0 ⎟− ln C ( x,t ) = ln ⎜ ⎜ π D t ⎟ 4 Dd td d d ⎠ ⎝ Now, rearranging and solving for x leads to 1/2

⎧ ⎡ ⎤ ⎫⎪ Q0 ⎪ ⎥⎬ x = ⎨(4 Dd td )ln ⎢ ⎢⎣ C (x,t ) π Dd td ⎥⎦ ⎪ ⎪⎩ ⎭

24

3

Now, incorporating values for Q0 and Dd determined above and taking C(x,t) = 10 P atoms/m yields 1/2

⎧ ⎡ ⎤⎫ 9.14 × 1018 ⎪ ⎥ ⎪⎬ x = ⎨(4)(2.58 × 10 −16 )(9000) ln ⎢ 24 −16 ⎢ ⎥ ⎪⎩ ⎣ (10 ) (π )(2.58 × 10 )(9000) ⎦ ⎪⎭ = 3.36 × 10 −6 m = 3.36 μm Reserve Problem 13

Aluminum atoms are to be diffused into a silicon wafer using both predeposition and drive-in heat 19 treatments; the background concentration of Al in this silicon material is known to be 3 × 10 3 atoms/m . The drive-in diffusion treatment is to be carried out at 1050°C for a period of 4.0 h, which gives a junction depth xj of 3.0 μm. Compute the predeposition diffusion time at 950°C if the surface 25 3 concentration is maintained at a constant level of 2 × 10 atoms/m . For the diffusion of Al in Si, -4 2 values of Qd and D0 are 3.41 eV and 1.38 × 10 m /s, respectively.

Solution

This problem asks that we compute the time for the predeposition heat treatment for the diffusion of Al in Si. In order to do this, it is necessary to determine the value of Q0 from Equation 6.13.

However, before doing this we must first calculate Dd, using Equation 6.8. Therefore

⎛ Q ⎞ Dd = D0 exp ⎜⎜ − d ⎟⎟ ⎝ kTd ⎠ ⎡ ⎤ 3.41 eV = (1.38 × 10−4 m2 /s)exp ⎢− ⎥ −5 ⎣⎢ (8.62 × 10 eV/atom-K)(1050°C + 273 K) ⎦⎥ = 1.43 × 10−17 m2 /s Now, solving for Q0 in Equation 6.13 leads to

⎛ x2 ⎞ j ⎟ Q0 = CB πDd td exp⎜ ⎜ 4Dd td ⎟ ⎝ ⎠

(

)

In the problem statement we are given the following values: 19

3

CB = 3 × 10 atoms/m td = 4 h (14,400 s) -6 xj = 3.0 μm = 3.0 × 10 m Therefore, incorporating these values into the above equation yields ⎡ ⎤ (3.0 × 10 −6 m)2 Q0 = ⎡ (3 × 1019 atoms/m3 ) (π )(1.43 × 10 −17 m2 /s)(14,400 s) ⎤ exp ⎢ ⎥ −17 2 ⎢⎣ ⎥⎦ ⎢⎣ (4)(1.43 × 10 m / s )(14, 400 s) ⎥⎦ = 1.34 × 1018 atoms/m 2

We may now compute the value of tp using Equation 6.12. However, before this is possible it is necessary to determine Dp (at 950°C) using Equation 6.8. Thus ⎡ ⎤ 3.41 eV D p = (1.38 × 10 −4 m 2 /s) exp ⎢ − ⎥ −5 ⎣⎢ (8.62 × 10 eV/atom-K)(950°C + 273 K) ⎦⎥ = 1.24 × 10 −18 m 2 /s

Now, solving for tp in Equation 6.12 we get

tp =

πQ02 4Cs2 Dp

25

3

And incorporating the value of Cs provided in the problem statement (2 × 10 atoms/m ) as well as values for Q0 and Dp determined above, leads to

tp =

(

π 1.34 × 1018 atoms/m2

(

)

2

)

2

(4) 2 × 1025 atoms/m3 (1.24 × 10−18 m2 /s) = 2.84 × 103 s = 47.4 min

Reserve Questions and Problems Solutions Manual CHAPTER 7 MECHANICAL PROPERTIES

Reserve Problem 01a-1 (question pool) Each of the rods depicted below were machined from same stock metal. If the same force is applied axially to each rod, which one will experience the highest stress?

Solution: (a) Since the force applied to each bar is the same, the bar with the smallest cross-sectional area will endure the highest stress, since stress equals force divided by the area over which it acts.

Reserve Problem 01b-1 (question pool) Each of the rods depicted below were machined from same stock metal. If the same force is applied axially to each rod, which one will experience the highest stress?

Solution: (a) Since the force applied to each bar is the same, the bar with the smallest cross-sectional area will endure the highest stress, since stress equals force divided by the area over which it acts.

Reserve Problem 01c-1 (question pool) Each of the rods depicted below were machined from same stock metal. If the same force is applied axially to each rod, which one will experience the highest stress?

Solution: (c) Since the force applied to each bar is the same, the bar with the smallest cross-sectional area will endure the highest stress, since stress equals force divided by the area over which it acts.

Reserve Problem 01d-1 (question pool) Each of the rods depicted below were machined from same stock metal. If the same force is applied axially to each rod, which one will experience the highest stress?

Solution: (d) Since the force applied to each bar is the same, the bar with the smallest cross-sectional area will endure the highest stress, since stress equals force divided by the area over which it acts.

Reserve Problem 01e-1 (question pool) Each of the rods depicted below were machined from same stock metal. If the same force is applied axially to each rod, which one will experience the highest stress?

Solution: (e) Since the force applied to each bar is the same, the bar with the smallest cross-sectional area will endure the highest stress, since stress equals force divided by the area over which it acts.

Reserve Problem 01f-1 (question pool) Each of the rods depicted below were machined from same stock metal. If the same force is applied axially to each rod, which one will experience the highest stress?

Solution: (f) Since the force applied to each bar is the same, the bar with the smallest cross-sectional area will endure the highest stress, since stress equals force divided by the area over which it acts.

Reserve Problem 02a (question pool) Each of the rods depicted below were machined from same stock metal. They were originally machined to be the same length, but their cross-sectional areas were different. Which plot correctly depicts the force required to elastically elongate these specimens?

(a)

(b)

(c)

(d)

(e)

(f)

Solution: (a) Since the rods were made from the same metal, their elastic moduli should be the same.

Comparing the case for which the rods are elongated by the same amount implies that the each exhibit the same elastic strain. Based on Hooke’s Law, if the axial strains and moduli are the same in each case, then each rod experienced the same stress, or ratio of force per area. The rod with the largest cross-sectional area therefore experienced the largest force.

Reserve Problem 03: Stress-strain I A specimen of some metal having a rectangular cross section 11.2 mm x 12.4 mm is pulled in tension with a force of 31200 N, which produces only elastic deformation. Given that the elastic modulus of this metal is 63 GPa, calculate the resulting strain.

Solution This problem calls for us to calculate the elastic strain that results for a specimen of some metal alloy stressed in tension. Combining Equations 7.1 and 7.5 and solving for strain yields

ε=

=

σ E

=

F A0 E

F A0 E

=

32,600N −3

−3

9

2

(10.9×10 mm)(13.3×10 mm)(77 ×10 N/m )

= 2.92 ×10−3

Reserve Problem 04: Maintaining plastic deformation For a bronze alloy, the stress at which plastic deformation begins is 277 MPa and the modulus of elasticity is 117 GPa. (a) What is the maximum load that may be applied to a specimen having a cross-sectional area 2 of 327 mm without plastic deformation? (b) If the original specimen length is 148 mm, what is the maximum length to which it may be stretched without causing plastic deformation?

Solution (a) This part of the problem asks for us to compute the maximum load that may be applied without yielding. From the definition of stress, Equation 7.1, the yield strength and load at yielding Fy are related as follows:

Fy = σ y A0 6

2

2

–6

2

Since, for this problem, σy = 271 MPa = 271 × 10 N/m and A0 = 320 mm = 320 ×10 m then

Fy = (271×106 N/m2 )(320 ×10−6 m2 ) = 86,700N (b) Now we are asked to compute the maximum length to which the specimen may be stretched without plastic deformation. At the point of yielding, Equation 7.5 takes on the following rearranged form:

εy =

σy E

Here ε y is the strain at yielding. The definition of strain from Equation 7.2 is

l − l0 l0

ε= i

Or, making this substitution for ε into the previous equation leads to

li − l0 l0

=

σy E

And solving for li, in this case the length to which the specimen may be elongated without plastic deformation gives ⎛σy ⎞ + 1⎟ li = l0 ⎜ ⎜ E ⎟ ⎝ ⎠

⎛ 271 × 106 N/m 2 ⎞ = (131 × 10 −3m) ⎜ + 1⎟ ⎜ 119 × 109 N/m 2 ⎟ ⎝ ⎠

0.13130 m = 131.30 mm

Reserve Problem 05 A steel bar 100 mm (4.0 in.) long and having a square cross section 20 mm (0.8 in.) on an edge is pulled in tension with a load of 89,000 N (20,000 lbf), and experiences an elongation of 0.10 mm (4.0 –3 × 10 in.). Assuming that the deformation is entirely elastic, calculate the elastic modulus of the steel.

Solution This problem asks us to compute the elastic modulus of steel. For a square cross-section, A0 = b02 , where b0 is the edge length. Combining Equations 7.1, 7.2, and 7.5 and solving for E, leads to

F A Fl0 σ = 0= E = Δl b 2 Δ l ε 0 l0 =

(89, 000 N) (100 × 10 − 3 m )

( 20 × 10 − 3 m ) 2 ( 0.10 × 10 −3 m ) 9

2

6

= 223 × 10 N/m = 223 GPa (31.3 × 10 psi) Reserve Problem 06 Figure 7.35 shows, for a gray cast iron, the tensile engineering stress–strain curve in the elastic region. Determine (a) the tangent modulus at 10.3 MPa (1500 psi), and (b) the secant modulus taken to 6.9 MPa (1000 psi).

Solution (a) This portion of the problem asks that the tangent modulus be determined for the gray cast iron, the stress-strain behavior of which is shown in Figure 7.34. In the figure below is shown a tangent draw on the curve at a stress of 10.3 MPa (1500 psi).

The slope of this line (i.e., Δσ/Δε), the tangent modulus, is computed as follows: Δσ 15 MPa − 5 MPa = = 1410 MPa = 1.41 GPa (2.04 × 105 psi) 0.0074 − 0.0003 Δε

(b) The secant modulus taken from the origin is calculated by taking the slope of a secant drawn from the origin through the stress-strain curve at 6.9 MPa (1000 psi). This secant is drawn on the curve shown below:

The slope of this line (i.e., Δσ/Δε), the secant modulus, is computed as follows:

Δσ 15 MPa − 0 MPa = = 3190 MPa = 3.19 GPa (4.63 × 105 psi) 0.0047 − 0 Δε

Reserve Question 07 Using the expression to which the modulus of elasticity is proportional ⎛ dF ⎞ ⎜ ⎟ =− ⎝ dr ⎠ r0

2A

+

( nBA )

3/(1 − n )

n ( n + 1) B

( nBA )

( n + 2)/(1 − n )

Rank the magnitudes of the moduli of elasticity for the following hypothetical X, Y, and Z materials from the greatest to the least. The appropriate A, B and n parameters (Equation 7.36) for these three materials are tabulated below; they yield EN in units of electron volts and r in nanometers: Material

A

X

2.5

Y

B 2.0 × 10

2.3

Z

n

8.0 × 10

3.0

1.5 × 10

–5

8

–6

10.5

–5

9

The highest modulus of elasticity is Metal The next highest modulus of elasticity is Metal The least modulus of elasticity is Metal

____ ____ ____

Solution This problem asks that we rank the magnitudes of the moduli of elasticity of the three hypothetical metals X, Y, and Z. From Reserve Problem 06, it was shown for materials in which the bonding energy is dependent on the interatomic distance r according to Equation 7.36 that the modulus of elasticity E is proportional to E ∝−

2A ⎛ A ⎞ ⎜ nB ⎟ ⎝ ⎠

3/(1 − n )

( n )( n + 1) B

+

⎛ A ⎞ ⎜ nB ⎟ ⎝ ⎠

( n + 2)/(1 − n )

–5

For metal X, A = 2.5, B = 2.0 × 10 , and n = 8. Therefore, E ∝ −

(2)(2.5) ⎡ ⎢ ⎢ ⎢ (8) ⎣

⎤ ⎥ ⎥ −5 ⎥ 2 × 10 ⎦

3/(1 − 8)

(8)(8 + 1 ) (2 × 10 −5 )

+

⎡ ⎤ 2.5 ⎢ ⎥ ⎢⎣ (8) (2 × 10 −5 ) ⎥⎦

2.5

(

)

(8 + 2)/(1 − 8)

= 1097 –6

For metal Y, A = 2.3, B = 8 × 10 , and n = 10.5. Hence E ∝ −

(2)(2.3) ⎡ 2.3 ⎢ ⎢ ⎢ (10.5) 8 × 10 − 6 ⎣

(

)

⎤ ⎥ ⎥ ⎥ ⎦

3/(1 − 10.5)

+

(10.5)(10.5 + 1) (8 × 10 −6 ) ⎡ ⎤ 2.3 ⎢ ⎥ −6 ⎢⎣ (10.5) (8 × 10 ) ⎥⎦

= 551

(10.5 + 2)/(1 − 10.5)

–5

And, for metal Z, A = 3.0, B = 1.5 × 10 , and n = 9. Thus E ∝−

(2)(3.0) ⎡ ⎢ ⎢ ⎢ (9) ⎣

⎤ ⎥ ⎥ −5 ⎥ 1.5 × 10 ⎦

3.0

(

)

3/(1 − 9)

+

(9)(9 + 1 ) (1.5 × 10 − 5 ) ⎡ ⎤ 3.0 ⎢ ⎥ ⎢⎣ (9) (1.5 × 10 − 5 ) ⎥⎦

(9 + 2)/(1 − 9)

= 1024 Therefore, metal X has the highest modulus of elasticity.

Reserve Problem 08a-1 (question pool) Each of the rods depicted below were machined from same stock metal. They were originally machined to be the same length, but their cross-sectional areas were different. If axial force is applied to each rod such that they all change length by the same amount, which rod experienced the largest force?

Solution: (a) The rods experience the same axial strain, since they featured the same initial length and were elongated by the same amount. Since the rods were made from the same metal, their elastic moduli should be the same.

Based on Hooke’s Law, if the axial strains and moduli are the same in each case, then each rod experienced the same stress, or ratio of force per area. The rod with the largest cross-sectional area therefore experienced the largest force.

Reserve Problem 08b-1 (question pool) Each of the rods depicted below were machined from same stock metal. They were originally machined to be the same length, but their cross-sectional areas were different. If axial force is applied to each rod such that they all change length by the same amount, which rod experienced the largest force?

Solution: (b) The rods experience the same axial strain, since they featured the same initial length and were elongated by the same amount. Since the rods were made from the same metal, their elastic moduli should be the same.

Based on Hooke’s Law, if the axial strains and moduli are the same in each case, then each rod experienced the same stress, or ratio of force per area. The rod with the largest cross-sectional area therefore experienced the largest force.

Reserve Problem 08c-1 (question pool) Each of the rods depicted below were machined from same stock metal. They were originally machined to be the same length, but their cross-sectional areas were different. If axial force is applied to each rod such that they all change length by the same amount, which rod experienced the largest force?

Solution: (c) The rods experience the same axial strain, since they featured the same initial length and were elongated by the same amount. Since the rods were made from the same metal, their elastic moduli should be the same.

Based on Hooke’s Law, if the axial strains and moduli are the same in each case, then each rod experienced the same stress, or ratio of force per area. The rod with the largest cross-sectional area therefore experienced the largest force.

Reserve Problem 08d-1 (question pool) Each of the rods depicted below were machined from same stock metal. They were originally machined to be the same length, but their cross-sectional areas were different. If axial force is applied to each rod such that they all change length by the same amount, which rod experienced the largest force?

Solution: (d) The rods experience the same axial strain, since they featured the same initial length and were elongated by the same amount. Since the rods were made from the same metal, their elastic moduli should be the same.

Based on Hooke’s Law, if the axial strains and moduli are the same in each case, then each rod experienced the same stress, or ratio of force per area. The rod with the largest cross-sectional area therefore experienced the largest force.

Reserve Problem 09: Poisson’s ratio II A cylindrical specimen of some metal alloy 10 mm in diameter and 150 mm long has a modulus of elasticity of 100 GPa. Does it seem reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen diameter of 0.08 mm? Assume that the deformation is totally elastic. a) Yes b) No Solution To solve this problem we compute the value of Poisson's ratio; if it is less than the maximum value for a metal (0.5) then this situation is possible. The axial strain ( ε z ) is determined from Equation 7.5— stress and elastic modulus values are provided in the problem statement. The lateral strain ( ε x ) is computed using Equation 7.2. Axial strain is first computed using a rearranged form of Equation 7.5:

εz =

σ E

=

200MPa 100 × 103 MPa

= 0.0020

We now determine the value of lateral strain using Equation 7.2 adapted to diameter reduction as follows:

εx =

Δd −0.08mm = = −0.0080 d0 10mm

Now, using Equation 7.8 the value of Poisson's ratio is

ε (−0.0080) = 4.0 ν =− x =− εz 0.0020 Inasmuch as this value of ν is greater than 0.5, such as situation is not possible.

Reserve Problem 12: Stress for tension test Consider the brass alloy the stress-strain behavior of which is shown in the Animated Figure 7.12. A cylindrical specimen of this alloy 21 mm in diameter and 224 mm long is to be pulled in tension.

Calculate the stress (in MPa) necessary to cause a 0.00832 mm reduction in diameter. Assume a value of 0.33 for Poisson's ratio.

Solution To solve this problem we first use Equation 7.5 to compute the axial strain ( ε z ) as follows:

εz =

σ E

We now express lateral strain (ε x ) using Equation 7.2 adapted to diameter reduction as follows:

εx =

Δd d0

Substitution of the above expressions for axial and lateral strain into Equation 7.8 leads to

Δd d0

ε EΔd ν =− x =− =− σ εz σ d0 E And solving for the applied stress (σ) yields

σ =−

EΔd νd0

From the Animated Figure 7.12 it is possible to determine the elastic modulus for this brass alloy.

From Equation 7.10

E = slope of line in elastic region =

σ 2 − σ1 ε 2 − ε1

Inasmuch as the linear elastic line segment passes through the origin it is convenient to take both

σ 1 and ε1 as zero. If we arbitrarily take σ2 = 150 MPa , then, from the inset plot of Animated Figure 7.12, the strain is 0.0018. Thus we compute the value of E as follows:

E=

150 MPa − 0 MPa 0.0018 − 0

= 83.3 × 103 MPa = 83.3 GPa

Incorporation of values for parameters on the right side of the above expression for the applied stress, we determine the value of stress as follows:

σ =−

EΔd (83.3 × 109 N/m2 )(−0.00714 × 10 −3 m) =− ν d0 (0.35)(17 × 10 −3 m)

= 100 ×106 N/m2 = 100 MPa Reserve Problem 13: Typical relationship between E and G For most metals, the relationship between elastic and shear moduli is approximately which of the following? a) b) c) d) e)

G = 0.1 E G = 0.2 E G = 0.3 E G = 0.4 E G = 0.5 E

Solution Elastic and shear moduli are related according to Equation 6.9 as follows: E = 2 G (1 + ν )

And solving for G gives G=

E 2(1 + ν )

Because the value of Poisson's ratio for most metals is approximately 0.30, then G=

Or, approximately 0.4E.

E E = = 0.38 E 2(1 + 0.30) 2.6

Reserve Problem 14 A cylindrical metal specimen 12.7 mm (0.5 in.) in diameter and 250 mm (10 in.) long is to be subjected to a tensile stress of 28 MPa (4000 psi); at this stress level, the resulting deformation will be totally elastic.

–3

(a) If the elongation must be less than 0.080 mm (3.2 × 10 in.), which of the metals in Table 6.1 are suitable candidates?

• Brass • Tungsten • Titanium • Magnesium • Copper • Steel • Nickel • Aluminum –3

–5

(b) If, in addition, the maximum permissible diameter decrease is 1.2 × 10 mm (4.7 × 10 in.) when the tensile stress of 28 MPa is applied, which of the metals that satisfy the criterion in part (a) are suitable candidates?

• Aluminum

• • • • • • •

Steel Copper Nickel Brass Magnesium Titanium Tungsten

Solution a) This part of the problem asks that we ascertain which of the metals in Table 7.1 experience an elongation of less than 0.080 mm when subjected to a tensile stress of 28 MPa. The maximum strain that may be sustained, (using Equation 7.2) is just

ε=

Δl 0.080 mm = = 3.2 × 10−4 l0 250 mm

Since the stress level is given (50 MPa), using Equation 7.5 it is possible to compute the minimum modulus of elasticity that is required to yield this minimum strain. Hence

E=

σ 28 MPa = = 87.5 GPa ε 3.2 × 10 −4

This means that those metals with moduli of elasticity greater than this value are acceptable candidates—namely, brass, Cu, Ni, steel, Ti and W.

(b) This portion of the problem further stipulates that the maximum permissible diameter decrease –3 is 1.2 × 10 mm when the tensile stress of 28 MPa is applied. This translates into a maximum lateral strain εx(max) as −1.2 × 10 −3 mm Δd = = − 9.45 × 10 −5 ε x (max) = d0 12.7 mm

But, since the specimen contracts in this lateral direction, and we are concerned that this strain be –5

less than 9.45 × 10 , then the criterion for this part of the problem may be stipulated as

−

Δd < 9.45 ×10−5 . d0

Now, Poisson’s ratio is defined by Equation 7.8 as

ε ν=− x εz For each of the metal alloys let us consider a possible lateral strain, ε x =

Δd . Furthermore, since the d0

deformation is elastic, then, from Equation 7.5, the longitudinal strain, εz is equal to

εz =

σ E

Substituting these expressions for εx and εz into the definition of Poisson’s ratio we have

Δd d0

ε ν = − x =− σ εz

E

which leads to the following: −

Δd νσ = d0 E

Using values for ν and E found in Table 7.1 for the six metal alloys that satisfy the criterion for part (a), and for σ= 28 MPa, we are able to compute a − Δd for each alloy as follows: d0

−

−

(0.34)(28 × 10 6 N/m 2 ) Δd (brass) = = 9.81 × 10 −5 9 2 d0 97 × 10 N/m

(0.34)(28 × 10 6 N/m 2 ) Δd (copper) = = 8.65 × 10 −5 9 2 d0 110 × 10 N/m

−

(0.34)(28 × 10 6 N/m 2 ) Δd (titanium) = = 8.90 × 10 −5 d0 107 × 10 9 N/m 2 (0.31)(28 × 10 6 N/m 2 ) Δd − (nickel) = = 4.19 × 10 −5 9 2 d0 207 × 10 N/m −

−

(0.30)(28 × 10 6 N/m 2 ) Δd (steel) = = 4.06 × 10 −5 9 2 d0 207 × 10 N/m

(0.28)(28 × 10 6 N/m 2 ) Δd (tungsten) = = 1.93 × 10 −5 d0 407 × 10 9 N/m 2

Thus, of the above six alloys, only brass will have a negative transverse strain that is greater than –5 9.45 × 10 . This means that the following alloys satisfy the criteria for both parts (a) and (b) of the problem: copper, titanium, nickel, steel, and tungsten.

Reserve Problem 15 A cylindrical rod 380 mm (15.0 in.) long and having a diameter of 10.0 mm (0.40 in.), is to be subjected to a tensile load. If the rod is to experience neither plastic deformation nor an elongation of more than 0.9 mm (0.035 in.) when the applied load is 24,500 N (5500 lbf), which of the four metals or alloys listed below are possible candidates?

Material

Modulus of Elasticity (GPa)

Yield Strength (MPa)

Tensile Strength (MPa)

70 100 110 207

255 345 250 450

420 420 290 550

Aluminum alloy Brass alloy Copper Steel alloy

• • • •

Aluminum alloy Brass alloy Copper Steel alloy

Solution This problem asks that we ascertain which of four metal alloys will not (1) experience plastic deformation, and (2) elongate more than 0.9 mm when a tensile load of 24,500 N is applied. It is first necessary to compute the stress using Equation 7.1; a material to be used for this application must necessarily have a yield strength greater than this value. Thus,

σ =

F = A0

24,500 N ⎛ 10.0 × 10 −3 m ⎞ ⎟ π⎜ ⎜ ⎟ 2 ⎝ ⎠

2

= 312 MPa

Of the metal alloys listed, only brass and steel have yield strengths greater than this stress.

Next, we must compute the elongation produced in both brass and steel using Equations 7.2 and 7.5 in order to determine whether or not this elongation is less than 0.9 mm. For brass

Δl =

σ l0 E

=

(312 MPa)(380 mm)

= 1.19 mm

100 × 103 MPa

Thus, brass is not a candidate. However, for steel

Δl =

σ l0 E

=

(312 MPa)(380 mm) 207 × 103 MPa

= 0.57 mm

Therefore, of these four alloys, only steel satisfies the stipulated criteria.

Reserve Problem 16: Load for elongation Consider the brass alloy the stress-strain behavior of which is shown in the Animated Figure 7.12. A cylindrical specimen of this material 13.7 mm in diameter and 142.1 mm long is pulled in tension after which the tensile load is released. After the load is released the total length has still increased to 142.4 mm. Calculate the magnitude of the load (in N) necessary to cause this elongation.

Solution It is first necessary to calculate the final strain of the specimen after the load release. From the definition of strain

l − l0 = 0.002 l0

ε= i

Next locate this strain value on the bottom horizontal axis on the inset plot, and then draw a straight line from this strain value parallel to the linear elastic region. (In this case, this line corresponds to the 0.002 strain offset line that is shown.) This line intersects the stress-strain curve at a stress of about 250 MPa. From the definition of stress for a cylindrical specimen

σ=

F = A0

F ⎛ d0 ⎞ ⎟ ⎝ 2 ⎠

π⎜

2

=

4F

π d0 2

it is possible to determine the applied load as ⎛ πd 2 ⎞ F = σ A0 = (250 MPa) ⎜ 0 ⎟ ⎜ 4 ⎟ ⎝ ⎠

Reserve Problem 17 A cylindrical specimen of aluminum having a diameter of 0.505 in. (12.8 mm) and a gauge length of 2.000 in. (50.800 mm) is pulled in tension. Use the load–elongation characteristics shown in the following table to complete parts (a) through (e).

Load in N

Load in lbf

Length in mm

Length in in.

0

0

50.800

2.000

7330

1650

50.851

2.002

15,100

3400

50.902

2.004

23,100

5200

50.952

2.006

30,400

6850

51.003

2.008

34,400

7750

51.054

2.010

38,400

8650

51.308

2.020

41,300

9300

51.816

2.040

44,800

10,100

52.832

2.080

46,200

10,400

53,848

2.120

47,300

10,650

54.864

2.160

47,500

10,700

55.880

2.200

46,100

10,400

56.896

2.240

44,800

10,100

57.658

2.270

42,600

9600

58.420

2.300

34,400

8200

59.182

2.330

Fracture

Fracture

Fracture

Fracture

Plot the data as engineering stress versus engineering strain. Based on your plot, (a) Compute the modulus of elasticity. (b) Determine the yield strength at a strain offset of 0.002. (c) Determine the tensile strength of this alloy. (d) What is the approximate ductility, in percent elongation? (e) Compute the modulus of resilience.

(a) ____ GPa (b) ____ MPa (c) ____ MPa (d) ____ %EL 3

(e) ____ J/m

Solution This problem calls for us to make a stress-strain plot for aluminum, given its tensile load-length data, and then to determine some of its mechanical characteristics.

The data are plotted below on two plots: the first corresponds to the entire stress-strain curve, while for the second, the curve extends to just beyond the elastic region of deformation.

(a) The elastic modulus is the slope in the linear elastic region (Equation 7.10) as

E=

Δσ

=

Δε

200 MPa − 0 MPa 0.0032 − 0

= 62.5 × 103 MPa = 62.5 GPa (9.1 × 106 psi)

(b) For the yield strength, the 0.002 strain offset line is drawn dashed. It intersects the stress-strain curve at approximately 285 MPa (41,000 psi). (c) The tensile strength is approximately 370 MPa (54,000 psi), corresponding to the maximum stress on the complete stress-strain plot. (d) The ductility, in percent elongation, is just the plastic strain at fracture, multiplied by onehundred. The total fracture strain at fracture is 0.165; subtracting out the elastic strain (which is about 0.005) leaves a plastic strain of 0.160. Thus, the ductility is about 16%EL. (e) From Equation 7.14, the modulus of resilience is just

Ur =

σ 2y 2E

which, using data computed above gives a value of Ur =

(285 MPa)2 3

(2) (62.5 × 10 MPa )

= 0.65 MN/m2 = 0.65 × 106 N/m2 = 6.5 × 105 J/m3 (93.8 in.-lb f /in.3 )

Reserve Problem 18 A specimen of ductile cast iron having a rectangular cross section of dimensions 4.8 mm × 15.9 mm (3/16 in. × 5/8 in.) is deformed in tension. Using the load–elongation data shown in the following table, complete problems (a) through (e).

Load in N

Load in lbf

Length in mm

Length in in.

0

0

75.000

2.953

4740

1065

75.025

2.954

9140

2055

75.050

2.955

12,920

2900

75.075

2.956

16,540

3720

75.113

2.957

18,300

4110

75.150

2.959

20,170

4530

75.225

2.962

22,900

5145

75.375

2.968

25,070

5635

75.525

2.973

26,800

6025

75.750

2.982

28,640

6440

76.500

3.012

30,240

6800

78.000

3.071

31,100

7000

79.500

3.130

31,280

7030

81.000

3.189

30,820

6930

82.500

3.248

29,180

6560

84.000

3.307

27,190

6110

85.500

3.366

24,140

5430

87.000

3.425

18,970

4265

88.725

3.493

Fracture

Fracture

Fracture

Fracture

Plot the data as engineering stress versus engineering strain. (a) Compute the modulus of elasticity. (b) Determine the yield strength at a strain offset of 0.002. (c) Determine the tensile strength of this alloy. (d) Compute the modulus of resilience. (e) What is the ductility, in percent elongation? (a) ____ GPa (b) ____ MPa (c) ____ MPa 3

(d) ____ J/m (e) ____ %EL

Solution This problem calls for us to make a stress-strain plot for a ductile cast iron, given its tensile loadlength data, and then to determine some of its mechanical characteristics.

The data are plotted below on two plots: the first corresponds to the entire stress-strain curve, while for the second, the curve extends just beyond the elastic region of deformation.

(a) The elastic modulus is the slope in the linear elastic region (Equation 7.10) as E=

Δσ Δε

=

100 MPa − 0 MPa 0.0005 − 0

= 200 × 103 MPa = 200 GPa (29 × 10 6 psi)

(b) For the yield strength, the 0.002 strain offset line is drawn dashed. It intersects the stress-strain curve at approximately 280 MPa (40,500 psi). (c) The tensile strength is approximately 410 MPa (59,500 psi), corresponding to the maximum stress on the complete stress-strain plot. (d) From Equation 7.14, the modulus of resilience is just

Ur =

σ 2y 2E

which, using data computed above, yields a value of Ur =

(280 × 10 6 N/m 2 ) 2 9

2

(2) (200 × 10 N/m )

= 1.96 × 10 5 J/m 3 (28.3 in.-lb f /in.3 )

(e) The ductility, in percent elongation, is just the plastic strain at fracture, multiplied by onehundred. The total fracture strain at fracture is 0.185; subtracting out the elastic strain (which is about 0.001) leaves a plastic strain of 0.184. Thus, the ductility is about 18.4%EL. Reserve Problem 19a-1 (question pool) Three hypothetical tensile engineering stress-strain curves through failure are depicted below.

Which response is considered to be the most brittle?

Solution: (a) The stress-strain response featuring the least strain at failure is considered to be most brittle.

Reserve Problem 19b-1 (question pool) Three hypothetical tensile engineering stress-strain curves through failure are depicted below.

Which response is considered to be the most brittle?

Solution: (c) The stress-strain response featuring the least strain at failure is considered to be most brittle.

Reserve Problem 19c-1 (question pool) Three hypothetical tensile engineering stress-strain curves through failure are depicted below.

Which response is considered to be the most brittle?

Solution: (a) The stress-strain response featuring the least strain at failure is considered to be most brittle.

Reserve Problem 19d-1 (question pool) Three hypothetical tensile engineering stress-strain curves through failure are depicted below.

Which response is considered to be the most brittle?

Solution: (c) The stress-strain response featuring the least strain at failure is considered to be most brittle.

Reserve Problem 19e-1 (question pool) Three hypothetical tensile engineering stress-strain curves through failure are depicted below.

Which response is considered to be the most brittle?

Solution: (a) The stress-strain response featuring the least strain at failure is considered to be most brittle.

Reserve Problem 19f-1 (question pool) Three hypothetical tensile engineering stress-strain curves through failure are depicted below.

Which response is considered to be the most brittle?

Solution: (c) The stress-strain response featuring the least strain at failure is considered to be most brittle.

Reserve Problem 20a (question pool) Tensile engineering stress-strain curves through failure are depicted below.

Which specimen features the most ductile response?

Options (These could be scrambled to produce additional versions). o

(a) 1340 Steel, Water-Quenched & Tempered at 370 C (b) Stainless Steel (18-8) (c) Aluminum Alloy 2024-T81 (d) Structural Steel (Mild Steel) (e) Magnesium

Solution: (d) The stress-strain response featuring the most strain at failure is considered to be most ductile.

Reserve Problem 21: True strain A tensile test is performed on a specimen of some metal alloy, and it is found that a true plastic strain of 0.12 is produced when a true stress of 280 MPa is applied. For this alloy, the value of the strain hardening exponent is 0.3. On the basis of these data what true plastic strain would be expected for a total true plastic stress of 330 MPa?

Solution We are asked to compute the true strain that results from the application of a true stress of 340 MPa; other true stress-strain data are also given. It first becomes necessary to solve for n in Equation 7.19.

Taking logarithms of this expression leads to

log σ T = log K + n log εT Next we rearrange this equation such that log K is the dependent variable:

log K = log σ T − n log εT We now solve for n using the following data given in the problem statement: sT = 260 MPa eT = 0.10 n = 0.2

Thus

log K = log(260MPa) − (0.2)[log(0.10)] = 2.615 Or

K = 102.615 = 412MPa We now rearrange Equation 7.19 such that εT is the dependent variable; we first divide both sides of the Equation 7.19 by K, which leads to the following expression:

εTn =

σT K

εT becomes the dependent variable by taking the 1/n root of both sides of this expression, as 1/n

⎛σ ⎞ εT = ⎜ T ⎟ ⎝ K ⎠

Finally, using values of K and n, we solve for the true strain at a true stress of 340 MPa: 1/n

⎛σ ⎞ εT = ⎜ T ⎟ ⎝ K ⎠

1/0.2

⎛ 340MPa ⎞ =⎜ ⎟ ⎝ 412MPa ⎠

= 0.383

Reserve Problem 22: True stress A cylindrical specimen of a metal alloy 48.8 mm long and 9.09 mm in diameter is stressed in tension.

A true stress of 327 MPa causes the specimen to plastically elongate to a length of 55 mm. If it is known that the strain-hardening exponent for this alloy is 0.3, calculate the true stress (in MPa)

necessary to plastically elongate a specimen of this same material from a length of 48.8 mm to a length of 57.6 mm.

Solution We are asked to compute the true stress that is required to elongate a cylindrical specimen from 54.8 mm to 64.7 mm; other true stress-strain data are also given. We first want to compute the value of K in Equation 7.19—the expression that relates true stress and true strain:

σT = KεTn Furthermore, substitution of the definition of true strain—namely Equation 7.16

⎛l ⎞

ε T = ln ⎜⎜ i ⎟⎟ ⎝ l0 ⎠

into Equation 7.19 leads to n

⎡ ⎛ l ⎞⎤ σT = K ⎢ln ⎜⎜ i ⎟⎟⎥ ⎣⎢ ⎝ l0 ⎠⎦⎥ And solving for K yields

K=

σT ⎡ ⎛ li ⎞ ⎤ ⎢ln ⎜ ⎟ ⎥ ⎣⎢ ⎝ l0 ⎠ ⎦⎥

n

We now solve for the value of K using the following data given in the problem statement: sT = 365 MPa l0 =54.8 mm li =61.8 mm n = 0.2 Thus K=

365MPa ⎡ ⎛ 61.8mm ⎞ ⎤ ⎢ ln ⎜ ⎟⎥ ⎣ ⎝ 54.8mm ⎠ ⎦

0.2

= 558MPa

It is now possible to compute the stress required to elongate the specimen from 54.8 mm to 64.7 mm using the modified Equation 7.19 cited above. Therefore,

n

⎡ ⎛ l ⎞⎤ σT = K ⎢ln ⎜⎜ i ⎟⎟⎥ ⎢⎣ ⎝ l0 ⎠⎥⎦ ⎡ ⎛ 64.7mm ⎞ ⎤ = (558MPa) ⎢ ln ⎜ ⎟⎥ ⎣ ⎝ 54.8mm ⎠ ⎦

0.2

= 390MPa

Reserve Problem 23

A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of 390 MPa (56,600 psi). If the specimen radius is 2.5 mm (0.10 in.) and the support point separation distance is 30 mm (1.2 in.), predict whether or not you would expect the specimen to fracture when a load of 620 N (140 lbf) is applied. Calculate the value of flexural strength for this test.

Solution

The problem asks that we determine whether or not a cylindrical specimen of aluminum oxide having a flexural strength of 390 MPa (56,600 psi) and a radius of 2.5 mm will fracture when subjected to a load of 620 N in a three-point bending test; the support point separation is given as 30 mm. Using Equation 12.7b we will calculate the value of σ; if this value is greater than σfs (390 MPa), then fracture is expected to occur. Employment of Equation 7.20b yields σ=

=

(620 N) (30 × 10 −3 m ) (π ) ( 2.5 × 10

−3

m)

3

FL

π R3

= 379 × 10 6 N/m 2 = 379 MPa (53,500 psi)

Since this value is less than the given value of σfs (390 MPa), then fracture is not predicted.

The certainty of this prediction is not 100% because there is always some variability in the flexural strength for ceramic materials, and since this value of σ (379 MPa) is relatively close to σfs (390 MPa) then there is some chance that fracture will occur.

Reserve Problem 24 Using the date in the table below, do the following:

(a) Determine the flexural strength for nonporous MgO assuming a value of 3.75 for n in Equation 7.22. (b) Compute the volume fraction porosity at which the flexural strength for MgO is 62 MPa (9000 psi).

Solution (a) This part of the problem asks us to determine the flexural strength of nonporous MgO assuming that the value of n in Equation 12.10 is 3.75. Taking natural logarithms of both sides of Equation 7.22 yields

ln σ fs = lnσ0 − nP In the table it is noted that for P = 0.05, σfs = 105 MPa. For the nonporous material P = 0 and, ln σ0 = ln σ . Solving for ln σ0 from the above equation and using these data gives

lnσ0 = lnσ fs + nP = ln (105 MPa) + (3.75)(0.05) = 4.841 4.841

or σ0 = e

= 127 MPa (18,100 psi).

(b) Now we are asked to compute the volume percent porosity to yield a σfs of 62 MPa (9000 psi).

Taking the natural logarithm of Equation 7.22 and solving for P leads to P=

=

ln σ 0 − ln σ fs

n

ln (127 MPa) − ln (62 MPa) 3.75

= 0.19 or 19 vol%

Reserve Problem 25 The flexural strength and associated volume fraction porosity for two specimens of the same ceramic material are as follows: σfs (MPa) 100 50

P 0.05 0.20

(a) Compute the flexural strength for a completely nonporous specimen of this material. (b) Compute the flexural strength for a 0.10 volume fraction porosity.

Solution (a) Given the flexural strengths at two different volume fraction porosities, we are asked to determine the flexural strength for a nonporous material. If the natural logarithm is taken of both sides of Equation 7.22, then

lnσ fs = lnσ0 − nP Using the data provided in the problem statement, two simultaneous equations may be written as

ln (100 MPa) = ln σ0 − (0.05) n ln (50 MPa) = ln σ 0 − (0.20) n Solving for n and σ0 leads to n = 4.62 and σ0 = 126 MPa. For the nonporous material, P = 0, and, from Equation 12.10, σ0 = σfs. Thus, σfs for P = 0 is 126 MPa. (b) Now, we are asked for σfs at P = 0.10 for this same material. Utilizing Equation 7.22 yields

σ fs = σ0 exp ( − nP) = (126 MPa) exp ⎡⎣ − (4.62)(0.10) ⎤⎦

= 79.4 MPa Reserve Problem 26a (question pool) Which polymer listed in the table below would feature the lowest tensile yield point?

Solution: The polymer whose molecular chains pack the most inefficiently will exhibit the weakest mechanical response.

All polymers are PP, so that is not a factor.

The polymer whose molecules are shortest on average (features the lowest degree of polymerization) will exhibit fewer secondary bonds across molecules. The polymer with the lowest degree of polymerization will therefore be the weakest, mechanically.

Branching tends to limit molecule proximity in the bulk polymer, which effectively weakens secondary bonds that exist between adjacent molecules. Therefore, branched structures will be weaker than linear structures.

Reserve Problem 27a (question pool) Which polymer listed in the table below would feature the highest tensile yield point?

Solution: The polymer whose molecular chains pack the most efficiently will exhibit the strongest mechanical response.

All polymers are PP, so that is not a factor.

The polymer whose molecules are longest on average (features the highest degree of polymerization) will exhibit more secondary bonds across molecules. The polymer with the highest degree of polymerization will therefore be the strongest, mechanically.

Branching tends to limit molecule proximity in the bulk polymer, which effectively weakens secondary bonds that exist between adjacent molecules. Therefore, linear structures will be stronger than branched structures.

The polymer that is most crystalline will feature chains that coil, ravel, pack, etc. effectively, thereby increasing the number of and/or the strength of secondary bonds across adjacent molecules. Thus, the higher the crystallinity, the stronger the polymer.

It is worth noting that in this problem, an option exists for which all favors maximize the strength of the polymer. Comparing options for which various factors are “competitive” in terms of enhanced or hindered mechanical response are not easy to discern without additional stress-strain data.

Reserve Problem 28a (question pool) Which polymer listed in the table below would feature the lowest elastic modulus?

Solution: The polymer whose molecular chains pack the most inefficiently will exhibit the weakest mechanical response.

All polymers are PP, so that is not a factor.

The polymer whose molecules are shortest on average (features the lowest degree of polymerization) will exhibit fewer secondary bonds across molecules. The polymer with the lowest degree of polymerization will therefore be the least stiff, mechanically.

Branching tends to limit molecule proximity in the bulk polymer, which effectively weakens secondary bonds that exist between adjacent molecules. Therefore, branched structures will be less stiff than linear structures.

The polymer that is least crystalline will feature chains that do not coil, ravel, pack, etc. effectively, thereby limiting the number of and/or the strength of secondary bonds across adjacent molecules.

Thus, the lower the crystallinity, the less stiff the polymer.

It is worth noting that in this problem, an option exists for which all favors minimize the strength of the polymer. Comparing options for which various factors are “competitive” in terms of enhanced or hindered mechanical response are not easy to discern without additional stress-strain data.

Reserve Problem 29a (question pool) Which polymer listed in the table below would feature the highest elastic modulus?

Solution The polymer whose molecular chains pack the most efficiently (long chains, chains in close proximity, crystalline arrangement of chains) will exhibit the strongest mechanical response.

All polymers are PP, so that is not a factor.

The polymer whose molecules are shortest on average (features the lowest degree of polymerization) will exhibit fewer secondary bonds across molecules. The polymer with the highest degree of polymerization will therefore be the most stiff, mechanically.

Branching tends to limit molecule proximity in the bulk polymer, which effectively weakens secondary bonds that exist between adjacent molecules. Therefore, linear structures will be more stiff than branched structures.

The polymer that is least crystalline will feature chains that do not coil, ravel, pack, etc. effectively, thereby limiting the number of and/or the strength of secondary bonds across adjacent molecules.

Thus, the higher the crystallinity, the more stiff the polymer.

It is worth noting that in this problem, an option exists for which all favors minimize the strength of the polymer. Comparing options for which various factors are “competitive” in terms of enhanced or hindered mechanical response are not easy to discern without additional stress-strain data.

Reserve Problem 30a (question pool) Which polymer listed in the table below would feature the lowest density?

Solution: The polymer whose molecular chains pack the least efficiently will exhibit the lowest density.

All polymers are PP, so that is not a factor.

The polymer whose molecules are shortest on average (features the lowest degree of polymerization) will tend to be less dense than a polymer with longer molecules.

Branching tends to limit molecule proximity in the bulk polymer, which reduces density. Therefore, branched structures will be less dense than linear structures.

The polymer that is least crystalline will feature chains that coil, ravel, pack, etc. ineffectively, thereby decreasing the density. Thus, the lower the crystallinity, the less dense the polymer.

It is worth noting that in this problem, an option exists for which all favors minimize the density of the polymer. Comparing options for which various factors are “competitive” in terms of increased or decreased density is not easy to discern without additional data.

Reserve Problem 31a (question pool) Which polymer listed in the table below would feature the highest density?

Solution: The polymer whose molecular chains pack the most efficiently will exhibit the highest density.

All polymers are PP, so that is not a factor.

The polymer whose molecules are longest on average (features the highest degree of polymerization) will tend to be more dense than a polymer with shorter molecules.

Branching tends to limit molecule proximity in the bulk polymer, which decreases density.

Therefore, linear structures will be more dense than branched structures.

The polymer that is most crystalline will feature chains that coil, ravel, pack, etc. effectively, thereby increasing the density. Thus, the higher the crystallinity, the more dense the polymer.

It is worth noting that in this problem, an option exists for which all favors maximize the density of the polymer.

Comparing options for which various factors are “competitive” in terms of increased or decreased density is not easy to discern without additional data.

Reserve Problem 32a (question pool) Which polymer listed in the table below would feature the lowest thermal expansion coefficient?

Solution: The polymer whose molecular chains pack the most efficiently will exhibit the lowest thermal expansion coefficient, since high-density configurations are associated with more / stronger secondary bonds, which act to bind the molecules together. Polymers with high-density configurations are therefore relatively hard (difficult to deform through mechanical force or thermal energy) Therefore, high-density configurations of a polymer are expected to correlate with lower thermal expansion coefficients.

The polymer whose molecular chains pack the most efficiently will exhibit the highest density.

All polymers are PP, so that is not a factor.

The polymer whose molecules are longest on average (features the highest degree of polymerization) will tend to be more dense than a polymer with shorter molecules.

Branching tends to limit molecule proximity in the bulk polymer, which decreases density.

Therefore, linear structures will be more dense than branched structures.

The polymer that is most crystalline will feature chains that coil, ravel, pack, etc. effectively, thereby increasing the density. Thus, the higher the crystallinity, the more dense the polymer.

It is worth noting that in this problem, an option exists for which all favors maximize the density of the polymer. Comparing options for which various factors are “competitive” in terms of increased or decreased density is not easy to discern without additional data.

Reserve Problem 33a (question pool) Which polymer listed in the table below would feature the highest thermal expansion coefficient?

Solution: The polymer whose molecular chains pack the least efficiently will exhibit the highest thermal expansion coefficient, since low-density configurations are associated with fewer / weaker secondary bonds, which act to bind the molecules together. Polymers with low-density configurations are therefore relatively soft (easy to deform through mechanical force or thermal energy) Therefore. lowdensity configurations of a polymer are expected to correlate with higher thermal expansion coefficients.

The polymer whose molecular chains pack the least efficiently will exhibit the lowest density.

All polymers are PP, so that is not a factor.

The polymer whose molecules are shortest on average (features the lowest degree of polymerization) will tend to be less dense than a polymer with longer molecules.

Branching tends to limit molecule proximity in the bulk polymer, which reduces density. Therefore, branched structures will be less dense than linear structures.

The polymer that is least crystalline will feature chains that coil, ravel, pack, etc. ineffectively, thereby decreasing the density. Thus, the lower the crystallinity, the less dense the polymer.

It is worth noting that in this problem, an option exists for which all favors minimize the density of the polymer. Comparing options for which various factors are “competitive” in terms of increased or decreased density is not easy to discern without additional data.

Reserve Problem 34a (question pool) Assuming the stress-strain curves provide below all correspond to the same polymer, which tensile test was most likely performed at the highest temperature above Tg ? Choose (e) if none tests likely meet this condition.

Solution:

Since the specimens for test a-d are all the same polymer, the specimen that was furthest above Tg is expected to exhibit the most ductile response.

Option (d) corresponds to the tensile test that was most likely performed at the highest temperature above Tg since it is the most ductile of the stress-strain responses, exhibiting the highest strain at failure.

Reserve Problem 35a (question pool) Assuming the stress-strain curves provide below all correspond to the same polymer tested at the same temperature, which tensile test was most likely performed with the highest strain rate?

Solution: Since the specimens for tests a-d are all the same polymer and each was tested at the same temperature, the specimen that endured the highest strain rate is expected to exhibit the least ductile response.

Reserve Problem 36 (a) Consider a thin-walled cylindrical tube having a radius of 65 mm that is to be used to transport pressurized gas. If inside and outside tube pressures are 100 and 1.0 atm (10.13 and 0.1013 MPa), respectively, compute the minimum required thickness for each of the following metal alloys.

Assume a factor of safety of 3.5.

σy (MPa)

ρ (g/cm )

Unit Mass Cost, c ($US/kg)

Steel (plain)

375

7.8

1.65

Steel (alloy)

1000

7.8

4.00

Cast iron

225

7.1

2.50

Aluminum

275

2.7

7.50

Magnesium

175

1.80

15.00

Alloy

• • • • •

Yield Strength,

Steel (plain) Steel (alloy) Cast iron Aluminum Magnesium

Density, 3

____ ____ ____ ____ ____

(b) A tube constructed of which of the alloys will cost the least amount?

• Cast iron • Aluminum • Magnesium • Steel (plain) • Steel (alloy) Solution (a) To solve this problem we use a procedure similar to that used for Design Example 6.2. Alloy yield strength, tube radius and thickness, and the inside-outside pressure difference are related according to Equation 7.31; furthermore, labels for the tube dimensions are represented in the sketch of Figure 7.23. For this problem all parameters in this equation are provided except for the tube thickness, t.

Rearranging Equation 7.31 such that t is the dependent variable leads to

t=

NrΔp

σy

(7.31b)

Tube wall thickness will vary from alloy to alloy since each alloy has a different yield strength, whereas N, r, and Δp are constant. Their values are provided in the problem statement as follows: N = 3.5 −3

r = 65 mm = 65 × 10 m Δp = 10.13 MPa − 0.1013 MPa = 10.029 MPa Using Equation 7.31b, we compute the tube wall thickness for these five alloys as follows:

t (plain steel) =

(3.5)(65 × 10 −3 m)(10.029 MPa) = 0.00608 m = 6.08 mm 375 MPa

t (alloy steel) =

(3.5)(65 × 10 −3 m)(10.029 MPa) = 0.00228 m = 2.28 mm 1000 MPa

t (cast iron) =

(3.5)(65 × 10 −3 m)(10.029 MPa) = 0.0101 m = 10.1 mm 225 MPa

t (aluminum) =

(3.5)(65 × 10 −3 m)(10.029 MPa) = 0.00830 m = 8.30 mm 275 MPa

t (magnesium) =

(3.5)(65 × 10 −3 m)(10.029 MPa) = 0.0130 m = 13.0 mm 175 MPa

(b) This portion of the problem asks that we determine at tube constructed of which of the alloys will cost the least amount. We begin by considering Equation 7.33, which gives tube volume as a function of t. Furthermore, when the expression for t, Equation 7.31b, is incorporated into Equation 7.33, the following equation for tube volume V results: V = π (2rt + t 2 ) L 2⎤ ⎡ ⎛ Nr Δp ⎞ ⎛ Nr Δp ⎞ ⎥ ⎢ ⎟+⎜ ⎟ L = π 2r ⎜ ⎢ ⎜ σ ⎟ ⎜ σ ⎟ ⎥ y y ⎠ ⎝ ⎠ ⎥⎦ ⎢⎣ ⎝

Now tube cost per unit length is dependent on volume, alloy density, and cost per unit mass values according to Equation 7.34. Substitution of the above expression for V into Equation 7.34 yields

⎛ Vρ ⎞ Cost = ⎜ ⎟ (c ) ⎝ 1000 ⎠

⎧ ⎡ 2⎤ ⎫ ⎪⎪ ⎢ ⎛ Nr Δ p ⎞ ⎛ Nr Δ p ⎞ ⎥ ⎪⎪ ⎟+⎜ ⎟ ⎥ Lρ ⎬ c ⎨π ⎢ 2r ⎜ ⎪ ⎢ ⎜⎝ σ y ⎟⎠ ⎜⎝ σ y ⎟⎠ ⎥ ⎪ ⎥⎦ ⎭⎪ ⎪ ⎢⎣ =⎩ 1000

Using this equation, let us now determine the cost per unit length for the plain steel.

Cost (plain steel) =

⎧ ⎡ ⎫ ⎛ (3.5)(65 × 10 −3 m)(10.029 MPa) ⎞ ⎤ ⎪ ⎢ (2)(65 × 10 −3 m) ⎜ ⎪ ⎟ +⎥ ⎜ ⎟ 375 MPa ⎪ ⎪ ⎝ ⎠ ⎥ 6 3 3 ⎪ ⎢⎢ 3 ⎪ ⎥ (10 cm /m )(1 m)(7.8 g/cm ) π ⎨ ⎬ (1.65 $US/kg) 2 ⎥ ⎛ (3.5)(65 × 10 −3 m)(10.029 MPa) ⎞ ⎪ ⎢ ⎪ ⎥ ⎜ ⎟ ⎪ ⎢ ⎪ ⎜ ⎟ 375 MPa ⎢ ⎥ ⎝ ⎠ ⎦ ⎩⎪ ⎣ ⎭⎪ 1000 g/kg

= $33.50

We will not present cost computations for the other alloys. However, the following table lists costs for all five alloys (along with wall thickness values calculated above.)

Alloy

t (mm)

Cost ($US)

Steel (plain) Steel (alloy) Cast iron Aluminum Magnesium

6.08 2.28 10.1 8.30 13.0

33.50 29.60 79.20 73.00 158.20

Hence, the alloy steel is the least expensive, the plain steel is next; the most expensive is the magnesium alloy.

Reserve Questions and Problems Solutions Manual CHAPTER 8 DEFORMATION AND STRENGTHENING MECHANISMS

Reserve Question 01: Atomic ordering with edge dislocations After an edge dislocation has passed through some region of a crystal, the atomic arrangement of that region is disordered. a) True b) False Solution False. Before and after an edge dislocation has passed through some region of a crystal, the atomic arrangement is ordered and perfect. It is only during the passage of the extra half-plane that there is disorder.

Reserve Question 02: Dislocation motion The process by which plastic deformation is produced by dislocation motion is called ____.

Solution The process by which plastic deformation is produced by dislocation motion is called slip.

Reserve Question 03: Edge dislocation movement Relative to the direction of an applied shear stress, the direction of motion of an edge dislocation’s line is a) perpendicular. b) parallel.

Solution The direction of motion of an edge dislocations’ line is parallel to the direction of applied shear stress.

Reserve Question 04: Screw dislocation movement Relative to the direction of an applied shear stress, the direction of motion of a screw dislocation’s line is a) perpendicular. b) parallel.

Solution The direction of motion of a screw dislocations’ line is perpendicular to the direction of applied shear stress.

Reserve Question 05: Edge dislocation stress In response to an applied shear stress, an edge dislocation moves in which direction relative to its line? a) Perpendicular b) Parallel Solution In response to an applied shear stress, the direction in which an edge dislocation moves is perpendicular to its line.

Reserve Question 06: Theoretical vs experimental strength How does the theoretical strength of a solid material compare with its experimental strength? a) Strengththeoretical < strengthexperimental b) Strengththeoretical = strengthexperimental c) Strengththeoretical > strengthexperimental Solution The experimental strength of a solid material is less than the theoretical strength. This behavior is explained by the presence dislocations.

Reserve Question 07: Edge dislocation strain The atoms surrounding an edge dislocation experience what kind(s) of strain(s)? a) Shear strains b) Tensile strains c) Compressive strains Solution The atoms surrounding an edge dislocation experience all three kinds of strains. Atoms below the dislocation experience tensile strains, while atoms above experience compressive strains. Shear strains also exist in the vicinity of the edge dislocation.

Reserve Question 08: Screw dislocation strain The atoms surrounding a screw dislocation experience what kind(s) of strain(s)? a) Shear strains b) Tensile strains c) Compressive strains Solution The atoms surrounding a screw dislocation experience only shear strains.

Reserve Question 09: Ductility vs slip systems A metal having a crystal structure with many operable slip systems will be relatively a) ductile. b) brittle.

Solution A metal having a crystal structure with many operable slip systems will be relatively ductile because considerable plastic deformation is normally possible when many slip systems are available.

Reserve Question 10: Slip direction For a particular crystal structure, the slip direction is that direction in the slip plane having the a) lowest linear density. b) highest linear density.

Solution For a particular crystal structure, the slip direction is that direction in the slip plane having the highest linear density.

Reserve Question 11: Slip plane – atomic packing For a particular crystal structure, the slip plane is that plane having the a) least dense atomic packing. b) most dense atomic packing.

Solution For a particular crystal structure, the slip plane is that plane having the most dense atomic packing.

Reserve Question 12: Slip planes Dislocations move with the same degree of ease on all crystallographic planes of atoms and in all crystallographic directions? a) True b) False Solution False. Dislocations do not move with the same degree of ease on all crystallographic planes of atoms and in all crystallographic directions. Crystallographic planes and directions along which dislocations move are those that are most closely packed with atoms.

Reserve Question 13: Slip systems The slip system for a particular crystal structure is that crystallographic plane-direction combination for which atomic distortion accompanying the motion of a dislocation is a) a minimum. b) a maximum.

Solution The slip system for a given crystal structure is that plane-direction combination for which the atomic distortion accompanying the motion of a dislocation is a minimum.

Reserve Problem 14 Determine the angles α, β, and γ that are listed in the cubic unit cell provided.

Enter the angles in degrees.

Note: You should be able to use basic facts about cube geometry and crystallographic convention to solve this, rather than elaborate direction cosine equations.

Solution:

Reserve Problem 15a (question pool) Each option below depicts a candidate slip plane and direction in a cubic crystal.

Which of the following slip system candidates is the least likely to exhibit slip?

Note: the loading axis is depicted with a BLACK dotted line and is always normal to a cube face. (a)

(b)

(c)

(d)

(e)

Solution: (e) is correct because the axis of load is parallel to the indicated plane, so slip is not expected since the corresponding Schmid factor is zero, making the resolved shear stress on this plane zero.

Reserve Problem 16a (question pool) Each option below depicts a candidate slip plane and direction in a cubic crystal.

Which of the following slip system candidates is the least likely to exhibit slip?

Note: the loading axis is depicted with a BLACK dotted line and is always normal to a cube face.

(a)

(b)

(c)

(d)

(e)

Solution: (a) is correct because the axis of load is parallel to the indicated direction, so slip is not expected since the corresponding Schmid factor is zero, making the resolved shear stress in this direction equal to zero.

Reserve Problem 17a (question pool) The figures below depict hypothetical room-temperature random polycrystalline grain structures (at the same magnification) for the same single-component metal. Which of these structures is expected to provide the lowest yield stress? (a)

(b)

(c)

(d)

Solution: (a) is correct because a large grain microstructure tends to be less strong and more ductile compared to structures with smaller grains. A higher grain density results in a higher yield stress because there is a higher probability numerous grains will be unfavorably oriented for slip for an arbitrary loading axis. These grains that are unfavorably oriented tend to hinder slip in adjacent grains (which are more favorably oriented grains). In other words, slip is expected to occur only when a sufficiently high stress is applied to the bulk specimen, such that the resolved stress in every grain exceeds the critical resolved shear stress.

Reserve Question 18: Twinning and crystal structures Mechanical twinning occurs in metals having which type(s) of crystal structure(s)? a) BCC b) FCC c) HCP Solution Mechanical twinning occurs in metals having BCC and HCP crystal structures.

Reserve Question 19: Grain size vs mechanical properties How does grain size influence strength of a polycrystalline material? a) Strengthfine-grained < strengthcourse-grained b) Strengthfine-grained = strengthcourse-grained c) Strengthfine-grained > strengthcourse-grained Solution A fine-grained material is stronger than a coarse-grained material.

Reserve Question 20: Grain size vs toughness Reducing the grain size of metal improves toughness. a) True b) False Solution True. Reducing the grain size of a metal improves its toughness.

Reserve Question 21: Dislocation density As dislocation density increases, the resistance to dislocation movement a) increases. b) decreases.

Solution As dislocation density increases, the resistance to dislocation movement increases. This phenomenon is responsible for cold working.

Reserve Question 22: Dislocation interactions On the average, dislocation-dislocation strain interactions are a) repulsive. b) attractive.

Solution On the average, dislocation-dislocation strain interactions are repulsive.

Reserve Question 23: Ductility vs strain hardening As a metal is strain hardened, its ductility a) increases b) decreases Solution As a metal is strain hardened, its ductility decreases.

Reserve Question 24: Strain hardening Most metals strain harden at room temperature. a) True b) False

Solution True. Most metals strain harden at room temperature.

Reserve Problem 25 It is necessary to select a metal alloy for an application that requires a yield strength of at least 345 MPa (50,000 psi) while maintaining a minimum ductility (%EL) of 20%. If the metal may be cold worked, decide which of the following are candidates: copper, brass, and a 1040 steel.

• • •

Brass Copper 1040 steel

Solution For each of these alloys, the minimum cold work necessary to achieve the yield strength may be determined from Figure 8.19a, while the maximum possible cold work for the ductility is found in Figure 8.19c. These data are tabulated below.

Yield Strength (> 345 MPa) Any %CW > 20%CW > 54%CW

Steel Brass Copper

Ductility (> 20%EL) < 5%CW < 23%CW < 15%CW

Thus, both the 1040 steel and brass are possible candidates since for these alloys there is an overlap of percents cold-work to give the required minimum yield strength and ductility values.

Reserve Problem 26 The average grain diameter and yield strength for a brass material were measured as a function of time at 650°C. Given the following yield strengths for the two specimens, compute the heat treatment time required at 650°C to give a yield strength of 100 MPa. Assume a value of 2 for n, the grain diameter exponent.

Time (min)

Yield Strength (MPa)

Grain Diameter (mm)

30 90

90 75

3.9 × 10 -2 6.6 × 10

-2

Solution Let us first calculate the percent cold work and attendant yield strength and ductility if the drawing is carried out without interruption. From Equation 8.8

2

2

⎛d ⎞ ⎛d ⎞ π⎜ 0 ⎟ −π⎜ d ⎟ 2 ⎝ 2 ⎠ ×100 %CW = ⎝ ⎠ 2 ⎛ d0 ⎞ π⎜ ⎟ ⎝ 2 ⎠ 2

⎛ 16.0 mm ⎞ ⎛ 11.3 mm ⎞ ⎟ −π⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠

π⎜ =

⎛ 16.0 mm ⎞ π⎜ ⎟ 2 ⎝ ⎠

2

2

× 100 = 50%CW

At 50%CW, the copper will have a yield strength on the order of 330 MPa (48,000 psi), Figure 8.19a, which is adequate; however, the ductility will be about 4%EL, Figure 8.19c, which is insufficient.

Instead of performing the drawing in a single operation, let us initially draw some fraction of the total deformation, then anneal to recrystallize, and, finally, cold work the material a second time in order to achieve the final diameter, yield strength, and ductility.

Reference to Figure 8.19a indicates that 21%CW is necessary to give a yield strength of 250 MPa. Similarly, a maximum of 23%CW is possible for 12%EL [Figure 7.19c]. The average of these two values is 22%CW, which we will use in the calculations. Thus, to achieve both the specified yield strength and ductility, the copper must be deformed to 22%CW. If the final diameter after the first drawing is d0′ , then, using Equation 8.8 2

2

⎛ 11.3 mm ⎞ ⎛ d′ ⎞ π ⎜ 0 ⎟ − π ⎜⎜ ⎟⎟ 2 ⎝ 2 ⎠ ⎝ ⎠ × 100 22%CW = 2 ⎛ d′ ⎞ π⎜ 0 ⎟ ⎝ 2 ⎠ And, solving for d0′ from the above expression yields d0′ =

11.3 mm 22%CW 1− 100

= 12.8 mm (0.50 in.)

Reserve Question 27: Recovery During the recovery of a cold-worked material, which of the following statement(s) is (are) true?

a) Some of the internal strain energy is relieved. b) All of the internal strain energy is relieved. c) There is some reduction in the number of dislocations.

d) There is a significant reduction in the number of dislocations, to approximately the number found in the precold-worked state. e) The electrical conductivity is recovered to its precold-worked state. f) The thermal conductivity is recovered to its precold-worked state. g) The metal becomes more ductile, as in its precold-worked state. h) Grains with high strains are replaced with new, unstrained grains.

Solution During the recovery process the following take place: • Some of the internal strain energy is relieved.

• There is some reduction in the number of dislocations.

• The electrical conductivity is recovered to its precold-worked state.

• The thermal conductivity is recovered to its precold-worked state.

Reserve Question 28: Recrystallization During the recrystallization of a cold-worked material, which of the following statement(s) is (are) true?

a) Some of the internal strain energy is relieved. b) All of the internal strain energy is relieved. c) There is some reduction in the number of dislocations. d) There is a significant reduction in the number of dislocations, to approximately the number found in the precold-worked state. e) The electrical conductivity is recovered to its precold-worked state. f) The thermal conductivity is recovered to its precold-worked state. g) The metal becomes more ductile, as in its precold-worked state. h) Grains with high strains are replaced with new, unstrained grains.

Solution During the recovery process the following take place:

• •

All of the internal strain energy is relieved.

There is significant reduction in the number of dislocations.

• •

The metal becomes more ductile, as in its precold-worked state.

Grains with high strains are replaced with new, unstrained grains.

Reserve Question 29: Grain growth requirements Grain growth must always be preceded by recovery and recrystallization.

a) True b) False Solution False. Grain growth does not always need to be preceded by recovery and recrystallization; it may occur in materials that have not been cold worked.

Reserve Problem 30 –2

A hypothetical metal alloy has a grain diameter of 2.4 × 10 mm. After a heat treatment at 575°C for –2 500 min, the grain diameter has increased to 7.3 × 10 mm. Compute the time required for a –2 –2 specimen of this same material (i.e., d0 = 2.4 × 10 mm) to achieve a grain diameter of 5.5 × 10 mm while being heated at 575°C. Assume the n grain diameter exponent has a value of 2.2.

Solution

To solve this problem requires that we use Equation 8.9 with n = 2.2. It is first necessary to solve for 2 the parameter K in this equation, using values given in the problem statement of d0 (2.4 × 10− mm) 2 and the grain diameter after the 500-min heat treatment (7.3 × 10− mm). The computation for K using a rearranged form of Equation 8.9 in which K becomes the dependent parameter is as follows:

d2.2 − d02.2 t (7.3 × 10 −2 mm)2.2 − (2.4 × 10 −2 mm)2.2 = 500 min = 5.77 × 10 −6 mm2.2 /min K=

It is now possible to solve for the time required to yield a value of 5.5 × 10− mm for d using a rearranged form of Equation 8.9: 2

t=

=

d 2.2 − d02.2 K

(5.5 × 10 −2 mm)2.2 − (2.4 × 10 −2 mm)2.2 5.77 × 10 −6 mm2.2 /min

= 246 min

Reserve Problem 31: Tensile strength vs number average molecular weight I Tensile strengths and number-average molecular weights for two polymers are as follows: Tensile strength

Number average molecular weight

(MPa)

(g/mol)

37.7

36800

131

62400

Estimate the tensile strength (in MPa) for a number-average molecular weight of 51500 g/mol.

Solution Tensile strength is related to the number-average molecular weight according to TS = TS∞ −

A Mn

Using the two data sets given, two equations having the above form may be written as follows:

A Mn1 A TS2 = TS∞ − Mn2 TS1 = TS∞ −

where each of the “1” and “2” subscripts refers to one set of data given in the problem statement.

Using these data, simultaneous solution of these equations yields values for TS∞ and A. The tensile strength called for may be calculated when values for these constants and the specified molecular weight are incorporated into the same equation.

Reserve Problem 32: Tensile strength vs number average molecular weight II Tensile strengths and number-average molecular weights for two polymers are as follows

Tensile strength

Number average molecular weight

(MPa)

(g/mol)

138

12600

184

28100

Estimate number average molecular weight (in g/mol) at a tensile strength of 141 Mpa.

Solution Tensile strength is related to the number-average molecular weight according to TS = TS∞ −

A Mn

Using the two data sets given, two equations having the above form may be written as follows: A Mn1 A TS2 = TS∞ − Mn2 TS1 = TS∞ −

where each of the “1” and “2” subscripts refers to one set of data given in the problem statement.

Using these data, simultaneous solution of these equations yields values for TS∞ and A. The numberaverage molecular weight required for the tensile strength specified in the problem statement may be determined using the following rearranged form of the above equation and incorporating the values of TS∞ and A just determined.

Mn =

A TS∞ − TS

Reserve Question 33: Adhesive bonding The bonding forces between adhesive and adherend surfaces are thought to be

a) Electrostatic b) Covalent c) Chemical Solution The bonding forces between adhesive and adherend surfaces are thought to be electrostatic.

Reserve Question 34: Anisotropy of drawing Deformation of a semicrystalline polymer by drawing produces which of the following?

a) b) c) d)

Increase in strength in the direction of drawing.

Decrease in strength in the direction of drawing.

Increase in strength perpendicular to the direction of drawing.

Decrease in strength perpendicular to the direction of drawing.

Solution Deformation of a semicrystalline polymer by drawing produces an increase in strength in the direction of drawing, and a decrease in strength perpendicular to the direction of drawing.

Reserve Question 35: Tensile strength – Deformation How does deformation by drawing of a semicrystalline polymer affect its tensile strength?

a) Increases b) Decreases Solution Deformation by drawing increases the tensile strength of a semicrystalline polymer. This effect is due to the highly oriented chain structure that is produced by drawing, which gives rise to higher interchain secondary bonding forces.

Reserve Question 36: Tensile strength – Degree of crystallinity How does increasing the degree of crystallinity of a semicrystalline polymer affect its tensile strength?

a) Increases b) Decreases Solution Increasing the degree of crystallinity of a semicrystalline polymer leads to an increase in its tensile strength. This is due to enhanced interchain bonding and forces in crystalline regions; in response to applied stresses, interchain motions become more restrained as degree of crystallinity increases.

Reserve Question 37: Tensile strength – Molecular weight How does increasing the molecular weight of a semicrysatlline polymer affect its tensile strength?

a) Increases b) Decreases Solution The tensile strength of a semicrystalline polymer increases with increasing molecular weight. This effect is explained by the increased chain entanglements at higher molecular weights.

Reserve Questions and Problems Solutions Manual CHAPTER 9 FAILURE

Reserve Question 01: Intergranular/transgranular fracture types Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?

• •

Intergranular Transgranular

____ ____

Solution Both intergranular and transgranular fractures are brittle. For transgranualar, crack propagation is through the grains, whereas for intergranular, crack propagation is along grain boundaries.

Reserve Problem 02: Altering crack radii The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter the surface crack geometry (i.e. reduce crack length and increase tip radius). Calculate the ratio of the etched and original crack tip radii if the fracture strength is increased by a factor of 7 when 28% of the crack length is removed.

Solution This problem asks that we compute the crack tip radius ratio before and after etching. Let

ρt = unetched crack tip radius,and ρt′ = etched crack tip radius The maximum stress, σm in Equation 9.1 will be the same for both unetched and etched specimens, that is ′ = σm σm

Also, as noted in the problem statement, the etched crack length (a') and unetched crack length (a) are related as a ' = (1 − 0.35)a = 0.65a

And the relationship between the etched and unetched fractured strengths (denoted

σ 0′ and σ 0 , respectively) is

σ0′ =7σ0 Using the above notations, we write the following expression for Equation 9.1: 1/2

⎛a⎞ σ m = 2σ 0 ⎜⎜ ⎟⎟ ⎝ ρt ⎠ We now solve for the

1/2

⎛ a' ⎞ ′ = 2σ 0′ ⎜ ⎟ = σm ⎜ ρ′ ⎟ ⎝ t⎠

ρt′ ratio, which yields the following: ρt 2

ρt′ ⎛ σ 0′ ⎞ ⎛ a ' ⎞ =⎜ ⎟ ρt ⎜⎝ σ 0 ⎟⎠ ⎜⎝ a ⎟⎠ And when we incorporate the above a − a′ and σ 0 − σ 0′ relationships into this expression, the following results: 2

ρt′ ⎛ 7σ 0 ⎞ ⎛ 0.65a ⎞ =⎜ =32 ⎟ ρt ⎜⎝ σ 0 ⎟⎠ ⎜⎝ a ⎟⎠ Reserve Problem 03: Crack length in plane strain A structural component in the shape of a flat plate 27.3 mm thick is to be fabricated from a metal alloy for which the yield strength and plane strain fracture toughness values are 535 MPa 1/2 and 31.3 MPa-m , respectively. For this particular geometry, the value of Y is 1.8. Assuming a design stress of 0.5 times the yield strength, calculate the critical length of a surface flaw.

Solution In order to solve this problem it is necessary to use Equation 8.7, with

σ = (0.4)σ y = (0.4)(533 MPa) = 213 MPa . Therefore, the value of ac is computed as follows: 2

2

1⎛K ⎞ 1 ⎡ 22.0 MPa-m1/2 ⎤ −3 ac = ⎜ Ic ⎟ = ⎢ ⎥ = 2.0 × 10 m = 2.0 mm π ⎝ σ Y ⎠ π ⎢⎣ (213 MPa)(1.3) ⎦⎥

Reserve Question 04: Effect of temp on fracture toughness How would the plane strain fracture toughness of a metal be expected to change with rising temperature?

a) Increase b) Decrease c) Remain constant Solution

The magnitude of the plane strain fracture toughness, KIc, diminishes with decreasing temperature; therefore as temperature increases, KIc should also increase.

Reserve Question 05: Factors in brittle polymers Which of the following factor(s) favor(s) brittle fracture in polymers?

a) b) c) d)

Increasing in temperature.

Increasing in strain rate.

The presence of a sharp notch.

Decreasing specimen thickness.

Solution Increasing strain rate, the presence of a sharp notch, and decreasing temperature all favor brittle fracture of polymers.

Reserve Question 06: Fracture test – Charpy The results of a laboratory test that is used to assess the mechanical or failure characteristics of metals are shown below. Select labels for both of the plot axes.

X-axis: ____ Y-axis: ____

Solution The plot represents the results of a typical impact test on a material that experiences a ductile-to-brittle transition. The label for the x-axis is temperature, whereas impact energy is the label for the y-axis.

Reserve Question 07: Plane strain plate fracture 1/2

A plate of an alloy steel has a plane-strain fracture toughness of 50 MPa-m . If it is known that the largest surface crack is 0.5 mm long, and that the value of Y is 1.1, which of the following can be said about this plate when a tensile stress of 1200 MPa is applied? a) The plate will definitely fracture. b) The plate will definitely not fracture. c) It is not possible to determine whether or not the plate will fracture.

Solution From Equation 9.5,

KIc = Yσ πa and for this problem, we determine the value of the right-hand side of the equation. If the value of Y σ π a is less than K Ic for this material, then fracture will not occur; if greater than K Ic , then

fracture will occur. For this problem: 1/2

KIc= 50 MPa-m σ = 1200 MPa Y = 1.1

–3

a = 0.5 mm = 0.5 × 10 m Therefore, Y σ π a = (1.1)(1200MPa) (π )(0.5 × 10 −3 m)

= 52.3 MPa m 1/2

Since this value is greater than KIc (50 MPa-m ), fracture will occur.

Reserve Question 08: Stress raisers The effect of a stress raiser is more significant for which of the following types of materials?

a) Brittle materials b) Ductile materials

Solution The effect of a stress raiser is more significant for brittle materials. For a ductile material, plastic deformation ensues when the maximum stress exceeds the yield strength. This leads to a more uniform distribution of stress in the vicinity of the stress raiser and to the development of a maximum stress concentration factor less than the theoretical value. Such yielding and stress redistribution do not occur to any appreciable extent around flaws and discontinuities in brittle materials; therefore, essentially the theoretical stress concentration will result.

Reserve Problem 09 A structural component is fabricated from an alloy that has a plane strain fracture toughness of 45 MPa. It has been determined that this component fails at a stress of 300 MPa when the maximum length of a surface crack is 0.95 mm. What is the maximum allowable surface crack length (in mm) without fracture for this same component exposed to a stress of 300 MPa and made from another alloy with a plane strain fracture toughness of 100.0 MPa? The geometry factor Y is the same in both cases.

Solution

This problem asks us to determine the maximum allowable surface crack length without fracture for a structural component for a specified fracture toughness (57.5 MPa m ) , given that fracture occurs for the same component using the same stress level (300 MPa) and another fracture toughness (45 MPa m ) . (Note: Because the cracks surface cracks, their lengths are equal to a.)

The maximum crack length without fracture ac may be calculated using Equation 9.7—i.e.,

ac =

1 ⎛K ⎞ π ⎜ Ic ⎟ ⎝ σY ⎠

2

where KIc is the plane strain fracture toughness, σ is the applied stress, and Y is a design parameter.

4 Using data for the first alloy [ K Ic = 45 MPa m , ac = 0.95 mm (9.5 × 10− m), σ = 300 MPa], it is possible to calculate the value of Y using a rearranged form of the above equation as follows: Y=

=

K Ic

σ π ac

45 MPa m (300 M Pa) (π )(9.5 × 10 − 4 m )

= 2.75

Using this value of Y, the maximum crack length for the second alloy is determined, again using Equation 9.7 as

ac =

1 ⎛K ⎞ π ⎜ Ic ⎟ ⎝ σY ⎠

2

1 ⎡ 57.5 MPa m ⎤ = ⎢ ⎥ π ⎣⎢ (300 MPa)(2.75) ⎦⎥

2

= 0.00155 m = 1.55 mm

Reserve Problem 10 Following are tabulated data that were gathered from a series of Charpy impact tests on a ductile cast iron.

Temperature (°C)

Impact Energy (J)

–25

124

–50

123

–75

115

–85

100

–100

73

–110

52

–125

26

–150

9

–175

6

(a) Determine a ductile-to-brittle transition temperature as that temperature corresponding to the average of the maximum and minimum impact energies. (b) Determine a ductile-to-brittle transition temperature as that temperature at which the impact energy is 80 J.

Solution The plot of impact energy versus temperature is shown below.

(a) The average of the maximum and minimum impact energies from the data is Average =

124 J + 6 J 2

= 65 J

As indicated on the plot by the one set of dashed lines, the ductile-to-brittle transition temperature according to this criterion is about –105°C. (b) Also, as noted on the plot by the other set of dashed lines, the ductile-to-brittle transition temperature for an impact energy of 80 J is about –95°C.

Reserve Problem 11 Following are tabulated data that were gathered from a series of Charpy impact tests on a tempered 4140 steel alloy.

Temperature (°C)

Impact Energy (J)

100

89.3

75

88.6

50

87.6

25

85.4

0

82.9

–25

78.9

–50

73.1

–65

66.0

–75

59.3

–85

47.9

–100

34.3

–125

29.3

–150

27.1

–175

25.0

(a) Determine a ductile-to-brittle transition temperature as that temperature corresponding to the average of the maximum and minimum impact energies. (b) Determine a ductile-to-brittle transition temperature as that temperature at which the impact energy is 70 J.

Solution The plot of impact energy versus temperature is shown below.

(a) The average of the maximum and minimum impact energies from the data is Average =

89.3 J + 25 J 2

= 57.2 J

As indicated on the plot by the one set of dashed lines, the ductile-to-brittle transition temperature according to this criterion is about –75°C. (b) Also, as noted on the plot by the other set of dashed lines, the ductile-to-brittle transition temperature for an impact energy of 70 J is about –55°C.

Reserve Problem 12 What is the maximum carbon content possible for a plain carbon steel that must have an impact energy of at least 150 J at 0°C? Solution From the curves in Figure 9.23, for only the 0.22 wt% C, 0.11 wt%C and 0.01 wt% C steels are impact energies (and therefore, ductile-to-brittle temperatures) greater than 150. Therefore, 0.22 wt% is the maximum carbon concentration.

Reserve Problem 13 Compute the minimum value of plane-strain fracture toughness required of a material to satisfy the leak-before-break criterion for a cylindrical pressure vessel similar to that shown in Figure 9.11. The vessel radius and wall thickness values are 250 mm and 10.5 mm, respectively, and the fluid pressure is 3.0 MPa. Assume a value of 3.5 for the factor of safety.

Solution

To solve this problem we first substitute the expression for the hoop stress (σh), Equation 9.8, into Equation 9.10, which leads to the following expression ⎛ ⎞ 1 ⎛ K Ic ⎞ 1 ⎜ K Ic ⎟ cc = ⎜ ⎟ ⎜ ⎟ = π N 2 ⎜⎝ σ h ⎟⎠ π N 2 ⎜ pr ⎟ ⎜ ⎟ ⎝ t ⎠

2

2

=

1 ⎛ tK Ic ⎞ ⎜ ⎟ π N 2 ⎝ pr ⎠

2

In the above expression we now substitute t for cc (inasmuch as t = cc) as follows: 1 ⎛ tK Ic ⎞ t= ⎜ ⎟ π N 2 ⎝ pr ⎠

2

It is now possible to solve this expression for KIc; this results in the following equation: K Ic = Npr

π t

Incorporation of values for N (3.5), p (3.0 MPa), r (250 mm = 0.25 m), and t (10. 5 mm = 10.5 × 10− m) into the above equation yields the following:

3

K Ic = (3.5)(3.0 MPa)(0.25 m) = 45.4 MPa m

π (10.5 × 10 −3 m)

Reserve Problem 14a (question pool) The figure below depicts a brittle material featuring four elliptical flaws, each featuring the same radius of curvature at their tips. Based on the loading type and the loading axis illustrated, which flaw is most likely to propagate?

Solution: (d) is correct because flaws whose major axis is normal to tensile loads tend to propagate.

Also, the magnified stress is higher at the tip of a longer flaw.

Therefore, flaw (d) is more likely to propagate compared to flaw (c).

Reserve Question 15: Fracture test – Tensile The results of a laboratory test that is used to assess the mechanical or failure characteristics of metals are shown below. Select labels for both of the plot axes.

X-axis ____ Y-axis ____

Solution The plot represents the results of a typical stress-strain test conducted on a material that exhibits the yield point phenomenon. The label for the x-axis is strain, whereas stress is the label for the y-axis.

Reserve Question 16: Fracture test – Fatigue The results of a laboratory test that is used to assess the mechanical or failure characteristics of metals are shown below. Select labels for both of the plot axes.

X-axis ____ Y-axis ____ Solution The plot represents the results of a typical fatigue test. The label for the x-axis is cycles to failure (logarithmic scale), stress amplitude is the label for the y-axis.

Reserve Problem 17 A cylindrical rod of diameter 9.5 mm fabricated from a 2014-T6 aluminum alloy is subjected to rotating-bending load cycling; test results (as S-N behavior) are shown in Figure 9.27. If the maximum and minimum loads are +400 N and –400 N, respectively, determine its fatigue life. Assume that the separation between loadbearing points is 72.5 mm.

Solution

In order to solve this problem we compute the maximum stress using Equation 9.16, and then determine the fatigue life from the curve in Figure 9.27 for the aluminum alloy. In Equation 9.16, F is the maximum applied load, which for this problem is +400 N. Values for L and d0 are provided in –3 –3 the problem statement—viz. 72.5 mm (72.5 × 10 m) and 9.5 mm (9.5 × 10 m), respectively.

Therefore the maximum stress σ is equal to

σ=

=

16FL

π d03

(16)(400 N)(72.5 × 10 −3 m) (π )(9.5 × 10 −3 m)3

= 1 7 2 × 1 0 6 N /m 2 = 1 7 2 M P a

From Figure 9.27 and the curve for the aluminum alloy, the logarithm of the fatigue life (log Nf) at 172 MPa is about 7.3, which means that the fatigue life is equal to

N f = 107.3 cycles = 2 × 107 cycles

Reserve Problem 18 (a) Compare the fatigue limits for polystyrene (Figure 9.29) and the cast iron for which fatigue data are given in the table below: Stress Amplitude [MPa (ksi)]

Cycles to Failure

248 (36.0)

1 × 10

236 (34.2)

3 × 10

224 (32.5)

1 × 10

213 (30.9)

3 × 10

201 (29.1)

1 × 10

193 (28.0)

3 × 10

193 (28.0)

1 × 10

193 (28.0)

3 × 10

5 5 6 6 7 7 8 8

6

(b) Compare the fatigue strengths at 10 cycles for poly(ethylene terephthalate) (PET, Figure 92.9) and 70Cu–30Zn brass (Figure 9.27).

Solution (a) The fatigue limits for polystyrene and the ductile cast iron are 10.5 MPa (1500 psi) and 193 MPa (28,000 psi), respectively.

6

(b) At 10 cycles, the fatigue strengths of PET and 70Cu-30Zn brass are 15 MPa (2175 psi) and 160 MPa (23,200 psi), respectively.

Reserve Question 19: Creep behavior Match each of the three stages of creep behavior with the manner in which creep rate changes with increasing time.

Primary (transient creep) • Decreasing creep rate • Constant creep rate • Increasing creep rate

Secondary creep • Decreasing creep rate • Constant creep rate • Increasing creep rate Tertiary creep • Decreasing creep rate • Constant creep rate • Increasing creep rate Solution During primary (transient) creep the creep rate decreases as time increases.

During secondary creep the creep rate is constant.

During tertiary creep the creep rate increases as time increases.

Reserve Question 20: Fracture test – creep The results of a laboratory test that is used to assess the mechanical or failure characteristics of metals are shown below. Select labels for both of the plot axes.

X-axis: ____ Y-axis: ____ Solution The plot represents the results of a typical creep test. The label for the x-axis is time, whereas strain is the label for the y-axis.

Reserve Question 21: Creep elongation A specimen 667 mm long of a low carbon-nickel alloy is to be exposed to a tensile stress of 37 MPa at 538°C. Using the Animated Figure 9.40, determine its elongation after 5300 hours. Assume that the total of both instantaneous and primary creep elongations is 1.5 mm.

Solution

Using the plot given in the problem statement, it is possible to calculate the steady-state creep rate at the specified stress level and from the 538°C line. This value of the strain rate is in units of %/1000 h, 5 –1 which must be divided by 10 for units of h . The steady state creep strain, εs, is the product of the steady-state creep rate and time-that is

εs = ε st From the definition of strain

l − l Δl ε= i 0 = l0 l0 the steady state elongation is equal to Δls = ε s l0 = ε s tl0

Finally, the total elongation is just the sum of this Δls and the value of instantaneous and primary creep elongations specified in the problem statement.

Reserve Question 22: Creep tensile load For a cylindrical low carbon-nickel alloy specimen originally 11 mm in diameter and 503 mm long, what tensile load (in N) is necessary to produce a total elongation of 4.8 mm after 19000 hours at 538°C? Assume that the sum of instantaneous and primary creep elongations is 0.86 mm. The logarithm stress versus logarithm of steady-state creep rate plot for this alloy is in the Animated Figure 9.40.

Solution The steady state creep elongation is equal to the difference between the total elongation, and the sum of the instantaneous and primary creep elongation-that is

Δls = Δltotal − Δli/p Furthermore, the steady-state creep strain is defined as

εs =

Δ ls l0

and the steady-state creep rate is equal to ε s =

εs t

In order to get this value into units of %/1000 h, for t in h, the value calculated using the previous 5 equation must be multiplied by a factor of 10 . Using the plot given in the problem statement and the 538°C line, it is possible to determine the stress required for this steady-state creep rate. And, finally, for a cylindrical specimen, the load is related to stress and cross-sectional diameter as 2

⎛d ⎞ σπd02 F = σA0 = σπ ⎜ 0 ⎟ = 4 ⎝ 2 ⎠ Reserve Question 23: Factors affecting creep Which of the following affect(s) the creep characteristics of metals?

a) b) c) d) e) f)

Melting temperature Grain size Yield strength Ductility Elastic modulus Resilience

Solution The creep characteristics of metals are affected by melting temperature, elastic modulus, and grain size.

Reserve Question 24: Increasing stress or temperature If either stress or temperature is increased, indicate which of the following consequences will result.

a) The instantaneous strain at the time of stress application decreases. b) The steady-state creep rate increases. c) The rupture lifetime is diminished.

Solution If either stress or temperature is increased, the steady-state creep rate increases, and the rupture lifetime is diminished. [The instantaneous strain at the time of stress application increases (i.e., it does not decrease.)]

Reserve Question 25: Temperature for creep Above about what temperature does the creep become an important failure mechanism for a metal?

a) 0.1Tm b) 0.2Tm c) 0.3Tm d) 0.4Tm e) 0.5Tm

Solution

Creep become an important failure mechanism for a metal above a temperature of about 0.4Tm, where Tm is the absolute melting temperature.

Reserve Problem 26 A cylindrical component 75 mm long constructed from an S-590 alloy (Figure 9.40) is to be exposed to a tensile load of 20,000 N. What minimum diameter is required for it to experience an elongation of no more than 10.2 mm after an exposure for 1250 h at 815°C? Assume that the sum of instantaneous and primary creep elongations is 0.7 mm.

Solution

It is first necessary to calculate the steady state creep rate so that we may utilize Figure 9.40 in order to determine the tensile stress. The steady state elongation, Δls, is just the difference between the total elongation and the sum of the instantaneous and primary creep elongations; that is, Δls = 10.2 mm − 0.7 mm = 9.5 mm

Now the steady state creep rate, ε s is just

ε s =

Δε Δt

Δls 9.5 mm l = 0 = 75 mm 1250 h Δt = 1.01 × 10 −4 h −1 –4

–1

From the 815°C line in Figure 9.40, a steady state creep rate of 1.01 × 10 h corresponds to a stress σ of about 115 MPa. Now from Equation 7.1 and for a cylindrical specimen having a diameter of d0

σ=

F = A0

F

⎛d ⎞ π⎜ 0 ⎟ ⎝ 2 ⎠

2

6

2

Solving for d0 and realizing that σ = 115 MPa (115 × 10 N/m ) and F = 20,000 N leads to d0 = =

4F

πσ

(4)(20,000 N) (π )(115 × 106 N/m2 )

= 0.0150 m = 15.0 mm

Reserve Problem 27

A cylindrical specimen 7.5 mm in diameter of an S-590 alloy is to be exposed to a tensile load of 9000 –2 –1 N. At approximately what temperature will the steady-state creep be 10 h ?

Solution Let us first determine the stress imposed on this specimen using values of cross-section diameter and applied load; this is possible using Equation 7.1 as

σ=

F = A0

F

⎛d ⎞ π⎜ 0 ⎟ ⎝ 2 ⎠

2

Inasmuch as d0 = 7.5 mm (7.5 × 10− m) and F = 9000 N, the stress is equal to 3

σ=

9000 N ⎛ 7.5 × 10 −3 m ⎞ π⎜ ⎟⎟ ⎜ 2 ⎝ ⎠

2

= 204 × 106 N/m2 = 204 MPa Or approximately 200 MPa. From Figure 9.40 the point corresponding 10− h− and 200 MPa is approximately 815°C. 2

1

Reserve Problem 28 If a component fabricated from an S-590 alloy (Figure 9.39) is to be exposed to a tensile stress of 300 MPa (43,500 psi) at 650°C (1200°F), estimate its rupture lifetime.

Solution This problem asks us to calculate the rupture lifetime of a component fabricated from an S-590 alloy exposed to a tensile stress of 300 MPa at 650°C. All that we need do is read from the 650°C line in Figure 9.39 the rupture lifetime at 300 MPa; this value is about 600 h.

Reserve Problem 29 A cylindrical component constructed from an S-590 alloy (Figure 9.39) is to be exposed to a tensile load of 10,000 N. What minimum diameter is required for it to have a rupture lifetime of at least 10 h at 730°C? Solution From Figure 9.39, at 730°C the stress corresponding to a rupture lifetime of 10 h is approximately 300 MPa.). In order to determine the minimum diameter, we use the expression used to compute stress, as given in Equation 7.1, which for a cylindrical specimen takes the form:

σ=

F = A0

F

⎛d ⎞ π⎜ 0 ⎟ ⎝ 2 ⎠

2

6

2

Solving for d0 and realizing that σ = 300 MPa (300 × 10 N/m ) and F = 10,000 N leads to d0 =

=

4F

πσ

(4)(10,000 N) (π )(300 × 106 N/m2 ) = 0.0065 m = 6.5 mm

Reserve Problem 30 (a) Using Figure 9.39, compute the rupture lifetime for an S-590 alloy that is exposed to a tensile stress of 100 MPa at 925°C.

(b) Compare this value to the one determined from the Larson-Miller plot of Figure 9.41, which is for this same S-590 alloy.

Solution (a) From Figure 9.39 using the 925°C the rupture lifetime at 100 MPa is about 10 h.

(b) Using Figure 9.41, at a stress of 100 MPa, the value of the Larson-Miller parameter is about m = 3 25.0 × 10 , which is equal to T(20 + log tr) for T in K and tr in h. We take the temperature to be 925°C + 273 = 1198 K. Therefore, we may write the following: 25.0 × 103 = (1198 K)(20 + log tr ) Or

25.0 × 103 = 20.87 = 20 + log tr 1198 K Which leads to log tr = 20.87 − 20 = 0.87 And, solving for tr we obtain tr = 10 0.87 h = 7.4 h

Which value is relatively close to the 10 h that was obtained using Figure 9.39.

Reserve Questions and Problems Solutions Manual CHAPTER 10 PHASE DIAGRAMS

Reserve Problem 01 The mineral olivine is a solid solution of the silicate compounds forsterite (Mg2SiO4) and fayalite (Fe2SiO4).

How many chemical components are there in a sample of olivine? (a) 1 (b) 2 (c) 3 (d) 4

Solution: The components of a system are defined to be the smallest set of independently variable chemical constituents that are necessary to describe the composition of each phase that is present in a system.

Components are chemically distinct. In this case, we only need to specify two things. It is convenient to simply use the compounds themselves as the components, but other options exist. For instance, specifying the amount of Mg and Fe in the system automatically fixes the Si and O content, assuming the compounds are present in their stoichiometric ratios (assuming they are defect free).

Reserve Problem 02a (question pool) Which of the following depictions is consistent with a single-component, single-phase 2D crystalline system? (a)

(b)

(c)

(d)

(e)

(f)

Solution: (f) is correct since only one color of atom is present (implying single-component), and only one crystal structure is depicted (simple square 2d lattice). Note that there are two crystallites separated by a grain boundary, so the system is polycrystalline and therefore not multiphase (since the crystal structure is the same in each grain).

Reserve Problem 03 Seawater, which covers the majority of the earth, is composed primarily of molecules of H2O and + equal numbers of Na ions and Cl ions. Suppose we have a thoroughly mixed solution (containing these species only) at 25°C. How many components and how many phases are in such a system? (a) 1 component, 1 phase (b) 1 component, 2 phase (c) 1 component, 3 phase (d) 1 component, 4 phase (e) 2 component, 1 phase (f) 2 component, 2 phase (g) 2 component, 3 phase (h) 2 component, 4 phase (i) 3 component, 1 phase (j) 3 component, 2 phase (k) 3 component, 3 phase

(l) 3 component, 4 phase (m) 4 component, 1 phase (n) 4 component, 2 phase (o) 4 component, 3 phase (p) 4 component, 4 phase

Solution: The components of a system are defined to be the smallest set of independently variable chemical constituents that are necessary to describe the composition of each phase that is present in a system.

Components are chemically distinct. In this case, we only need to specify two things. It is convenient to simply use the compounds H2O and NaCl themselves as the components, but other options exist. + For instance, specifying the amount of Na ions and H atoms in the system automatically fixes the Cl ion and O atom content, assuming the compounds are present in their stoichiometric ratios (assuming they are defect free).

It is worth noting that even though NaCl dissolves into its constituent ions in the water, the ions are present in the same ratio as their parent compound. In other words, the Na and Cl ion concentrations are coupled and not independent. If they were independent, the system would be described as three components, since knowing one ion concentration would not inherently imply the value of the other ion’s concentration.

Reserve Problem 04 Select all of the two-phase material systems from the list below. Select all that apply. (a) Solid ductile cast iron (ferrite solid solution + embedded graphite spheres) (b) Solid sodium chloride (salt, NaCl) (c) Liquid bronze (Cu + Sn liquid solution) (d) Solid gray cast iron (ferrite solid solution + embedded graphite flakes) (e) Solid aluminum featuring dissolved silicon (f) Solid lead-tin solder (a mixture of Pb-rich and Sn-rich solid solutions) (g) Partially melted aluminum (h) Frozen water with trapped air bubbles (i) Epoxy embedded with carbon fibers

Solution: (a), (d), (f), (h), (i) are two-phase systems because there is a physical boundary beyond the particle level (atom, ions, molecules, etc.) that separates chemically & structurally distinct volumes. In the case of (g), the physical boundary merely separates volumes that are structurally distinct (crystalline and liquid). (b) is a single-phase compound, not a solution. The ratio of the two ions is in a fixed stoichiometry and they adopt an orderly arrangement in the crystal rather than random substitution or random placement in interstitial sites. (c) is a liquid solution, and solutions – themselves – are single-phase by definition. (e) is a solid solution since silicon is implied to substitute for aluminum. Solutions are single-phase by definition.

Reserve Problem 05 Complete the following statements regarding conditions that must be satisfied in order for a solid solution to exhibit extensive solubility.

The solute and host species must have a very [w] sizes.

The solute and host species must attempt to pack with [x] crystal structure.

The solute and host species must feature [y] valence electron configuration.

The solute and host species must feature [z] ability to attract electrons (electronegativity). w = similar, different x = a similar (or the same), a different y = a similar (or the same), a different z = a similar, a different

Solution: The Hume-Rothery rules state that two species must have similar size, the same (or similar) crystal structure, the same (or similar) valence, and a similar electronegativity in order to have extensive solid-solubility.

Reserve Question 06: Solvent/solute For a solution, which of the following is present in the higher concentration? a) Solvent b) Solute Solution Solvent. By definition, solvent is the element/compound that is present in a solution in the greatest amount.

Reserve Problem 07 Consider a specimen of ice that is at −10°C and 1 atm pressure. Using Figure 10.2, the pressure– temperature phase diagram for H2O, determine the pressure to which the specimen must be raised or lowered to cause it (a) to melt, and (b) to sublime.

Solution The figure below shows the pressure-temperature phase diagram for H2O, Figure 10.2; a vertical line has been constructed at –10°C, and the location on this line at 1 atm pressure (point B) is also noted.

(a) Melting occurs, (by changing pressure) as, moving vertically (upward) at this temperature, we cross the Ice-Liquid phase boundary. This occurs at approximately 570 atm; thus, the pressure of the specimen must be raised from 1 to 570 atm. (b) In order to determine the pressure at which sublimation occurs at this temperature, we move vertically downward from 1 atm until we cross the Ice-Vapor phase boundary. This intersection occurs at approximately 0.0023 atm.

Reserve Problem 08 At a pressure of 0.01 atm, determine (a) the melting temperature for ice, and (b) the boiling temperature for water.

Solution The melting temperature for ice and the boiling temperature for water at a pressure of 0.01 atm may be determined from the pressure-temperature diagram for this system, Figure 10.2, which is shown below; a horizontal line has been constructed across this diagram at a pressure of 0.01 atm.

The melting and boiling temperatures for ice at a pressure of 0.01 atm may be determined by moving horizontally across the pressure-temperature diagram at this pressure. The temperature corresponding to the intersection of the Ice-Liquid phase boundary is the melting temperature, which is approximately 1°C. On the other hand, the boiling temperature is at the intersection of the horizontal line with the Liquid-Vapor phase boundary--approximately 16°C. Reserve Question 09: Liquidus line A liquidus line separates which of the following combinations of phase fields? a) b) c) d)

Liquid and Liquid + α α and Liquid + α α and α+ β Liquid + α and α+ β

Solution A liquidus line separates Liquid and Liquid + α phase fields.

Reserve Question 10: Solvus line A solvus line separates which of the following pairs of phase fields? a) b) c) d)

Liquid and Liquid + α α and Liquid + α α and α+ β Liquid + α and α+ β

Solution A solvus line separates α and α + β phase fields.

Reserve Problem 11 How many kilograms of nickel must be added to 5.66ks of copper to yield a liquidus temperature of 1200°C? Use Animated Figure 10.3a.

Solution According to the copper-nickel phase diagram (Figure 10.3a), the composition at the 1200°C liquidus temperature is about 20 wt% Ni. In order to determine the amount of nickel that must be added to 5.66 kg Cu, it is necessary to use Equation 5.6a. If we let mNi represent the mass of nickel required, then Equation 5.6a takes the form

20 wt% Ni =

mNi × 100 5.66 kg + mNi

And solving this expression for mNi leads to mNi = 1.42 kg.

Reserve Question 12: Information on phase diagrams Which of the following kinds of information may be determined with the aid of a phase diagram? a) The phase(s) present at a specified temperature and composition. b) The composition(s) of phase(s) present at a specified temperature and composition. c) The fraction(s) of phase(s) present at specified temperature and composition.

Solution With the aid of a phase diagram the following may be determined: • • •

The phase(s) present at a specified temperature and composition.

The composition(s) of phase(s) present at a specified temperature and composition.

The fraction(s) of phase(s) present at a specified temperature and composition.

Reserve Problem 13: Pb-Sn phase diagram – composition In the Animated Figure 10.8 is shown the lead-tin phase diagram. For an alloy of composition 25 wt% Sn – 75 wt% Pb, select the phase(s) present and their composition(s) for each of the temperatures cited.

(a) 300°C? • • • •

L = 25.0 wt% Sn – 75.0 wt% Pb L = 25.0 wt% Sn – 75.0 wt% Pb; α = 10.0 wt% Sn – 90.0 wt% Pb L = 10.0 wt% Sn – 90.0 wt% Pb; α = 10.0 wt% Sn – 90.0 wt% Pb L = 90.0 wt% Sn – 10.0 wt% Pb; α = 10.0 wt% Sn – 90.0 wt% Pb

(b) 200°C? • • • •

α = 17.0 wt% Sn – 83.0 wt% Pb; L = 55.7 wt% Sn – 44.3 wt% Pb α = 17.0 wt% Sn – 83.0 wt% Pb; β = 55.7 wt% Sn – 44.3 wt% Pb α = 18.3 wt% Sn – 81.7 wt% Pb; β = 97.8 wt% Sn – 2.2 wt% Pb L = 25.0 wt% Sn – 75.0 wt% Pb; α = 25.0 wt% Sn – 75.0 wt% Pb

(c) 183°C? • • • •

α = 25.0 wt% Sn – 75.0 wt% Pb; β = 25.0 wt% Sn – 75.0 wt% Pb L = 25.0 wt% Sn – 75.0 wt% Pb; α = 25.0 wt% Sn – 75.0 wt% Pb; β = 25.0 wt% Sn – 75.0 wt% Pb α = 18.3 wt% Sn – 81.7 wt% Pb; β = 97.8 wt% Sn – 2.2 wt% Pb L = 61.9 wt% Sn – 38.1 wt% Pb; α = 18.3 wt% Sn – 81.7 wt% Pb; β = 97.8 wt% Sn – 2.2 wt% Pb

(d) 100°C? • • • •

α = 25.0 wt% Sn – 75.0 wt% Pb; β = 25.0 wt% Sn – 75.0 wt% Pb L = 25.0 wt% Sn – 75.0 wt% Pb; α = 25.0 wt% Sn – 75.0 wt% Pb; β = 25.0 wt% Sn – 75.0 wt% Pb α = 5.1 wt% Sn – 94.9 wt% Pb; β = 98.7 wt% Sn – 1.3 wt% Pb L = 25.0 wt% Sn – 75.0 wt% Pb; α = 5.1 wt% Sn – 94.9 wt% Pb; β = 98.7 wt% Sn – 1.3 wt% Pb

Solution

(a) 300°C The answers for this problem are determined using the horizontal and vertical cursor lines in Animated Figure 10.8. (a) At 300°C only the liquid phase is present. Its composition is the same as that of the alloy (i.e., 25 wt% Sn – 75 wt% Pb).

(b) 200°C The answers for this problem are determined using the horizontal and vertical cursor lines in Animated Figure 10.8. (b) At 200°C the α and liquid phases are present. A tie line drawn across the α + L phase field intersects the α – α + L phase boundary at about 17.0 wt% Sn – 83.0 wt% Pb (which is the composition of the α phase) and the α + L – L phase boundary at about 55.7 wt% Sn – 44.3 wt% Pb (which is the composition of the liquid phase)

(c) 183°C The answers for this problem are determined using the horizontal and vertical cursor lines in Animated Figure 10.8. (c) At 183°C, which is the eutectic temperature, all three phases are present; the liquid phase has a composition of 61.9 wt% Sn – 38.1 wt% Pb, the α phase has a composition of 18.3 wt% Sn – 81.7 wt% Pb, while the β phase has a composition of 97.8 wt% Sn – 2.2 wt% Pb.

(d) 100°C The answers for this problem are determined using the horizontal and vertical cursor lines in Animated Figure 10.8. (d) At 100°C the α and β phases are present. A tie line drawn across the α + β phase field intersects the α – α + β phase boundary at about 5.1 wt% Sn – 94.9 wt% Pb (which is the composition of the α phase) and the α + β – β phase boundary at about 98.7 wt% Sn – 1.3 wt% Pb (which is the composition of the β phase).

Reserve Problem 14: Pb- Sn phase diagram – Phases In Animated Figure 10.8 is shown the lead-tin phase diagram. Using this diagram determine which of the following phase(s)/phase combination(s) (alpha, beta, liquid, alpha + liquid, beta + liquid, alpha + beta, and alpha + beta + liquid) will be present for an alloy of composition 61.9 wt% Sn – 38.1 wt% Pb that is at equilibrium at 185°C? Solution Using the cursor lines in Animated Figure 10.8 we note that the point corresponding to a composition 61.9 wt% Sn – 38.1 wt% Pb at equilibrium at 183°C is the eutectic point on the phase diagram—therefore, all three phases, alpha + beta + liquid phases will be present.

Reserve Question 15: Phases present at eutectic point At a eutectic point on a binary temperature-composition phase diagrams, how many phases are present when the system is at equilibrium? a) b) c) d)

0 1 2 3

Solution At a eutectic point on a binary phase diagram, three phases (Liquid, α, and β) are present when the system is at equilibrium.

Reserve Problem 16 A 45 wt% Pb–55 wt% Mg alloy is rapidly quenched to room temperature from an elevated temperature in such a way that the high-temperature microstructure is preserved. This microstructure is found to consist of the phase and Mg2Pb, having respective mass fractions of 0.65 and 0.35. Determine the approximate temperature from which the alloy was quenched.

Solution We are asked to determine the approximate temperature from which a 45 wt% Pb-55 wt% Mg alloy was quenched, given the mass fractions of α and Mg2Pb phases. We can write a lever-rule expression for the mass fraction of the α phase as Wα = 0.65 =

CMg Pb − C0 2

CMg Pb − Cα 2

The value of C0 is stated as 45 wt% Pb-55 wt% Mg, and CMg pb is 81 wt% Pb-19 wt% Mg, which is 2

independent of temperature (Figure 10.20); thus,

0.65 =

81 − 45 81 − Cα

which yields Cα = 25.6 wt% Pb The temperature at which the α–(α+ Mg2Pb) phase boundary (Figure 10.20) has a value of 25.6 wt% Pb is about 360°C (680°F).

Reserve Problem 17 Two intermetallic compounds, AB and AB2, exist for elements A and B. If the compositions for AB and AB2 are 34.3 wt% A–65.7 wt% B and 20.7 wt% A–79.3 wt% B, respectively, and element A is potassium, identify element B. • • • •

Phosphorus Sulfur Arsenic Antimony

Solution This problem gives us the compositions in weight percent for the two intermetallic compounds AB and AB2, and then asks us to identify element B if element A is potassium. Probably the easiest way to solve this problem is to first compute the ratio of the atomic weights of these two elements using Equation 5.9a; then, since we know the atomic weight of potassium (39.10 g/mol, per inside the front cover), it is possible to determine the atomic weight of element B, from which an identification may be made.

First of all, consider the AB intermetallic compound; inasmuch as it contains the same numbers of A and B atoms, its composition in atomic percent is 50 at% A-50 at% B. Equation 5.9a may be written in the form: CB′ =

CB AA × 100 CA AB + CB AA

where AA and AB are the atomic weights for elements A and B, and CA and CB are their compositions in weight percent. For this AB compound, and making the appropriate substitutions in the above equation leads to

50 at% B =

(65.7 wt% B)( AA ) (34.3 wt% A)( AB ) + (65.7 wt% B)( AA )

×100

Now, solving this expression yields, AB = 1.916 AA Since potassium is element A and it has an atomic weight of 39.10 g/mol, the atomic weight of element B is just AB = (1.916)(39.10 g/mol) = 74.92 g/mol Upon consultation of the period table of the elements (Figure 2.8) we note the element that has an atomic weight closest to this value is arsenic (74.92 g/mol). Therefore, element B is arsenic, and the two intermetallic compounds are KAs and KAs2.

Reserve Problem 18 An intermetallic compound is found in the magnesium-gallium system that has a composition of 41.1 wt% Mg – 58.9 wt% Ga. Specify the formula for this compound.

•

MgGa

• • •

Mg2Ga MgGa2 Mg3Ga2

Solution Probably the easiest way to solve this problem is to begin by computing concentrations of Mg and Ga in terms of atom percents; from these values, it is possible to determine the ratio of Mg to Ga in this

′ and CGa ′ ) are computed using Equation 5.9 as follows (incorporating compound. Atom percent (CMg atomic weight values of 24.31 g/mol and 69.72 g/mol, respectively for Mg and Ga):

′ = CMg =

CMg AGa CMg AGa + CGa AMg

×100

(41.1 wt%)(69.72 g/mol) × 100 (41.1 wt%)(69.72 g/mol) + (58.9 wt%)(24.31 g/mol)

= 67.7 at%

′ will equal 33.3 at%.

Because this intermetallic compound is composed of only Mg and Ga, CGa Finally, the formula for this intermetallic compound is Mg2Ga, since the magnesium-to-gallium atom percent ratio is 2:1.

Reserve Problem 19 Specify the liquidus, solidus, and solvus temperatures for the following alloy: 1.5 wt% C – 98.5 wt% Fe. If no temperature, plese put in “0.0”.

Liquidus temperature Solidus temperature Solvus temperature

____ ____ ____

Solution: Definitions of liquidus, solidus, and solvus lines are as follows: Liquidus—boundary between liquid and (liquid + solid) phase regions Solidus—boundary between solid and (liquid + solid) phase regions Solvus—boundary between a single solid phase region and a two solid phase region.

For this problem, a liquidus, solidus, or solvus temperature corresponds to the intersection between a vertical line (constructed at a specific composition) with the respective liquidus, solidus, or solvus line.

The liquidus temperature for 1.5 wt% C-98.5 wt% Fe are 950°C, 1260°C, and 1420°C, respectively as noted on the Fe-Fe3C phase diagram shown below.

Reserve Question 20: Eutectoid reaction A eutectoid reaction involves which of the following phases? a) b) c) d)

One liquid and one solid One liquid and two solid Two liquids and one solid Three solid

Solution A eutectoid reaction involves three solid phases.

Reserve Question 21: Peritectic reaction A peritectic reaction involves which of the following combinations of phase fields? a) b) c) d)

One liquid and one solid One liquid and two solid Two liquids and one solid Three solid

Solution A peritectic reaction involves one liquid and two solid phases.

Reserve Question 22: Congruent transformations For a congruent phase transformations there are a) no composition alterations. b) compositional alterations.

Solution There are no compositional alterations for a congruent phase transformation.

Reserve Problem 23 Specify the number of degrees of freedom for the following alloys: (a) 95 wt% Ag-5 wt% Cu at 780°C (b) 80 wt% Ni-20 wt% Cu at 1400°C (c) 44.9 wt% Ti-55.1 wt% Ni at 1310°C (d) 61.9 wt% Sn-38.1 wt% Pb at 183°C (e) 2.5 wt% C-97.5 wt% Fe at 1000°C Solution We are asked to specify the value of F for Gibbs phase rule for several alloys and, for each, at a specified temperature. Gibbs phase rule in general form is P+F=C+N For this system, the number of components, C, is 2, whereas N, the number of noncompositional variables, is 1—viz. temperature. Thus, the phase rule now becomes P+F=2+1=3 Or F=3–P

(10.17a)

where P is the number of phases present at equilibrium.

Therefore, for each of these alloys we need to ascertain the number of phases present at equilibrium at the specified temperature. This gives us the value of P in the preceding equation, from which it is possible to determine the value of F, the number of degrees of freedom.

(a) A 95 wt% Ag-5 wt% Cu alloy at 780°C is entirely the β phase, which means that only a single phase is present—i.e., P = 1; insertion of this value into Equation 10.17a yields F=3–P=3–1=2 (b) The location of an 80 wt% Ni-20 wt% Cu alloy at 1400°C on the Ni-Cu phase diagram is in the α + L phase region, which means that two phases (L and α) are in equilibrium and the value of P = 2; or that, from Equation 10.17a F=3–P=3–2=1 (c) The location of a 44.9 wt% Ti-55.1 wt% Ni alloy at 1310°C on the Ti-Ni phase diagram (Figure 10.22) is at a congruent melting point, which means that two phases (γ and liquid) are in equilibrium and the value of P = 2; or that, from Equation 10.17a F=3–P=3–2=1 (d) The location of a 61.9 wt% Sn-38.1 wt% Pb alloy at 183°C is at the eutectic point, which means that three phases (α, β, and liquid) are in equilibrium and that the value of P = 3; or that, from Equation 10.17a F=3–P=3–3=0 (e) The location of a 2.5 wt% C-97.5 wt% Fe alloy at 1000°C on the Fe-Fe3C phase diagram is in the γ + Fe3C phase region, or that two phases are present—i.e., P = 2. This means that Equation 10.17a is of the form F=3–P=3–2=1 Reserve Problem 24 The mass fractions of total ferrite and total cementite in an iron–carbon alloy are 0.88 and 0.12, respectively. Is this a hypoeutectoid or hypereutectoid alloy?

Solution In this problem we are given values of Wα and WFe C for an iron-carbon alloy (0.88 and 0.12, 3

respectively), and then are asked to specify whether the alloy is hypoeutectoid or hypereutectoid.

Employment of the lever rule for total α leads to

Wα = 0.88 =

CFe C − C0 3

CFe C − Cα 3

=

6.70 − C0 6.70 − 0.022

Now, solving for C0, the alloy composition, leads to C0 = 0.82 wt% C. Therefore, the alloy is hypereutectoid since C0 is greater than 0.76 wt% C.

Reserve Problem 25 A steel alloy is known to contain 93.8 wt% Fe, 6.0 wt% Ni, and 0.2 wt% C. Assume that there are no alterations in the positions of other phase boundaries with the addition of Ni. (a) What is the approximate eutectoid temperature of this alloy? (b) What is the proeutectoid phase when this alloy is cooled at temperature below the eutectoid? (c) Compute the relative amounts of the proeutectoid phase and pearlite.

Solution (a) From Figure 10.38, the eutectoid temperature for 6.0 wt% Ni is approximately 650°C (1200°F). (b) From Figure 10.39, the eutectoid composition is approximately 0.62 wt% C. Since the carbon concentration in the alloy (0.2 wt%) is less than the eutectoid (0.62 wt% C), the proeutectoid phase is ferrite. (c) Assume that the α–(α + Fe3C) phase boundary is at a negligible carbon concentration.

Modifying Equation 10.21 leads to

Wα ′ =

0.62 − C0′ 0.62 − 0

=

0.62 − 0.20 0.62

= 0.68

Likewise, using a modified Equation 10.20

Wp =

C0′ − 0 0.62 − 0

=

0.20 = 0.32 0.62

Reserve Questions and Problems Solutions Manual CHAPTER 11 PHASE TRANSFORMATIONS

Reserve Problem 01 It is known that the kinetics of some transformation obeys the Avrami equation and that the value of –8 k is 6.0 × 10 (for time in minutes). If the fraction transformed is 0.75 after 200 min, determine the rate of this transformation.

Solution First of all, we note that the rate of a reaction is defined by Equation 11.18 as rate =

1 t0.5

For this problem it is necessary to compute the value of t0.5, the time it takes for the reaction to progress to 50% completion—or for the fraction of reaction y to equal 0.50. Furthermore, we may determine t0.5 using the Avrami equation, Equation 11.17:

y = 1 − exp(−kt n )

(11.17)

The problem statement provides us with the value of y (0.75) at some time t (200 min), and also the –8 value of k (6.0 × 10 ) from which data it is possible to compute the value of the constant n. In order to perform this calculation some algebraic manipulation of Equation 11.17 is necessary. First of all, we rearrange this expression as follows:

exp (− kt n ) = 1 − y We then take natural logarithms of both sides, which leads to:

− kt n = ln(1 − y)

(11.17a)

Now solving for n requires further algebraic manipulation. Performing another rearrangement yields

tn =

ln (1 − y ) −k

And taking natural logarithms of both sides of this equation results in the following expression

⎛ ln (1 − y ) ⎞ n ln t = ln ⎜ ⎟ ⎜ −k ⎟ ⎝ ⎠

From which we solve for n as follows ⎡ ln (1 − y ) ⎤ ln ⎢ ⎥ ⎢⎣ − k ⎥⎦ n= ln t And incorporating values cited above for y, k, and t yields

⎡ ln (1 − 0.75) ⎤ ⎥ ln ⎢ ⎢⎣ − (6.0 × 10 −8 ) ⎥⎦ n= = 3.20 ln(200 min) At this point we want to compute t0.5, the value of t for y = 0.5, which means that it is necessary to establish a form of Equation 11.17 in which t is the dependent variable. This is accomplished using a rearranged form of Equation 11.17a as

tn = −

ln (1 − y ) k

From which we solve for t:

⎡ ln (1 − y ) ⎤ t = ⎢− ⎥ k ⎣⎢ ⎦⎥

1/n

And for t = t0.5, this equation becomes

⎡ ln (1 − 0.5) ⎤ t0.5 = ⎢ − ⎥ k ⎢⎣ ⎥⎦

1/ n

Now incorporation of the value of n determined above, as well as the value of k cited in the problem –8 statement (6.0 × 10 ), we calculate t0.5 as follows:

⎡ ln (1 − 0.5) ⎤ ⎥ t0.5 = ⎢ − ⎢ 6.0 × 10 −8 ⎥ ⎣ ⎦

1/3.20

= 161 min

And, finally, from Equation 11.18, the rate is equal to rate =

1 t0.5

=

1 = 6.21 ×10 −3 (min) −1 161 min

Reserve Question 02: Types of phase transformations Match each transformation description below with name of a transformation of this type.

Diffusion-dependent with change(s) in phase composition(s) • Martensitic transformation • Eutectoid reaction • Recrystallization Diffusionless • Recrystallization • Martensitic transformation • Eutectoid reaction Diffusion-dependent with non change(s) in phase composition(s) • Eutectoid reaction • Recrystallization • Martensitic transformation Solution The martensitic transformation is diffusionless.

Recrystallization is diffusion-dependent wherein there is no change in phase composition.

The eutectoid reaction is diffusion-dependent wherein there are changes in phase compositions.

Reserve Question 03: Equilibrium in solid Equilibrium structures are commonly achieved in solid systems.

a) True b) False Solution False. The rate of approach to equilibrium for most solid systems is so slow that true equilibrium structures are rarely achieved.

Reserve Question 04: Heat treatment Below is shown the isothermal transformation diagram for a 0.45 wt% C iron-carbon alloy. List the microconstituent(s) present for the heat treatment labeled (a) on this diagram. It is not necessary to state the proportion(s) of the microconstituents.

Solution

Heat treatment (a): Upon cooling to 725°C, the microstructure remains 100% austenite. While being 3 held at this temperature for about 10 seconds, proeutectoid ferrite begins to form. As it is rapidly cooled to room temperature, the remaining austenite begins to transform to martensite at about 300°C; all of this austenite will have transformed martensite at the end of this quench. Furthermore, there are no changes with the proeutectoid ferrite that has already formed; thus the final composition will consist of proeutectoid ferrite and martensite.

Heat treatment (b): Upon rapid cooling to and while at 100°C C for 20 s, all of the austenite is converted to martensite. During the subsequent rapid heating to about 400°C, no changes occur.

While being held at 400°C for about 104 s, all the martensite will transform to tempered martensite; and during the final cooling to about 100°C no additional changes will occur. Thus the final microconstituent is tempered martensite.

Heat treatment (c): There are no changes in the microstructure during the rapid cooling to 350°C (i.e., only austenite will be present). As the alloy is held at that this temperature for about 10 seconds, approximately 25% of the alloy transforms to bainite. During the rapid heating to 625°C 3 there are no microstructural changes; however, while maintaining the alloy for about 10 seconds at this temperature, all the remaining austenite will transform to proeutectoid ferrite and pearlite.

Since, by this time all of the austenite has transformed, no changes will occur during the final rapid cooling to room temperature. Thus, the final microstructure will consist of bainite, proeutectoid ferrite, and pearlite.

Reserve Problem 05 Suppose that a steel of eutectoid composition is cooled to 550°C (1020°F) from 760°C (1400°F) in less than 0.5 s and held at this temperature. (a) How long will it take for the austenite-topearlite reaction to go to 50% completion? To 100% completion?

50% completion: 100% completion:

____ ____

(b) estimate the hardness of the alloy that has completely transformed to pearlite.

Solution We are called upon to consider the isothermal transformation of an iron-carbon alloy of eutectoid composition.

(a) From Figure 11.23, a horizontal line at 550°C intersects the 50% and reaction completion curves at about 2.5 and 6 seconds, respectively; these are the times asked for in the problem statement. (b) The pearlite formed will be fine pearlite. From Figure 11.31a, the hardness of an alloy of composition 0.76 wt% C that consists of fine pearlite is about 265 HB (27 HRC).

Reserve Question 06: Fe microstructures I Schematic room temperature microstructures for four iron-carbon alloys are shown below. Rank these alloy microstructures (by letter) from most ductile to the least ductile.

a) b) c) d) e) f) g) h)

C, D, A, B B, D, A, C A, B, D, C D, B, C, A C, A, B, D B, A, D, C A, B, C, D D, C, B, A

Solution The following lists the microstructures for these four alloys:

A = hypoeutectoid alloy (proeutectoid ferrite) and fine pearlite B = hypereutectoid alloy (proeutectoid cementite) and coarse pearlite C = spheroidite D = hypereutectoid alloy (proeutectoid cementite) and fine pearlite Alloy C will be the most ductile, because its microstructure is spheroidite, which more ductile than coarse and fine pearlites (Figure 11.31b). The next most ductile alloy is A; its carbon content is lower than B and D (because it is a hypoeutectoid alloy)—as carbon content increases, ductility decreases (Figure 11.31b). The carbon contents of B and D are approximately the same—the proeutectoid phase for both is cementite—however, the pearlite for B is coarse, while for D it is fine; coarse pearlite is more ductile than fine pearlite (Figure 11.31b). Therefore, the ranking is C, A, B, D.

Reserve Questions and Problems Solutions Manual CHAPTER 12 ELECTRICAL PROPERTIES

Reserve Problem 01 Select the word combination that best completes this statement.

Electric current is a measure of [a] _____ passing per unit [b]_____, and the base units of current are [c]____ per [d]____, otherwise known as an [e]_____.

• • • • • • • • •

Time Ampere Second Charge Voltage Volts Resistance Ohm Coulomb

Solution: [a] charge [b] time [c] coulomb [d] second [e] ampere

Reserve Problem 02a (question pool) The current-voltage behavior of specimens [a] - [d] is depicted in the figure below.

Which of these specimens features the lowest resistance?

Solution: The specimen that passes the most current for a given voltage (largest I for a given V) will feature the lowest current.

According to Ohm’s law, the slope of these lines is the reciprocal of resistance. Therefore, the specimen associated with the largest slope will feature the lowest resistance.

Reserve Question 03: Conductivities Match each of the conductivity values/ranges with its associated class of materials.

–20

–10

7

–1

10

–1

to 10 (Ω-m) • Semiconductors • Metals • Insulators

10 (Ω-m) • Metals • Insulators • Semiconductors –6

4

–1

10 to 10 (Ω-m) • Insulators • Semiconductors • Metals

Solution 7

–1

Metals have conductivities on the order of 10 (Ω-m) .

–6

4

–1

Semiconductors have conductivities in the range of 10 to 10 (Ω-m) .

Insulators have conductivities in the range of 10

–20

to 10

–10

–1

(Ω-m) .

Reserve Question 04: Band structures Match each of the following energy band structures with the type of material it represents.

A

• • •

Semiconductor Metal Insulator

• • •

Metal Insulator Semiconductor

B

C

• • •

Insulator Semiconductor Metal

Solution An insulator has a relatively wide energy band gap, so A represents an insulator. A semiconductor has a relatively narrow energy band gap, so B represents a semiconductor. A metal has empty states above and adjacent to filled states, so C represents a metal.

Reserve Question 05: Electrons for conductivity All electrons present in a material are available to participate in conduction process. a) True b) False Solution False. The number of electrons available for electrical conduction in a particular material is related to the arrangement of electron states or levels with respect to energy and the manner in which these states are occupied by electrons.

Reserve Question 06: Energy name ____ energy is the name of energy corresponding to the highest filled electron state at 0 K. Solution The energy corresponding to the highest filled electron energy state at 0 K is called the Fermi energy.

Reserve Question 07: Free electrons 01 Are energies for electrons that participate in the conduction process (i.e., free electrons) greater or less than the Fermi energy? a) Greater than b) Less than Solution Energies of electrons that participate in the conduction process (i.e., free electrons), are greater than the Fermi energy.

Reserve Question 08: Holes Are energies for holes greater or less than the Fermi energy? a) Greater than b) Less than Solution Energies of holes are less than the Fermi energy.

Reserve Question 09: Insulator band gap An insulator has an energy band gap that is relatively a) wide. b) narrow.

Solution An insulator has a relatively wide energy band gap; thus, it is difficult for electrons to be promoted from the valence band to the conduction band.

Reserve Question 10: P band For a material that contains N atoms, how many electron states will there be in a p band?

Solution There will be 3N electron states in a p band.

Reserve Question 11: S band For a material that contains N atoms, how many electron states will there be in an s band?

Solution There will be N electron states in an s band.

Reserve Question 12: Semiconductor band gap A semiconductor has an energy band gap that is relatively a) wide b) narrow Solution A semiconductor has a relatively narrow energy band gap; thus, it is relatively easy for electrons to be promoted from the valence band to the conduction band.

Reserve Question 13: Free electrons in metals Electrons in metals do not require any excitation before becoming conduction electrons that are free. a) True b) False Solution False. Electrons in metals require some excitation before becoming conduction electrons that are free.

Reserve Question 14: Temperature vs band gap For nonmetallic materials, which of the following is true? a) The wider the band gap, the lower the electrical conductivity at a given temperature. b) The wider the band gap, the higher the electrical conductivity at a given temperature.

Solution The wider the band gap, the lower the electrical conductivity at a given temperature.

Reserve Question 15: Current in perfect crystal In a perfect crystal (i.e., without any imperfections), application of an electric field will result in a continuously increasing electric current with time. a) True b) False

Solution True. In a perfect crystal, application of an electric field will result in a continuously increasing electric current with time.

Reserve Question 16: Electric field induced electron movement When an electric field is applied, in which direction are the free electrons accelerated? a) Opposite to the direction of the electric field b) In the same direction as the electric field.

Solution When an electric field is applied, the free electrons are accelerated in the direction opposite to that of the field.

Reserve Question 17: Ion mobility At temperatures between 775°C and 1100°C, the activation energy and preexponential for the 2+ –7 2 diffusion coefficient of A in the metal oxide, AO, are 110 kJ/mol and 7.1 × 10 m /s, respectively.

2 2+ Compute the mobility [in m /(V-s)] of A at 908°C. Solution This solution provides a guideline to solving the problem, not the numerical answer.

The mobility of an ion may be calculated using the following equation:

μi =

ni eDi kT

However, the diffusion coefficient is function of temperature as

⎛ Q ⎞ D = D0 exp ⎜ − d ⎟ ⎝ RT ⎠ The value of μi may be determined by incorporation of the second equation (for D) into the first equation.

Reserve Problem 18: Calculating resistivity contributions Estimate the electrical resistivity (in Ω-m) of metal Y containing 1.3 at% of metal Z at 138°C, given the following data:

ρ0

a

A

(Ω-m)

(Ω-m/°C)

(Ω-m)

1.5 × 10

–7

6.7 × 10

–11

1.2 × 10

–6

Assume that all of the 1.3 at% of metal Z goes into solid solution in metal Y. Use scientific notation.

Solution This solution provides a guideline to solving the problem, not the numerical answer.

The total resistivity is equal to the sum of thermal, impurity, and deformation resistivity contributions as ρTotal = ρt + ρi + ρd Since there is no deformation of the material, the ρd term may be ignored. The thermal and impurity contributions to the resistivity may be determined using the following two equations ρt = ρ0 + aT ρi = Aci (1 − ci ) Concentration in the latter equation is expressed as atom fraction. And, from the first equation cited, the total resistivity is the sum of these two resistivity contributions.

Reserve Problem 19: Multi-component conductivity Some two-phase metal alloy is known to be composed of α and β phases; mass fractions of these phases are 0.74 and 0.26, respectively. Room-temperature electrical resistivity and density data for these phases are tabulated below. Using this information, calculate the electrical resistivity (in Ω-m) of the alloy at room temperature (use scientific notation).

Phase

Resistivity (Ω-m)

α

1.7 × 10

β

6.7 × 10

2

Density (g/cm )

–7

8.31

–8

8.18

Solution This solution provides a guideline to solving the problem, not the numerical answer.

For a two-phase alloy, the following rule-of-mixtures expression may be utilized to approximate the resistivity of the alloy:

ρi = ρα Vα + ρβ Vβ

However, since volume fractions are required, it is first necessary to convert the mass fractions into phase volume fractions using the following equation, and the phase mass fraction and density values provided in the problem statement.

Wi density i Vi = Wj Wi + density i density j

And, finally, the alloy resistivity is determined using the first equation above, incorporating the ρα and ρβ values given in the problem statement.

Reserve Question 20: Resistivity contributions Which of the following have a significant influence on a material’s electrical resistivity?

a) b) c) d) e)

impurity concentration temperature grain size cold work vacancy concentration

Solution A material’s electrical resistivity will depend on the following: • Impurity concentration • Vacancy concentration • Temperature • Cold work Reserve Problem 21

(a) Using the data in Figure 12.8, determine the values of ρ0 and a from Equation 12.10 for pure copper. Take the temperature T to be in degrees Celsius. (b) Determine the value of A in Equation 12.11 for nickel as an impurity in copper, using the data in Figure 12.8. (c) Using the results of parts (a) and (b), estimate the electrical resistivity of copper containing 1.75 at% Ni at 100°C.

Solution

(a) Perhaps the easiest way to determine the values of ρ0 and a in Equation 12.10 for “pure copper” in Figure 12.8, is to set up two simultaneous equations using two resistivity values (labeled ρt1 and ρt2) taken at two corresponding temperatures (T1 and T2). Thus,

ρt1 = ρ0 + aT1 ρt 2 = ρ0 + aT2 And solving these equations simultaneously lead to the following expressions for a and ρ0:

a=

ρ t 1 − ρt 2 T1 − T2

⎡ρ − ρ ⎤ t2 ⎥ ρ0 = ρt1 − T1 ⎢⎢ t1 ⎥ ⎢⎣

T1 − T2

⎥⎦

⎡ρ

⎤ t 1 − ρt 2 ⎥ ⎥ ⎢⎣ T1 − T2 ⎥⎦

= ρt 2 − T2 ⎢⎢

–8

From Figure 12.8, let us take T1 = –150°C, T2 = –50°C, which gives ρt1 = 0.6 × 10 (Ω-m), and ρt2 = –8 1.25 × 10 (Ω-m). Therefore

a=

ρ t 1 − ρt 2 T1 − T2

⎡(0.6 × 10 −8 ) − (1.25 × 10 −8 ) ⎤ ( Ω-m ) ⎦ = ⎣ −150°C − ( − 50°C) = 6.5 × 10

–11

(Ω-m)/°C

and ⎡ρ − ρ ⎤ t2 ⎥ ρ0 = ρt1 − T1 ⎢⎢ t1 ⎥ ⎢⎣

T1 − T2

⎥⎦

⎡(0.6 × 10 −8 ) − (1.25 × 10 −8 ) ⎤ ( Ω-m ) ⎦ = (0.6 × 10 ) − ( − 150) ⎣ −150°C − ( − 50°C) −8

–8

= 1.58 × 10 (Ω-m) (b) For this part of the problem, we want to calculate A from Equation 12.11

ρi = Aci (1 − ci )

In Figure 12.8, curves are plotted for three ci values (0.0112, 0.0216, and 0.0332). Let us find A for each of these ci’s by taking a ρtotal from each curve at some temperature (say 0°C) and then –8 subtracting out ρi for pure copper at this same temperature (which is 1.7 × 10 Ω-m). Below are tabulated values of A determined from these three ci values, and other data that were used in the computations. ρtotal (Ω-m)

1 – ci

ci 0.0112

0.989

3.0 × 10

0.0216

0.978

4.2 × 10

0.0332

0.967

5.5 × 10

ρ (Ω-m)

–8

1.3 × 10

–8

2.5 × 10

–8

3.8 × 10

–8 –8 –8

A (Ω-m) 1.17 × 10 1.18 × 10 1.18 × 10

–6 –6 –6

–6

The average of these three A values is 1.18 × 10 (Ω-m). (c) We use the results of parts (a) and (b) to estimate the electrical resistivity of copper containing 1.75 at% Ni (ci = 0.0175) at 100°C. The total resistivity is just

ρtotal = ρt + ρi And incorporating the expressions for ρt and ρi from Equations 18.10 and 18.11, and the values of ρ0, a, and A determined above, leads to

ρtotal = (ρ0 + aT )+ Aci (1 − ci ) = {1.58 × 10 −8 (Ω-m) + [6.5 × 10 + { [1.18 × 10

−6

−11

(Ω-m) / °C](100°C)}

(Ω-m)](0.0175) (1 − 0.0175)} –8

= 4.25 × 10 (Ω-m) Reserve Problem 22 Conductivity in a metal is almost always reduced by the introduction of defects into the lattice.

The factor primarily affected by defects is: [a] free electron concentration [b] electron charge [c] electron mobility [d] electron spin

Solution:

The conductivity of a metal is given as the product of free electron concentration, charge of one 29 electron, and electron mobility. The free electron concentration is essentially constant at around 10 electrons per cubic meter for a typical metal. Changes of this value are typically insignificant. The charge of an electron is a fundamental constant. Electron spin is a quantum number that is not related to electrical conduction within the scope of this text. As more defects are introduced into a crystalline lattice there is an increased probability of electron scattering. As the scattering frequency increases, the electron drift velocity decreases and the electron mobility decreases.

Reserve Problem 23a (question pool) Compute the number of electrons that each aluminum atom donates, on average, to a bulk piece of aluminum metal. Room temperature data for aluminum: –8

The resistivity of aluminum is 2.63x10 Ω-m 2

The electron mobility of aluminum is 0.0012 m /(V-s) 3

The mass density of aluminum is 2.7 g/cm

The atomic weight of aluminum is 27 g/mol Solution: The atom concentration of aluminum at room temperature is:

n ρ N A (2.7 g/cm3 )(6.022 × 1023 atom/mol) = = = 6.022 × 1022 atom/cm3 V A 27 g/mol The free electron concentration at room temperature is:

n=

σ 1 1 = = = 1.98 × 1029 electron/m3 − − 8 19 2 e μe ρ e μe (2.63 × 10 Ω m)(1.6 × 10 C)(0.0012 m )/(V s))

Therefore, the number of free electrons donated by each atom, on average, is: 1.98 × 1029 electron/m3 6.022 × 1028 atom/m3

= 3.29 electron/atom

Reserve Problem 24 Compute the number of electrons that each atom donates, on average, to a bulk piece of hypothetical metal. Room temperature data for the metal:

The resistivity of the metal is [r] Ω-cm 2

The electron mobility of the metal is [m] cm /(V-s) 3

The mass density of the metal is [d] g/cm

The atomic weight of the metal is [w] g/mol Solution: The atom concentration of aluminum at room temperature is:

ρmass N A A The free electron concentration at room temperature is: 1 ρ e μe

Therefore, the number of free electrons donated by each atom, on average, is:

1 ρ e μe

ρmass N A

=

A [ w] = − 19 (ρ e μe )(ρmass N A ) ([r ][1.6 × 10 ][m])([d][6.022 ×23 ])

A Reserve Question 25: Electric field induced hole movement When an electric field is applied, in which direction are the free electrons accelerated?

a) Opposite to the direction of the electric field. b) In the same direction as the electric field.

Solution When an electric field is applied, the free electrons are accelerated in the direction opposite to that of the field.

Reserve Question 26: Intrinsic semiconductor The electrical conductivity of an intrinsic semiconductor is

a) characteristic of the high-purity metal. b) due to the presence of impurities.

Solution The electrical conductivity of an intrinsic semiconductor is characteristic of the high-purity material.

Reserve Question 27: Metals vs semiconductors How do the electrical conductivities of metals compare with those of semiconductors?

a) σmetals > σsemiconductors b) σmetals = σsemiconductors c) σmetals < σsemiconductors Solution The electrical conductivities of metals are greater than those of semiconductors.

Reserve Question 28: Temperature vs electron/hole concentrations – Intrinsic How does increasing temperature affect the concentration of both electrons and holes in an intrinsic semiconductor?

a) Increases the concentration. b) Decreases the concentration. c) May increase and/or decrease the concentration, depending on the temperature range.

Solution Increasing temperature increases the concentration of both electrons and holes in an intrinsic semiconductor.

Reserve Question 29: Acceptor impurity Which type of charge carrier will be introduced into a semiconductor by the presence of an acceptor impurity?

a) Electron b) Hole Solution The presence of an acceptor impurity introduces holes into a semiconductor.

Reserve Question 30: Donor impurity Which type of charge carrier will be introduced into a semiconductor by the presence of a donor impurity?

a) Impurity b) Hole

Solution The presence of a donor impurity introduces electrons into a semiconductor.

Reserve Question 31: Extrinsic semiconductor The electrical conductivity of an extrinsic semiconductor is

a) Characteristic of the high-purity material. b) due to the presence of impurities.

Solution The electrical conductivity of an extrinsic semiconductor is due to the presence of impurities.

Reserve Question 32: n-type semiconductor – Conduction For an n-type semiconductor, which type of charge carrier is present in the greater concentration?

a) Hole b) Electron Solution For an n-type semiconductor, electrons (i.e., negative charge carriers) are present in a greater concentration than holes.

Reserve Question 33: n-type semiconductor – electrons/holes For an n-type semiconductor

a) Concentrationelectrons < concentrationholes b) Concentrationelectrons = concentrationholes c) Concentrationelectrons < concentrationholes Solution For an n-type semiconductor, the concentration of electrons is much greater than the concentration of holes.

Reserve Question 34: p-type semiconductor – conduction For a p-type semiconductor, which type of charge carrier is present in the greater concentration?

a) Holes b) Electrons

Solution For a p-type semiconductor, holes, (i.e., positive charge carriers) are present in a greater concentration than electrons.

Reserve Question 35: p-type semiconductor – electrons/holes For a p-type semiconductor

a) Concentrationelectrons < concentrationholes b) Concentrationelectrons = concentrationholes c) Concentrationelectrons < concentrationholes Solution For a p-type semiconductor, the concentration of electrons is much lower than the concentration of holes.

Reserve Question 36: Required impurity concentrations In order for a semiconductor to exhibit extrinsic electrical characteristics, relatively high impurity concentrations are required.

a) True b) False Solution 12

False. Even when minute impurity concentrations (e.g., 1 atom in 10 ) are present in a semiconductor, its electrical characteristics will be extrinsic.

Reserve Problem 37 The following electrical characteristics have been determined for both intrinsic and n-type extrinsic indium phosphide (InP) at room temperature: σ (Ω·m)

Intrinsic

2.5 × 10

Extrinsic (n-type)

3.6 × 10

−1

−6

−5

−3

n (m )

3.0 × 10 4.5 × 10

13

14

Calculate electron and hole mobilities.

μe = ____ μh = ____

−3

p (m )

3.0 × 10 2.0 × 10

13

12

Solution In order to solve for the electron and hole mobilities for InP, we must write conductivity expressions for the two materials, of the form of Equation 12.13—i.e.,

σ = n| e | μe + p| e | μh For the intrinsic material

2.5 × 10-6 (Ω-m) −1 = (3.0 × 1013 m−3 )(1.602 × 10−19 C) μe

+ (3.0 × 1013 m−3 )(1.602 × 10−19 C) μh which reduces to

0.52 =μe + μh Whereas, for the extrinsic InP

3.6 × 10-5 (Ω-m)−1 = (4.5 × 1014 m−3 )(1.602 × 10−19 C) μe + (2.0 × 1012 m−3 )(1.602 × 10 −19 C) μh which may be simplified to

112.4 = 225 μe + μh Thus, we have two independent expressions with two unknown mobilities. Upon solving these 2 2 equations simultaneously, we get μe = 0.50 m /V-s and μh = 0.02 m /V-s.

Reserve Question 38: Semiconducting devices Which of the following are preferred for semiconducting devices?

a) Single crystals b) Polycrystalline materials Solution Single crystals are preferred for semiconducting devices because grain boundaries are deleterious to the performance of electronic phenomena.

Reserve Problem 39 Which of the following current-voltage plots is consistent with the behavior of a p-n junction?

a)

b)

c)

d)

e)

Solution: A p-n junction acts as a one-way “valve” for conduction. In forward bias, it ideally acts as an ohmic device since charge carriers (electrons and holes) migrate towards the junction and recombine, thus completing the electric circuit. In reverse bias, charge carriers migrate away from the junction and an insulative (low current) response results until the reverse bias is high enough to induce dielectric breakdown.

Reserve Question 40: Temperature vs conductivity As temperature increases, the electrical conductivities of polymers and ionic ceramics

a) Increase b) Decrease Solution As temperature increases, the electrical conductivities of polymers and ionic ceramics increase.

Reserve Question 41: Band gap of ionic ceramics/polymers Most polymers and ionic ceramics have energy band gap structures that are most similar to those of

a) Insulators b) Semiconductors c) Metals Solution Most polymers and ionic ceramics have wide energy band gaps; thus, they are most like those of insulators.

Reserve Question 42: Charge carriers in ionic ceramics/polymers The charge carriers in ionic ceramics and polymers can be

a) b) c) d)

Electrons Holes Anion Cations

Solution The charge carriers in ionic ceramics and polymers can be electrons, holes, anions, and/or cations.

Reserve Problem 43: Capacitor calculations I 2

Consider a parallel-plate capacitor having an area of 2300 mm and a plate separation of 3.6 mm, and with a material of dielectric constant 5.8 positioned between the plates. Also, the value of ε0 is –12 8.85 × 10 F/m. a) What is the capacitance of this capacitor?

–9

b) Compute the electric field that must be applied for a charge of 8.8 × 10 C to be stored on each plate.

Solution This solution provides a guideline to solving the problem, not the numerical answer. (a) The capacitance of a parallel-plate capacitor containing a dielectric material may be calculated using the following equation:

C =ε

A l

where the dielectric constant is equal to

εr =

ε ε0

Thus, the capacitance in terms of the dielectric constant is

C=

Aεr ε 0 l

(b) The capacitance is related to the charge and voltage as C=

Q V

V=

Q C

or

The electric field is a function of the voltage and separation distance according to E=

V l

And, substitution of the expression for V from the next to last equation yields E=

Q Cl

Reserve Problem 44: Capacitor spacing

A parallel-plate capacitor using a dielectric material having an εr of 2.8 has a plate spacing of 2.6 mm.

If another material having a dielectric constant of 4.5 is used and the capacitance is to not change, –12 what is the new spacing between the plates? The value of ε0 is 8.85 × 10 F/m.

Solution This solution provides a guideline to solving the problem, not the numerical answer.

The capacitance of a parallel-plate capacitor containing a dielectric material may be calculated using the following equation: C =ε

A l

where the dielectric constant is equal to

εr =

ε ε0

Since the capacitances of capacitors 1 and 2 must be the same, the capacitance may be written as

C=

Aεr ,1ε0 l1

=

Aεr ,2ε0 l2

And, solving for the spacing of the second capacitor leads to

ε l l2 = r ,2 1 εr ,1

Reserve Questions and Problems Solutions Manual CHAPTER 13 TYPES AND APPLICATIONS OF MATERIALS

Reserve Question 01: Alloy designations I Which type of steel has the designation 4330? a) Plain carbon steel b) Alloy steel Solution A steel with the 4330 designation is an alloy steel.

Reserve Question 02: Alloy designations II Which type of steel has the designation 1015? a) Plain carbon steel b) Alloy steel Solution A steel with the 1015 designation is a plain carbon steel.

Reserve Question 03: Alloy designations III What is the carbon concentration of a steel having designation 1050? a) b) c) d) e)

0.01 wt% C 0.05 wt% C 0.10 wt% C 0.50 wt% C Impossible to say

Solution The carbon concentration of a 1050 steel is 0.50 wt%.

Reserve Question 04: Alloying elements – stainless steels Which three elements in the list below are primary alloying elements for the stainless steels? a) b) c) d) e) f) g)

Copper Vanadium Nickel Molybdenum Chromium Tungsten Silicon

Solution Although chromium is the primary alloying element for stainless steels, nickel and molybdenum are also used to improve corrosion resistance.

Reserve Question 05: Carbon constituent in white cast iron In what form is carbon found in white cast irons? a) Cementite b) Graphite Solution Because of the low silicon content of white cast irons, carbon is present as cementite.

Reserve Question 06: Carbon steel strength Which of the sequences below represents the various steels types in order of decreasing hardness? a) b) c) d) e) f)

High-carbon > Medium-carbon > Low-carbon High-carbon > Low-carbon > Medium-carbon Medium-carbon > High-carbon > low-carbon Medium- carbon > Low-carbon > High-carbon Low-carbon > High- carbon > Medium-carbon Low-carbon > Medium-carbon > High-carbon

Solution The sequence of steel types in order of decreasing hardness is as follows: High-carbon >Mediumcarbon >Low-carbon.

Reserve Question 07: Cast irons – Minimum C Which of the following is the minimum carbon content for cast irons? a) b) c) d) e) f)

1.0 wt% C 1.56 wt% C 2.00 wt% C 2.14 wt% C 3.42 wt% C 4.96 wt% C

Solution The minimum carbon content for cast irons is 2.14 wt%.

Reserve Question 08: Cast irons – Typical C Which of the following is the typical range of carbon concentrations for cast iron? a) b) c) d) e) f)

1.0 wt% – 1.5 wt% C 1.0 wt% – 2.0 wt% C 2.0 wt% – 3.0 wt % C 2.0 wt% – 3.5 wt% C 3.0 wt% – 4.0 wt% C 3.0 wt% – 4.5 wt% C

Solution The typical range of carbon concentrations for cast irons is 3.0 wt% – 4.5 wt%C. Reserve Question 09: Cold worked stainless steels Which of the following stainless steel types may be strengthened/hardened only by cold working? a) Ferritic b) Martensitic c) Austenitic Solution Ferritic and austenitic stainless steels may be strengthened/hardened only by cold working.

Reserve Question 10: Conversion of gray to nodular irons Which two of the following elements may be added to gray irons before casting so as to produce ductile (or nodular) irons? a) b) c) d)

Silicon Magnesium Serium Molybdenum

Solution Magnesium and cerium may be added to gray irons before casting to produce ductile (or nodular) irons.

Reserve Question 11: Corrosion resistance low-carbon vs HSLA steels Which type of steel has the greater resistance to corrosion? a) plain low-carbon b) high-strength, low-alloy Solution High-strength, low-alloy steels have greater resistances to corrosion than plain low-carbon steels.

Reserve Question 12: Disadvantages of ferrous alloys Which of the following factors restrict(s) the use of ferrous alloys? a) b) c) d)

Poor corrosion resistance.

Poor mechanical properties.

Ores containing iron are rare.

Costly and difficult to process.

Solution Poor corrosion resistance is the factor that restricts the use of ferrous alloys.

Reserve Question 13: Ferrous alloy structures All ferrous alloys have similar microstructures. a) True b) False

Solution False. Ferrous alloys have a wide range of microstructures that depend on composition and processing.

Reserve Question 14: Ferrous alloys ____ is the primary constituent of ferrous alloys.

Solution Iron (Fe) is the primary constituent of ferrous alloys.

Reserve Question 15: Gray iron structures Which of the following microconstituents/phases is (are) most commonly found in gray cast irons? a) b) c) d) e)

Graphite Pearlite Ferrite Austenite Martensite

Solution Graphite, pearlite, and ferrite are the microconstituents/phases most commonly found in gray cast irons.

Reserve Question 16: Heat treatable stainless steels 01 Which of the following types of stainless steels may be heat treated to improve their mechanical properties? a) b) c) d) e)

Ferritic Pearlitic Martensitic Austenitic Precipitation-hardenable

Solution Martensitic and precipitation-hardenable stainless steels may be heat treated to improve their mechanical properties.

Reserve Question 17: Heat treatment of low-carbon steels The strengths of typical low-carbon steels are often improved by heat treatment. a) True b) False Solution False. The strengths of typical low-carbon steels are not generally improved by heat treatment since they do not easily transform to martensite.

Reserve Question 18: High-carbon steels What is the range of carbon concentrations for high-carbon steels? a) b) c) d) e)

0.40 wt% – 1.0 wt% C 0.50 wt% – 1.1 wt% C 0.50 wt% – 1.4 wt% C 0.65 wt% – 1.4 wt% C 0.65 wt% – 1.7 wt% C

Solution The range of carbon concentrations for high-carbon steels is 0.65 wt% – 1.4 wt% C. Reserve Question 19: Low-carbon steel structures Which of the following microconstituent(s)/phase(s) is (are) typically found in a low-carbon steel? a) b) c) d) e)

Ferrite Pearlite Martensite Austenite Tempered martensite

Solution Ferrite and pearlite are the microconstituents/phases typically found in a low-carbon steel.

Reserve Question 20: Low-carbon steels Which of the following is the typical carbon concentration for a low-carbon steel? a) b) c) d) e) f)

0.05 wt% C 0.10 wt% C 0.25 wt% C 0.50 wt% C 0.75 wt% C 1.00 wt% C

Solution The typical carbon concentration for a low-carbon steel is 0.25 wt%.

Reserve Question 21: Magnetic stainless steels Which of the following stainless steel types may be magnetized? a) Ferritic b) Martensitic c) Austenitic Solution Ferritic and martensitic stainless steels may be magnetized.

Reserve Question 22: Mechanical properties of white cast irons Which of the following characteristics best describe the mechanical properties of white cast irons? a) b) c) d)

Hard Weak Ductile Brittle

Solution White cast irons are very hard and very brittle.

Reserve Question 23: Medium-carbon steel structures Which of the following microconstituents/phases is typically found in a medium-carbon steel? a) b) c) d) e)

Ferrite Pearlite Martensite Austenite Tempered martensite

Solution Tempered martensite is the microconstituent typically found in a medium-carbon steel.

Reserve Question 24: Medium-carbon steels What is the typical carbon concentration range for medium-carbon steels? a) b) c) d) e)

0.05 wt% – 1.00 wt% C 0.10 wt% – 0.50 wt% C 0.25 wt% – 0.65 wt% C 0.25 wt% – 1.00 wt% C 0.50 wt% – 0.75 wt% C

Solution The typical carbon concentration range for medium-carbon steels is 0.25 wt% – 0.65 wt% C. Reserve Question 25: Melting Which of the following ferrous alloy types is most easily melted? a) b) c) d) e)

Cast irons Low-carbon steels Medium-carbon steels High-carbon steels Stainless steels

Solution Cast irons are most easily melted, which makes them amenable to casting.

Reserve Question 26: Metal alloy production Which metal alloy type is produced in the greatest quantities? a) b) c) d)

Ferrous Copper Aluminum Titanium

Solution Ferrous alloys are produced in greater quantities than any other alloy type.

Reserve Question 27: Most common steels Which of the following steel types is produced in the greatest quantities? a) Low-carbon b) Medium-carbon c) High-carbon

Solution Low-carbon steels are produced in the greatest quantities.

Reserve Question 28: Nodular iron structures Which of the following microconstituents/phases are generally found in nodular irons? a) b) c) d) e)

Graphite Pearlite Ferrite Austenite Martensite

Solution Graphite, pearlite, and ferrite microconstituents/phases are generally found in nodular irons.

Reserve Question 29: Types of steels Which type of steel contains only residual amounts of alloying elements? a) Plain carbon steel b) Alloy steel Solution A plain carbon steel contains only residual amounts of alloying elements.

Reserve Question 30: white cast irons vs malleable irons Which of the following changes occur(s) when white cast irons are converted to malleable irons? a) b) c) d) e)

The cementite is converted into graphite.

The graphite is converted into cementite.

The ductility increases.

The ductility remains about the same.

The ductility decreases.

Solution When white cast irons are converted to malleable irons, the cementite is converted to graphite and the ductility increases.

Reserve Question 31: Alloy selection For each of the following applications, match the metal or alloy that is most suitable.

Milling machine base • Titanium alloy • Plain carbon steel • Tool steel • Zinc • Tungsten • Aluminum • Platinum • Gray cast iron • Stainless steel • Brass Walls of steam boiler • Plain carbon steel • Tool steel • Zinc • Tungsten • Aluminum • Platinum • Gray cast iron • Stainless steel • Brass • Titanium alloy High-speed aircraft • Brass • Stainless steel • Gray cast iron • Platinum • Aluminum • Tungsten • Zinc • Tool steel • Plain carbon steel • Titanium alloy

Drill bit • Titanium alloy • Brass • Stainless steel • Gray cast iron • Platinum • Aluminum • Tungsten • Zinc • Tool steel • Plain carbon steel Cryogenic (very low temperature) container • Aluminum • Platinum • Gray cast iron • Stainless steel • Brass • Titanium alloy • Plain carbon steel • Tool steel • Zinc • Tungsten Pyrotechnic (i.e., flares or fireworks) • Tungsten • Zinc • Tool steel • Plain carbon steel • Titanium alloy • Brass • Stainless steel • Gray cast iron • Platinum • Aluminum

High-temperature furnace elements to be used in oxidizing environments • Plain carbon steel • Tool steel • Zinc • Tungsten • Aluminum • Platinum • Gray cast iron • Stainless steel • Brass • Titanium alloy Solution • Gray cast iron would be the best choice for a milling machine base because it effectively absorbs vibrations and is inexpensive.

• Stainless steel would be the best choice for the walls of a steam boiler because it is corrosion resistant to the steam and condensate.

• Titanium alloys are the best choice for high-speed aircraft because they are light weight, strong, and easily fabricated. However, one drawback is their cost.

• A tool steel would be the best choice for a drill bit because it is very hard and wear resistant, and, thus, will retain a sharp cutting edge.

• For a cryogenic (low-temperature) container, an aluminum alloy would be the best choice; aluminum alloys have an FCC crystal structure, and therefore, are ductile down to very low temperatures.

• As a pyrotechnic in flares and fireworks, magnesium is the best choice because it ignites easily and burns readily in air.

• Platinum is the best choice for high-temperature furnace elements to be used in oxidizing atmospheres because it is very ductile, has a relatively high melting temperature, and is highly resistant to oxidation.

Reserve Question 32: Beryllium copper alloys Which of the following elements, when alloyed with copper, results in an alloy that is precipitation hardenable? a) b) c) d) e)

Beryllium Zinc Gold Tin Lead

Solution Copper-beryllium alloys are precipitation hardenable.

Reserve Question 33: Brass Which of the following are primary constituents of brass? a) b) c) d) e)

Copper and zinc Copper and iron Copper and lead Copper and zinc Tin and zinc

Solution Copper and zinc are the primary constituents of brass.

Reserve Question 34: Bronze compositions Which four of the following elements are alloyed with copper to produce bronze alloys? a) b) c) d) e) f)

Zinc Tin Aluminum Silicon Nickel Lead

Solution Tin, aluminum, silicon, and nickel are alloyed with copper to produce bronze alloys.

Reserve Question 35: Cast alloys 01 Which type of nonferrous alloy is amenable to mechanical deformation? a) Cast b) Wrought Solution Wrought alloys are amenable to mechanical deformation.

Reserve Question 36: Cast alloys 02 Which type of nonferrous alloy is not often amenable to mechanical deformation? a) Cast b) Wrought Solution Cast alloys are normally too brittle to be shaped by mechanical deformation.

Reserve Question 37: Magnesium alloys Which of the following is (are) desirable characteristics of magnesium and its alloys? a) b) c) d)

Very carrion resistant in a variety of environments.

Low melting temperatures.

Easily work-hardenend.

Very high specific strengths.

Solution Magnesium and its alloys have very high specific strengths. Undesirable characteristics are as follows: they have low melting temperatures, and, because of their brittleness, are not easily work hardened; also, while they are reasonably corrosion resistant in "normal" environments, they extremely susceptible corrosion in marine environments.

Reserve Question 38: Mg alloys replacing plastics For which of the following reasons have magnesium alloys replaced engineered plastics having comparable densities? a) Magnesium alloys are stiffer. b) Magnesium alloys are more recyclable. c) Magnesium alloys are cheaper to produce.

Solution Magnesium alloys have replaced engineered plastics having comparable densities because they (magnesium alloys) are stiffer, are more recyclable, and are cheaper to produce.

Reserve Question 39: Nickel alloys In which of the following environments do nickel and its alloys perform extremely well? a) Acidic b) Basic Solution Nickel and its alloys perform very well in basic environments.

Reserve Question 40: Noble metal advantages Which of the following are desirable characteristics of the noble metals? a) b) c) d)

High strengths.

High ductilities.

Oxidation resistance.

High electrical conductivities.

Solution Noble metals are very ductile, oxidation resistant, and have high electrical conductivities.

Reserve Question 41: Noble metals Which of the following are noble metals? a) b) c) d) e) f) g) h) i) j) k)

silver gold platinum palladium rhodium ruthenium iridium osmium plutonium zirconium lead

Solution Silver, gold, platinum, palladium, rhodium, ruthenium, iridium, and osmium are noble metals.

Reserve Question 42: Refractory alloys Which of the following are refractory metals? a) b) c) d) e) f) g) h)

niobium molybdenum tungsten tantalum copper titanium iron aluminum

Solution Niobium, molybdenum, tungsten, and tantalum are refractory metals.

Reserve Question 43: Superalloys Which of the following are the primary superalloy metals? a) b) c) d) e) f)

Irons Nickel Cobalt Titanium Copper Aluminum

Solution The primary superalloy metals are iron, nickel, and cobalt.

Reserve Question 44: Titanium alloys Which of the following are desirable characteristics of titanium and its alloys? a) b) c) d) e)

Corrosion resistant Low melting temperatures.

Easily processed, even at high temperatures.

High strengths.

High ductilities.

Solution Desirable characteristics of titanium and its alloys are corrosion resistant, and they have high strengths and high ductilities.

Reserve Question 45: Zinc content of brass At room temperature, what is the approximate maximum zinc content of α-brass? a) b) c) d) e)

5 wt% Zn 15 wt% Zn 25 wt% Zn 35 wt% Zn 45 wt% Zn

Solution At room temperature, the maximum zinc content of α-brass is about 35 wt%.

Reserve Problem 46a (question pool) Select T/F for each of the following statements regarding aluminum / aluminum alloys: [a] Aluminum alloys are generally not viable as lightweight structural materials in humid environments because they are highly susceptible to corrosion by water vapor. [b] Aluminum alloys are generally superior to pure aluminum, in terms of yield strength, because their microstructures often contain precipitate phases that strain the lattice, thereby hardening the alloy relative to pure aluminum. [c] Aluminum is not very workable at high temperatures in air, in terms of extrusion and rolling, because a non-protective oxide grows and consumes the metal, converting it to a hard and brittle ceramic. [d] Compared to most other metals, like steel, pure aluminum is very resistant to creep deformation. [e] The relatively low melting point of aluminum is often considered a significant limitation for structural applications.

Solution [a] F [b] T [c] F [d] F [e] T

Reserve Problem 47a (question pool) Select T/F for each of the following statements regarding aluminum / aluminum alloys: [a] Aluminum alloys are generally viable as lightweight structural materials in humid environments because they are not very susceptible to corrosion by water vapor. [b] Aluminum is not very workable at high temperatures in air, in terms of extrusion and rolling, because a non-protective oxide grows and consumes the metal, converting it to a hard and brittle ceramic. [c] Aluminum alloys are generally superior to pure aluminum, in terms of yield strength, because their microstructures often contain precipitate phases that strain the lattice, thereby hardening the alloy relative to pure aluminum. [d] Compared to other metals, like steel, pure aluminum is very resistant to failure via fatigue. [e] Aluminum exhibits one of the highest melting points of all metals, which makes it difficult and expensive to cast.

Solution: These questions could be scrambled to create additional versions. [a] T [b] F [c] T [d] F [e] F

Reserve Problem 48a (question pool) Select T/F for each of the following statements regarding copper & copper alloys: [a] Copper has a higher elastic modulus than aluminum. [b] The density of copper is closer to that of aluminum than it is to iron. [c] Bronze is an alloy of copper and zinc. [d] Copper and its alloys form a green tarnish over time, consisting of sulfides and carbonates. [e] Copper is relatively resistant to corrosion by neutral and even mildly basic water, making it useful for freshwater plumbing applications.

Solution: [a] T [b] F [c] F [d] T [e] T

Reserve Problem 49a (question pool) Select T/F for each of the following statements regarding copper & copper alloys: [a] Copper is much more abundant in the earth's crust compared to iron or aluminum. [b] Copper is one of just a few metals that can be found in metallic form in nature. [c] Pure and/or annealed copper is more difficult to machine compared to its work-hardened form or its alloys. [d] Copper is a minor component (by weight) of most brass & bronze alloys. [e] Amongst metals and alloys copper is one of the best conductors of heat.

Solution: [a] F [b] T [c] T [d] F [e] T

Reserve Problem 50a (question pool) Select T/F for each of the following statements regarding various metals & alloys: TM

[a] Nickel is majority component (by mass) in certain superalloys such as Waspaloy . [b] Tungsten is the lowest density metal that has structural use. [c] Tantalum offers extremely good corrosion resistance, especially at low temperatures.

[d] Magnesium metal is very similar to aluminum, in terms of its physical and mechanical properties. [e] Beryllium metal is commonly used as an alloying agent in copper metal.

Solution: [a] T [b] F [c] T [d] T [e] T

Reserve Question 51: Alumina content of fireclays Increasing the alumina (Al2O3) content of fireclays results in a) an increase in maximum service temperature. b) a decrease in maximum service temperature.

Solution Increasing the alumina (Al2O3) content of fireclays results in an increase in maximum service temperature.

Reserve Question 52: Alumina in silica refractories The high-temperature performance of silica refractories is compromised by the presence of even small concentrations of alumina (Al2O3). a) True b) False Solution True. The presence of even small amounts of alumina (Al2O3) in silica refractory ceramics compromises their high-temperature performance.

Reserve Question 53: Porosity effects As the porosity of refractory ceramic bricks increases, a) b) c) d) e) f)

Strength decreases.

Strength increases.

Chemical resistance decreases.

Chemical resistance increases.

Thermal insulation decreases.

Thermal insulation increases.

Solution As the porosity of refractory ceramic bricks increases, strength decreases, chemical resistance decreases, and thermal insulation increases.

Reserve Question 54: Silica in basic refractories The presence of silica (SiO2) in basic refractory ceramics is beneficial to their high-temperature performance. a) True b) False Solution False. The presence of silica (SiO2) in basic refractory ceramics is deleterious on their hightemperature performance.

Reserve Question 55: Slag application of basic refractories Basic refractory ceramics are often used for the containment of slags that are rich in a) silica b) CaO c) MgO Solution Basic refractory ceramics are often used for the containment of slags that are rich in CaO and/or MgO. Reserve Question 56: Slag application of silica refractories Silica refractory ceramics are often used for the containment of slags that are rich in a) silica b) CaO c) MgO

Solution Silica refractory ceramics are often used for the containment of slags that are rich in silica.

Reserve Problem 57 Find the maximum temperature to which the following two magnesia–alumina refractory materials may be heated before a liquid phase will appear. (a) A spinel-bonded alumina material of composition 95 wt% Al2O3–5 wt% MgO. (b) A magnesia–alumina spinel of composition 65 wt% Al2O3–35 wt% MgO. Consult Figure 10.24.

Solution (a) From Figure 10.24, for a spinel-bonded magnesia material (95 wt% Al2O3-5 wt% MgO), the maximum temperature without a liquid phase corresponds to the temperature of the eutectic isotherm on the Al2O3-rich side of the phase diagram, which is approximately 2000°C (3630°F). (b) The maximum temperature without the formation of a liquid phase for a magnesia-alumina spinel (65 wt% Al2O3-35 wt% MgO) lies at the phase boundary between MgAl2O4(ss)-MgAl2O4 + Liquid phase fields (just slightly to the left of the congruent melting point at which the two phase boundaries become tangent); this temperature is approximately 2030°C (3685°F).

Reserve Question 58: Abrasive materials Materials that are used as abrasive ceramics include which of the following? a) b) c) d) e) f)

Diamond Silicon carbide Tungsten carbide Aluminum oxide Silica sand Titanium dioxide

Solution Diamond, silicon carbide, tungsten carbide, aluminum oxide, and silica sand are materials used as abrasive ceramics.

Reserve Question 59: Abrasive properties Which of the following are properties required for abrasive ceramics? a) b) c) d) e)

High wear resistance High temperature stability Low temperature stability High toughness High ductility

Solution The important properties required for abrasive ceramics are high wear resistance, high temperature stability, and high toughness.

Reserve Question 60: Cement consumption Which of the following cement materials is consumed in the largest tonnages? a) Portland cement b) Plaster of paris c) Lime Solution Portland cement is consumed in the largest tonnages.

Reserve Question 61: Hardening mechanism of lime The hardening of lime is associated with a) a drying process. b) A chemical reaction involving water (i.e. hydration). c) A chemical reaction involving compound other than water.

Solution The hardening of lime involves a chemical reaction between the lime and a compound other than water.

Reserve Question 62: Hardening mechanism of Portland cement The hardening of portland cement is associated with a) a drying process. b) A chemical reaction involving water (i.e. hydration). c) A chemical reaction involving a compound other than water.

Solution The hardening of portland cement involves a chemical reaction between the cement and water (i.e. hydration).

Reserve Question 63: Fluorocarbons Which of these polymers are most resistant to attack by chemicals, and, as such, are often used as coatings? a) b) c) d)

Fluorocarbons Polystyrene Polyethylene Rubber

Solution Fluorocarbons are typically used as coatings that need to be extremely resistant to attack by chemicals.

Reserve Question 64: Optical transparency For applications that require optical transparency, a polymer must be which of the following: a) b) c) d)

Amorphous.

Semicrystalline with very small crystallites.

Semicrystalline with very large crystallites.

Crystalline.

Solution For applications that require optical transparency, a polymer must be either amorphous or semicrystalline with very small crystallites.

Reserve Question 65: Plastics Which of the following polymers are classified as plastics? a) b) c) d) e) f) g) h) i)

Polyethylene Polypropylene Poly(vinyl chloride) Polystyrene The epoxies The polyesters Rubber Polyisoprene Polychloroprene

Solution Polyethylene, polypropylene, poly(vinyl chloride), polystyrene, the epoxies, and the polyesters are all classified as plastics.

Reserve Question 66: Crosslinking of elastomers Elastomers must have some crosslinking. a) True b) False Solution True. Elastomers must have some crosslinking.

Reserve Question 67: Fiber molecular weight For fiber applications, the molecular weight of a polymer should be a) Relatively high. b) Relatively low.

Solution For fiber applications, the molecular weight of a polymer should be relatively high.

Reserve Questions and Problems Solutions Manual CHAPTER 14 SYNTHESIS, FABRICATION, AND PROCESSING OF MATERIALS

Reserve Question 01: Advantages of hot vs cold working Which type of forming operation produces a higher quality surface finish, better mechanical properties, and closer dimensional control of the finished piece? a) Cold working b) Hot working Solution Cold working forming operations produce a higher quality surface finish, better mechanical properties, and closer dimensional control of the finished piece.

Reserve Question 02: Cold working Cold working takes place at a temperature that is below a metal's a) b) c) d)

Melting temperature Recrystallization temperature Eutectoid temperature Glass transition temperature

Solution Cold working takes place at a temperature that is below a metal's recrystallization temperature.

Reserve Question 03: Deformation in hot vs cold working Which type of forming operation has lower deformation energy requirements? a) Cold working b) Hot working Solution Hot working forming operations have lower deformation energy requirements.

Reserve Question 04: Forging Forging operations normally take place at a) Low temperature b) Room temperature c) High temperature Solution Forging operations normally take place at high temperatures.

Reserve Question 05: Hot working Hot working takes place at a temperature that is above a metal's a) b) c) d)

melting temperature recrystallization temperature eutectoid temperature glass transition temperature

Solution Hot working takes place at a temperature that is above a metal's recrystallization temperature.

Reserve Question 06: Metal forming Which of the following are forming operations? a) b) c) d) e) f) g) h)

Forging Rolling Extrusion Drawing Powder metallurgy Welding Continuous casting Die casting

Solution Forging, rolling, extrusion, and drawing are all forming operations.

Reserve Question 07: Annealing Which of the following may occur during annealing? a) Stresses may be relieved. b) Ductility may be increased.

c) Toughness may be increased. d) A specific microstructure may be produced.

Solution During annealing • • • •

Stresses may be relieved.

Ductility may be increased.

Toughness may be increased.

A specific microstructure may be produced.

Reserve Question 08: Annealing temperature How does increasing the annealing temperature influence the rate of an annealing process? a) Increases the rate b) Decreases the rate Solution Increasing the annealing temperature increases the rate of an annealing process because it involves diffusion, the rate of which is accelerated by increasing the temperature.

Reserve Question 09: Full annealing Full annealing is used on which types of steels? a) Low-carbon steels b) Medium-carbon steels c) High-carbon steels Solution Full annealing is used on low-carbon and medium-carbon steels.

Reserve Question 10: Full annealing temperature Full annealing is accomplished by heating at a temperature approximately how many degrees Celsius above either A3 line (for compositions less than the eutectoid) or A1 line (for compositions greater than the eutectoid) a) b) c) d) e)

25°C 40°C 50°C 55°C 70°C

Solution Full annealing is accomplished by heating at a temperature approximately 50°C above the upper critical temperature.

Reserve Question 11: Full annealing cooling A full annealing heat treatment is terminated by cooling in a) Air b) A furnace Solution A full annealing heat treatment is terminated by cooling in a furnace.

Reserve Question 12: Metals for powder metallurgy For which type of metals is powder metallurgy particularly suited? a) Metals having high melting temperatures b) Metals having low melting temperatures Solution Powder metallurgy is particularly suited for metals having high melting temperatures.

Reserve Question 13: Normalizing Normalizing of a ferrous alloy causes the average grain size to a) Increase b) Decrease Solution Normalizing of a ferrous alloy causes the average grain size to decrease.

Reserve Question 14: Normalizing temperature Normalizing is accomplished by heating at a temperature at least how many degrees Celsius above the upper critical temperature? a) b) c) d) e)

25°C 40°C 50°C 55°C 70°C

Solution Normalizing is accomplished by heating at a temperature approximately 55°C above the upper critical temperature.

Reserve Question 15: Normalizing cooling A normalizing heat treatment is terminated by cooling in a) Air b) A furnace Solution A normalizing heat treatment is terminated by cooling in air.

Reserve Question 16: Powder metallurgy The density of a piece produced by powder metallurgy may have the same density as the parent material. a) True b) False Solution True. The density of a piece produced by powder metallurgy may have the same density as the parent material.

Reserve Question 17: Spheroidizing A spheroidizing heat treatment is normally used on which type(s) of steels? a) Low-carbon steels b) Medium-carbon steels c) High-carbon steels Solution The spheroidizing annealing process is used on medium-carbon and high-carbon steels.

Reserve Problem 18 Give the approximate minimum temperature at which it is possible to austenitize each of the following iron–carbon alloys during a normalizing heat treatment: (a) 0.20 wt% C (b) 0.76 wt% C (c) 0.95 wt% C.

Solution (a) For 0.20 wt% C, heat to at least 905°C (1660°F) since the A3 temperature is 850°C (1560°F). (b) For 0.76 wt% C, heat to at least 782°C (1440°F) since the A3 temperature is 727°C (1340°F). (c) For 0.95 wt% C, heat to at least 840°C (1545°F) since the Acm temperature is 785°C (1445°F).

Reserve Problem 19 Give the approximate temperature at which it is desirable to heat each of the following iron–carbon alloys during a full anneal heat treatment: (a) 0.25 wt% C (b) 0.45 wt% C (c) 0.85 wt% C (d) 1.10 wt% C. Solution (a) For 0.25 wt% C, heat to about 880°C (1510°F) since the A3 temperature is 830°C (1420°F). (b) For 0.45 wt% C, heat to about 830°C (1525°F) since the A3 temperature is 780°C (1435°F). (c) For 0.85 wt% C, heat to about 777°C (1430°F) since the A1 temperature is 727°C (1340°F). (d) For 1.10 wt% C, heat to about 777°C (1430°F) since the A1 temperature is 727°C (1340°F).

Reserve Question 20: Cooling rate – Geometry I The smaller the ratio of surface are to the mass of steel specimen, a) b) c) d)

The slower the cooling rate, and consequently, the shallower the hardening effect The slower the cooling rate, and consequently, the deeper the hardening effect The faster the cooling rate, and consequently, the shallower the hardening effect The faster the cooling rate, and consequently, the deeper the hardening effect

Solution The smaller the ratio of surface area to the mass of a steel specimen, the slower the cooling rate, and consequently, the shallower the hardening effect.

Reserve Question 21: Cooling rate – Geometry II Based upon a specimen's surface area-to-mass ratio, which geometrical shape is more amenable to hardening by quenching? a) Irregular with edges and corners b) Regular and rounded

Solution Based upon a specimen's surface area-to-mass ratio, irregular shapes with edges and corners are more amenable to hardening by quenching.

Reserve Question 22: Jominy end-quench test composition An alloy steel (containing Ni, Cr, Mo, etc.) may be hardened to a greater degree than an unalloyed steel. This behavior is due to the presence of the alloying elements that delay austenite-to-pearlite and austenite-to-bainite reactions. a) True b) False Solution True. The presence of alloying elements, such as nickel, chromium, and molybdenum in steels delays austenite transformation reactions; this results in harder steels.

Reserve Question 23: Jominy end-quench test cooling rate The rate of heat transfer from the specimen during a Jominy end-quench test is nearly independent of composition. a) True b) False Solution True. The rate of heat transfer from the specimen during a Jominy end-quench test is nearly independent of composition.

Reserve Question 24: Martensitic content vs strength I For parts that are to be used in relatively high stress applications, what percentage of the part's interior should be martensite? a) b) c) d) e) f) g)

40% 50% 60% 70% 80% 90% 100%

Solution For parts that are to be used in relatively high stress applications, 80% of the part's interior should be martensite.

Reserve Question 25: Martensitic content vs strength II For parts that are to be used in moderately stressed applications, what percentage of the part's interior should be martensite? a) b) c) d) e) f) g)

40% 50% 60% 70% 80% 90% 100%

Solution For parts that are to be used in moderately stressed applications, 50% of the part's interior should be martensite.

Reserve Question 26: Quenching During a quenching treatment, it is possible to cool the specimen at a uniform rate throughout the entire piece. a) True b) False Solution False. During a quenching treatment, it is not possible to cool the entire specimen at a uniform rate throughout the piece — the surface will always cool more rapidly than interior regions.

Reserve Question 27: Quenching media Which of the following sequences correctly indicates the severity of quench order of the three quenching media form least severe to most severe? a) b) c) d) e)

Water; oil; air Water; air; oil Oil; water; air Air; oil; water Air; water; oil

Solution The correct sequence is air; oil; water—i.e. air offers the least severe quench, followed by oil and then water (which is the most severe).

Reserve Question 28: Quenching media – steel Which of the three quenching media is most commonly used for alloy steels? a) Oil b) Water c) Air Solution Oil is the most commonly used quenching medium for alloy steels.

Reserve Question 29: Severity of quench The more “severe” a quench, a) The more rapid the quench. b) The less rapid the quench.

Solution The more "severe" a quench, the more rapid the quench.

Reserve Question 30: Glass forming methods Match each product with the glass forming method that is used to produce it.

Dishes • Drawing • Blowing • Pressing • Fiber forming Light bulbs • Blowing • Pressing • Fiber forming • Drawing

Rods • • • •

Pressing Fiber forming Drawing Blowing

Fibers • Blowing • Drawing • Fiber forming • Pressing Solution Dishes—pressing; light bulbs—blowing; rods—drawing; fibers—fiber forming.

Reserve Question 31: Glass phases Which of the following represents the correct phase transformation sequence as a glass material is heated? a) b) c) d) e) f)

Solid; supercooled liquid; liquid Solid; liquid; supercooled liquid Supercooled liquid; solid; liquid Supercooled liquid; liquid; solid Liquid; supercooled liquid; solid Liquid; solid; supercooled liquid

Solution With increasing temperature, the following is the correct phase transformation sequence for a glass: solid, supercooled liquid, and liquid.

Reserve Question 32: Temperature points relative to glasses Match the viscosity values with the names of their corresponding temperature points.

100 Pa-s • Melting point • Working point • Softening point • Annealing point • Strain point

3

10 Pa-s • Softening point • Annealing point • Strain point • Working point • Melting point 6

4 × 10 Pa-s • Working point • Softening point • Annealing point • Strain point • Melting point 12

10 Pa-s • Softening point • Melting point • Working point • Strain point • Annealing point 13

3 × 10 Pa-s • Annealing point • Strain point • Working point • Melting point • Softening point Solution Melting point:

100 Pa-s

Working point:

10 Pa-s

Softening point:

4 × 10 Pa-s

Annealing point:

10 Pa-s

Strain point:

3 × 10 Pa-s

3

6

12

13

Reserve Question 33: Thermal stress reduction Once thermal stresses have been introduced into a ceramic piece, it is impossible to remove them. a) True b) False Solution False. Once thermal stresses have been introduced into a ceramic piece, they may reduced or eliminated by an annealing heat treatment.

Reserve Question 34: Thermal tempering The strength of a glass piece may be improved by inducing thermal compressive residual surface stresses. a) True b) False Solution True. The strength of a glass piece may be improved by inducing thermal compressive residual surface stresses; this is called thermal tempering.

Reserve Question 35: Thermal tempering stresses Thermal tempering results in the introduction of a) b) c) d)

compressive stresses on the surface compressive stresses internally tensile stresses on the surface tensile stresses internally.

Solution Thermal tempering results in the introduction of compressive stresses on the surface and tensile stresses internally.

Reserve Question 36: Thermal tempering temperature During thermal tempering, a glass piece is heated to a temperature a) b) c) d)

below the glass transition temperature above the glass transition temperature below the softening point above the softening point

Solution During thermal tempering, a glass piece is heated to a temperature above the glass transition temperature and below the softening point.

Reserve Question 37: Particle size of clay products The larger the initial particle size of a clay-based ceramic body that has been dried, a) the lower the shrinkage b) the greater the shrinkage Solution The larger the initial particle size of a clay-based ceramic body that has been dried, the lower the shrinkage.

Reserve Question 38: Vitrification As the vitrification of clay-based products increases, which of the following also increase(s)? a) b) c) d)

Shrinkage Strength Durability Density

Solution As the vitrification of clay-based products increases, shrinkage, strength, durability, and density also increase.

Reserve Question 39: Water content of clay products During drying, the greater the initial water content of a clay-based ceramic body, a) the greater the shrinkage b) the lower the shrinkage Solution During drying, the greater the initial water content of a clay-based ceramic body, the greater the shrinkage.

Reserve Question 40: Plastic deformation in powder pressing During the powder pressing of ceramic pieces, plastic deformation of the particles occurs. a) True b) False

Solution False. During the powder pressing of ceramic pieces, no plastic deformation of the particles occurs.

This may be contrasted to powder metallurgical processing wherein the particles experience some plastic deformation.

Reserve Question 41: Powder pressing Which of the following ceramic products are normally produced using powder pressing? a) b) c) d) e)

Clay ceramics Electronic ceramics Magnetic ceramics Refractory ceramics Glass ceramics

Solution Clay, electronic, magnetic, and refractory ceramics are normally produced using powder pressing.

Reserve Question 42: Powder pressing techniques If a ceramic piece is desired that is dense and, in addition, is to experience very little or no grain growth during its processing, which of the following powder pressing techniques should be used? a) Uniaxial pressing, followed by firing. b) Isostatic pressing, followed by firing. c) Hot pressing.

Solution If a ceramic piece is desired that is dense and, in addition, is to experience very little or no grain growth during its processing, then it should be hot pressed.

Reserve Questions and Problems Solutions Manual CHAPTER 15 COMPOSTIES

Reserve Question 01: Hardness matrix vs dispersed For composite materials, which is phase is normally harder? a) The matrix phase b) The dispersed phase Solution The dispersed phase is normally harder than the matrix phase.

Reserve Question 02: Particle content As particle content is increased, how does the strength a particle-reinforced composite change? a) it increases b) it decreases Solution As the particle content is increased, the strength of the composite increases.

Reserve Question 03: Cermets In order to produce very hard cermet (ceramic-metal) composites, what volume percent of the particulate phase is normally used? a) b) c) d) e)

> 50 vol% > 60 vol% > 70 vol% > 80 vol% > 90 vol%

Solution In order to produce very hard cermet (ceramic-metal) composites, greater than 90 volume percent of the particulate phase is normally used.

Reserve Question 04: Dispersion-strengthened composite vs precipitation hardened alloy II Which material retains its strength better at elevated temperatures? a) A dispersion-strengthened composite b) A precipitation-hardened alloy Solution A dispersion-strengthened composite retains its strength better at elevated temperatures.

Reserve Question 05: Continuous fibers The length of continuous fibers is typically a) b) c) d) e)

l > 5 lc l > 15 lc l > 25 lc l > 35 lc l > 45 lc

Solution The length of continuous fibers is typically greater than 15 lc .

Reserve Question 06: Critical fiber length What is the order of magnitude critical length of glass and carbon fibers in composites? a) b) c) d) e)

0.01 mm 0.1 mm 1.0 mm 10.0 mm 100 mm

Solution The critical length of glass and carbon fibers in composites is on the order of 1.0 mm.

Reserve Problem 07a (question pool) Such that the bond strength across the fiber-epoxy interface is [s] MPa, and the shear yield strength of the epoxy is [y] MPa, compute the minimum fiber length, in millimeters, to guarantee that the fibers are conveying an optimum fraction of force that is applied to the composite. The tensile strength of these carbon fibers is [f] MPa.

Solution:

τ fiber − matrix < τ y , matrix SO τ c = τ fiber − matrix l=

σ *f d 2τ c

=

σ *f d 2τ fiber − matric

=

[ f ][d] 1 2[s] 1000

Reserve Problem 08: Reinforcement efficiency I For an aligned fibrous composite, when a stress is applied in a direction that is parallel to the fibers, what is the reinforcement efficiency? a) 0 1 5 3 c) 8 3 d) 4 e) 1

b)

Solution For an aligned fibrous composite when the stress is applied parallel to the fibers, the reinforcement efficiency is 1.

Reserve Problem 09: Reinforcement efficiency II For an aligned fibrous composite, when a stress is applied perpendicular to the fibers, what is the reinforcement efficiency? a) 0 1 5 3 c) 8 3 d) 4 e) 1

b)

Solution For an aligned fibrous composite when a stress is applied perpendicular to the fibers, the reinforcement efficiency is 0.

Reserve Problem 10: Reinforcement efficiency III For a fibrous composite with fibers that are randomly and uniformly oriented within a specific plane, when a stress is applied in any direction within the plane of the fibers, what is the reinforcement efficiency? a) 0 1 5 3 c) 8 3 d) 4 e) 1

b)

Solution For a fibrous composite with fibers that are randomly and uniformly oriented within a specific plane, when a stress is applied in any direction within the plane of the fibers, the reinforcement efficiency is

3 .

8

Reserve Problem 11: Reinforcement efficiency IV For a fibrous composite with fibers that are uniformly distributed and randomly oriented in all directions, when a stress is applied in any direction, what is the reinforcement efficiency? a) 0 1 5 3 c) 8 3 d) 4 e) 1 b)

Solution For a fibrous composite with fibers that are uniformly distributed and randomly oriented in all directions, when a stress is applied in any direction, the reinforcement efficiency is 1 5 .

Reserve Question 12: Applied load A stress-strain test is performed on an aligned fibrous composite such that the force is applied in the longitudinal direction. During the initial stage of the test, which phase bears most of the load? a) Fibers b) Matrix

Solution During a stress-strain test that is performed on an aligned fibrous composite, the fibers bear more of the applied load than the matrix in the initial stage of the test.

Reserve Question 13: Composite failure Once the fibers fail in a fibrous composite, catastrophic failure of the piece takes place. a) True b) False Solution False. Once the fibers fail in a composite, catastrophic failure of the piece does not take place. Since the broken fibers are still embedded within the matrix, they are still capable of sustaining a diminished load.

Reserve Question 14: Continuous fiber orientation How are continuous fibers typically oriented in fibrous composites? a) Aligned b) Partially oriented c) Randomly oriented Solution Continuous fibers are typically aligned in fibrous composites.

Reserve Question 15: Discontinuous fiber orientation How are discontinuous fibers typically oriented in fibrous composites? a) Aligned b) Partially oriented c) Randomly oriented Solution Discontinuous fibers may be aligned, partially oriented, and randomly oriented in fibrous composites.

Reserve Question 16: Discontinuous vs continuous fiber use If the fiber orientation is random, which type of fibers is normally used? a) Discontinuous b) Continuous

Solution Discontinuous fibers are normally used when the fiber orientation is random.

Reserve Problem 17a (question pool) Select T/F for each of the following statements: [a] Composites are single-phase materials by definition. [b] The term “composite” applies to materials that feature polymeric materials only. [c] Structural composites are, in general, highly regarded for their specific strengths. [d] Composites featuring continuous and aligned fibers for reinforcement generally offer properties that are highly isotropic compared to most metals (random polycrystals).

Solution: [a] F [b] F [c] T [d] F

Reserve Problem 18a (question pool) The figure below depicts a continuous aligned fiber-reinforced composite featuring carbon fiber and epoxy. If the elastic moduli of the composite components are [m] GPa and [f] GPa, what is the expected modulus, in GPa, for the depicted loading scenario if the fibers comprise [v] vol% of the composite?

Solution: The schematic scenario depicts longitudinal loading of the composite, so the maximum elastic modulus is expected along this loading axis.

Eclong = E f V f + EmVm = ([ f ])([v]/100) + ([m]) (1 − [v]/100)

Reserve Problem 19a (question pool) The figure below depicts the elastic modulus rules-of-mixtures for an aligned and continuous fiberreinforced epoxy composite. What is the composite modulus, measured parallel to the fiber alignment axis, for a sample containing 80 vol% fiber? Note that the volume fraction axis is intentionally unlabelled. You should be able to discern this vol% label for yourself based on the logic for producing a structural composite.

(a) 5 GPa (b) 7.5 GPa (c) 10 GPa (d) 15 GPa (e) 20 GPa (f) 25 GPa (g) 30 GPa

Solution: The fiber phase has a higher elastic modulus than epoxy, so the x-axis label is vol% epoxy.

Therefore, a composite comprised of 80 vol% fiber would feature 20 vol% epoxy.

The elastic modulus measured parallel to the fiber alignment will be a maximum value.

Thus, reading the upper plot for 20% indicates an elastic modulus of 25 GPa.

Reserve Problem 20a (question pool) The figure below depicts the elastic modulus rules-of-mixtures for an aligned and continuous fiberreinforced epoxy composite. What is the composite modulus, measured parallel to the fiber alignment axis, for a sample containing 80 vol% fiber? Note that the volume fraction axis is intentionally unlabelled. You should be able to discern this vol% label for yourself based on the logic for producing a structural composite.

(a) 5 GPa (b) 7.5 GPa (c) 10 GPa (d) 15 GPa (e) 20 GPa (f) 25 GPa (g) 30 GPa

Solution: The fiber phase has a higher elastic modulus than epoxy, so the x-axis label is vol% fiber.

The elastic modulus measured parallel to the fiber alignment will be a maximum value.

Thus, reading the upper plot for 80% indicates an elastic modulus of 25 GPa.

Reserve Question 21: Fiber materials Which of the following materials are typically used as fibers? a) b) c) d) e) f) g) h) i)

Graphite/carbon Silicon carbide Silicon nitride Aluminum oxide Glass Boron Steel Tungsten Molybdenum

Solution Graphite, silicon carbide, glass, boron, and aluminum oxide are typically used as fibers.

Reserve Question 22: Fiber types Match each fiber type with its description.

Whiskers • Polycrystalline or amorphous materials with small diameters • Single crystals with extremely large length-to-diameter ratios • Metals having relatively large diameters Fibers • Single crystals with extremely large length-to-diameter ratios • Metals having relatively large diameters • Polycrystalline or amorphous materials with small diameters Wires • Metals having relatively large diameters • Polycrystalline or amorphous materials with small diameters • Single crystals with extremely large length-to-diameter ratios

Solution Whiskers are single crystals with extremely large length-to-diameter ratios. Fibers are polycrystalline or amorphous materials with small diameters. Wires are large-diameter metals.

Reserve Question 23: Fiber properties For a composite material, which phase normally has the higher elastic modulus? a) Fiber phase b) Matrix phase Solution The fiber phase normally has a higher elastic modulus than the matrix phase.

Reserve Question 24: Matrix properties For a composite material, how does the ductility of the matrix phase normally compare with the ductility of the dispersed phase? a) more ductile b) less ductile Solution A composite’s matrix phase is normally more ductile than the dispersed phase.

Reserve Question 25: Aramid fiber composites Aramid fiber-reinforced composites have very high tensile strengths and relatively low compressive strengths. a) True b) False Solution True. Aramid fiber-reinforced composites have very high tensile strengths and relatively low compressive strengths.

Reserve Question 26: Aramid fibers Which of aramid and metal fibers have higher strength-to-weight ratios? a) Aramid fibers b) Metal fibers

Solution Aramid fibers have higher strength-to-weight ratios than metal fibers.

Reserve Question 27: Carbon fiber composites Carbon fiber-reinforced composites have which of the following properties? a) Relatively high strengths b) Relatively high stiffnesses c) High service temperatures (>200°C) Solution Carbon fiber-reinforced composites have relatively high strengths, relatively high stiffnesses, and high service temperatures (> 200°C).

Reserve Question 28: Whisker materials Which of the following materials are typically used as whiskers? a) b) c) d) e) f) g) h) i)

graphite/carbon silicon carbide silicon nitride aluminum oxide glass boron steel tungsten molybdenum

Solution Graphite, silicon carbide, silicon nitride, and aluminum oxide are typically used as whiskers.

Reserve Question 29: Wire materials Which of the following materials are typically used as wires in composites? a) b) c) d) e) f) g) h) i)

graphite/carbon silicon carbide silicon nitride aluminum oxide glass boron steel tungsten molybdenum

Solution Steel, tungsten, and molybdenum are typically used as wires in composites.

Reserve Question 30: Ceramic-matrix composites Compared to other ceramic materials, ceramic-matrix composites have better/higher a) fracture toughnesses b) oxidation resistance c) stability at elevated temperatures Solution Ceramic-matrix composites have higher fracture toughnesses than other ceramic materials.

Reserve Question 31: Transformation toughened composites Which of the following materials are typically used as stabilizers in transformation-toughened ceramic-matrix composites? a) CaO b) MgO c) Y2O3 d) CeO e) Al2O3 f) SiC Solution CaO, MgO, Y2O3, and CeO are typically used as stabilizers in transformation-toughened ceramicmatrix composites.

Reserve Question 32: Carbon-carbon composites Carbon-carbon composites exhibit which of the following properties/characteristics? a) b) c) d) e) f) g) h)

High tensile moduli at elevated temperatures High tensile strengths at elevated temperatures Resistance to creep Large fracture toughness values High thermal conductivities Low coefficients of thermal expansion Resistance to oxidation at elevated temperatures Low cost

Solution Carbon-carbon composites have high tensile moduli at elevated temperatures, high tensile strengths at elevated temperatures, large fracture toughness values, high thermal conductivities, and low coefficients of thermal expansion, and they are highly resistant to creep.

Reserve Question 33: Laminar composites Laminar composites have high strengths in all directions (in three dimensions). a) True b) False

Solution False. Laminar composites have high strengths in all directions only in their two-dimensional planes.

Reserve Problem 34a (question pool) The figure below depicts four equal-sized composite plates created by laminating separate sheets of unidirectional Kevlar reinforced polyester. Which of the following laminate configurations would exhibit the lowest strain if elastically loaded with an axial force F, as depicted below? Select (e) if the strains for each of the laminate configurations are equivalent. a)

b)

c)

d)

Solution: The laminated configuration featuring the most alignment of the fibers should exhibit the highest modulus, specifically along the alignment axis. One of the options features loading along this axis of highest modulus, so this will be the scenario featuring the minimum strain, according to Hooke’s Law. Loading the completely aligned configuration parallel to the alignment axis maximizes the reinforcement offered by the aligned fibers. Therefore, the modulus follows the upper bound for the rule of mixtures.

Other configurations offer alignment axes that vary resulting in a lower modulus along the indicated loading axis.

Reserve Problem 35a (question pool) The figure below depicts four equal-sized composite plates created by laminating separate sheets of unidirectional Kevlar reinforced polyester. Which of the following laminate configurations is expected to exhibit the most anisotropic mechanical properties? Select (e) if the laminate configurations exhibit an equivalent degree of anisotropy. a)

b)

c)

d)

Solution: The laminated configuration featuring the most alignment of the fibers should exhibit the most anisotropy. The configuration whose unidirectional sheets are all aligned to the same angle (listed as o 0 ) should exhibit the most directionally dependent mechanical properties.

Reserve Question 36: Sandwich panels The strong outer sheets of sandwich panels are separated by a layer of material that is a) less dense than the outer sheet material b) more dense than the outer sheet material Solution The strong outer sheets of sandwich panels are separated by a layer of material that is less dense than the outer sheet material.

Reserve Questions and Problems Solutions Manual CHAPTER 16 CORROSION AND DEGRADATION OF MATERIALS

Reserve Problem 01: Electrochemical cell temperature An electrochemical cell is constructed such that on one side a pure lead electrode is in contact with a 2+ solution containing Pb ions at a concentration of 0.002 M. The other cell half consists of a pure 2+ nickel electrode that is immersed in a solution of Ni ions having a concentration of 0.5 M. Given that the standard electrode potentials for lead and nickel are –0.126 and –0.250 V, respectively, at what temperature will the potential between the two electrodes be –0.017 V? Solution This solution provides a guideline to solving the problem, not the numerical answer.

The cell potential between these two cell halves may be calculated from the following equation:

ΔV = (V20 − V10 ) −

n+ [Mn + ] RT [M1 ] 0.0592 ln n + = (V20 − V10 ) − log 1n + nℑ [M2 ] n [M2 ]

where the subscripts “1” and “2” denote the oxidized and reduced species, respectively. On the basis of the values of their standard electrode potentials, assume that Ni is oxidized and Pb is reduced.

Thus, the above equation takes the form 0 0 ΔV = (VPb )− − VNi

RT [Ni2+ ] 0.0592 [Ni2+ ] ln log = [( −0.126 − ( −0.250)] − nℑ [Pb2+ ] n [Pb2+ ]

Solving this equation for the temperature (K) yields

T=

0 0 −2ℑ [ΔV − (VPb − VNi )]

R ln

[Ni2+ ] [Pb2+ ]

Reserve Question 02: Chemical attack resistance Which material type is more resistant to attack by acidic and alkaline solutions?

a) Polymeric materials b) Metallic materials Solution Polymeric materials are more resistant to attack by acidic and alkaline solutions.

Reserve Question 03: EMF – least reactive Below are shown, for five metals, reduction reactions and standard electrode potential values. Which of these metals is the least reactive?

Electrode reaction

Standard electrode potential

(V) 3+

+ 3 e− --> Au

+1.420

2+

+ 2 e− --> Cu

+0.340

2+

+ 2 e− --> Ni

–0.250

2+

+ 2 e− --> Fe

–0.440

Na + e− --> Na

–2.924

Au

Cu Ni

Fe

+

a) b) c) d) e)

Au Cu Ni Fe Na

Solution Since Au has the most positive standard electrode potential, it is the least reactive metal.

Reserve Question 04: EMF – most reactive Below are shown, for five metals, reduction reactions and standard electrode potential values. Which of these metals is the most reactive?

Electrode reaction

Standard electrode potential

(V) 3+

+ 3 e− --> Au

+1.420

2+

+ 2 e− --> Cu

+0.340

2+

+ 2 e− --> Ni

–0.250

2+

+ 2 e− --> Fe

–0.440

Na + e− --> Na

–2.924

Au

Cu Ni

Fe

+

a) b) c) d) e)

Au Cu Ni Fe Na

Solution Since Na has the most negative standard electrode potential, it is the most reactive metal.

Reserve Question 05: Galvanic series I Metals near the top of the galvanic series are

a) b) c) d)

Cathodic Anodic Unreactive Reactive

Solution Metals near the top of the galvanic series are both cathodic and unreactive.

Reserve Question 06: Galvanic series II Metals near the bottom of the galvanic series are

a) b) c) d)

Cathodic Anodic Unreactive Reactive

Solution Metals near the bottom of the galvanic series are both anodic and reactive.

Reserve Question 07: Metal oxidation/reduction Metal ions in solution may only be oxidized.

a) True b) False Solution False. Metal ions in solution may either be oxidized or reduced. For example: 2+ 3+ − Oxidation aq --> Fe aq + e 2+ o Fe aq + 2 e− --> Fe s Reduction

Fe

Reserve Question 08: Multiple reduction reactions In a corrosion process, only one reduction reaction is possible.

a) True b) False

Solution False. It is possible for two or more reduction reactions to occur simultaneously.

Reserve Question 09: Overall electrochemical reaction An overall electrochemical reaction must consist of at least

a) A single reduction reaction b) A single oxidation reaction c) A single reduction reaction and a single oxidation reaction.

Solution An overall electrochemical reaction must consist of at least a single reduction reaction and a single oxidation reaction.

Reserve Question 10: Oxidation I Oxidation of an atom involves the

a) Loss of electrons b) Gain of electrons Solution Oxidation of an atom involves the loss of electrons.

Reserve Question 11: Oxidation II Oxidation takes place at the

a) Anode b) Cathode Solution Oxidation takes place at the cathode.

Reserve Question 12: Oxidation III Which of the following is/are oxidation reactions?

a) b) c) d)

2+

Fe Æ Fe + 2 e− 3+ Al Æ Al + 3 e− + 2 H + 2 e− Æ H2 + H2 Æ 2 H + 2 e−

Solution Oxidation of an atom involves the loss of electrons; thus, the following are oxidation reactions: 2+

Fe Æ Fe + 2 e− 3+ Al Æ Al + 3 e− + H2 Æ 2 H + 2 e− Reserve Question 13: Oxidation of metals Most metals and alloys are subject to oxidation.

a) True b) False Solution True. Most metals and alloys are subject to oxidation.

Reserve Question 14: Reduction I Reduction of an atom involves the

a) loss of electrons b) gain of electrons Solution Reduction of an atom involves the gain of electrons.

Reserve Question 15: Reduction II Reduction takes place at the

a) anode b) cathode Solution Reduction takes place at the cathode.

Reserve Question 16: Reduction III Which of the following are reduction reactions?

a) b) c) d)

2+

Fe Æ Fe + 2 e− 3+ Al Æ Al + 3 e− + 2 H + 2 e− Æ H2 + H2 Æ 2 H + 2 e−

Solution Reduction of an atom involves the gain of electrons; thus the following is a reduction reaction: +

2 H + 2 e− Æ H2 Reserve Question 17: Activation and concentration polarizations I Match the polarization conditions with their descriptions

Reaction rate is controlled by the slowest step in the electrochemical reaction.

• •

Activation polarization Concentration polarization

Reaction rate is controlled by the diffusion of ions in solution.

• •

Activation polarization Concentration polarization

Solution For activation polarization, the reaction rate is controlled by the slowest step in the electrochemical reaction. For concentration polarization, the reaction rate is controlled by the diffusion of ions in solution.

Reserve Question 18: Passivity – maintenance Once a metal has become passivated, it will always remain so.

a) True b) False Solution False. Once a metal has passivated, changes in the environment may cause it to become electrochemically active.

Reserve Question 19: Passivity – metals Alloys of which of the following metals may passivate?

a) b) c) d) e) f)

Chromium Iron Nickel Titanium Copper Gold

Solution Alloys of chromium, iron, nickel, and titanium may passivate.

Reserve Question 20: Environmental effects Which of the following factors may influence the corrosion rates of materials?

a) Fluid velocity b) Temperature c) Fluid composition Solution Fluid velocity, temperature, and fluid composition all influence the corrosion rates of materials.

Reserve Question 21: Forms of corrosion – crevice corrosion Which of the following describes crevice corrosion?

a) Oxidation and reduction reactions occur randomly over the surface. b) Two metals/alloys of different compositions are coupled while exposed to an electrolyte. c) Corrosion that results from a difference in concentration of ions or dissolved gases in the electrolyte. d) Localized corrosion that may be initiated at the surface defect. e) Corrosion that occurs preferentially along grain boundaries. f) One element is preferentially removed as a result of corrosion. g) Corrosion resulting from the combined action of chemical attach and mechanical abrasion or wear. h) Corrosion resulting from the combined action of an applied tensile stress and a corrosive environment.

Solution Crevice corrosion results from a difference in concentration of ions or dissolved gases in the electrolyte.

Reserve Question 22: Forms of corrosion – erosion corrosion Which of the following describes erosion-corrosion?

a) Oxidation and reduction reactions occur randomly over the surface. b) Two metals/alloys of different compositions are coupled while exposed to an electrolyte. c) Corrosion that results from a difference in concentration of ions or dissolved gases in the electrolyte. d) Localized corrosion that may be initiated at the surface defect. e) Corrosion that occurs preferentially along grain boundaries.

f) One element is preferentially removed as a result of corrosion. g) Corrosion resulting from the combined action of chemical attach and mechanical abrasion or wear. h) Corrosion resulting from the combined action of an applied tensile stress and a corrosive environment.

Solution Erosion-corrosion results from the combined action of chemical attack and mechanical abrasion or wear.

Reserve Question 23: Forms of corrosion – Galvanic corrosion Which of the following describes galvanic corrosion?

a) Oxidation and reduction reactions occur randomly over the surface. b) Two metals/alloys of different compositions are coupled while exposed to an electrolyte. c) Corrosion that results from a difference in concentration of ions or dissolved gases in the electrolyte. d) Localized corrosion that may be initiated at the surface defect. e) Corrosion that occurs preferentially along grain boundaries. f) One element is preferentially removed as a result of corrosion. g) Corrosion resulting from the combined action of chemical attach and mechanical abrasion or wear. h) Corrosion resulting from the combined action of an applied tensile stress and a corrosive environment.

Solution Galvanic corrosion involves two metals/alloys of different compositions that are coupled while exposed to an electrolyte.

Reserve Question 24: Forms of corrosion – Intergranular corrosion Which of the following describes intergranular corrosion?

a) Oxidation and reduction reactions occur randomly over the surface. b) Two metals/alloys of different compositions are coupled while exposed to an electrolyte. c) Corrosion that results from a difference in concentration of ions or dissolved gases in the electrolyte. d) Localized corrosion that may be initiated at the surface defect. e) Corrosion that occurs preferentially along grain boundaries. f) One element is preferentially removed as a result of corrosion.

g) Corrosion resulting from the combined action of chemical attach and mechanical abrasion or wear. h) Corrosion resulting from the combined action of an applied tensile stress and a corrosive environment.

Solution Intergranular corrosion occurs preferentially along grain boundaries.

Reserve Question 25: Forms of corrosion – Pitting Which of the following describes pitting?

a) Oxidation and reduction reactions occur randomly over the surface. b) Two metals/alloys of different compositions are coupled while exposed to an electrolyte. c) Corrosion that results from a difference in concentration of ions or dissolved gases in the electrolyte. d) Localized corrosion that may be initiated at the surface defect. e) Corrosion that occurs preferentially along grain boundaries. f) One element is preferentially removed as a result of corrosion. g) Corrosion resulting from the combined action of chemical attach and mechanical abrasion or wear. h) Corrosion resulting from the combined action of an applied tensile stress and a corrosive environment.

Solution Pitting involves localized corrosion that may be initiated at a surface defect.

Reserve Question 26: Forms of corrosion – Selective leaching Which of the following describes selective leaching?

a) Oxidation and reduction reactions occur randomly over the surface. b) Two metals/alloys of different compositions are coupled while exposed to an electrolyte. c) Corrosion that results from a difference in concentration of ions or dissolved gases in the electrolyte. d) Localized corrosion that may be initiated at the surface defect. e) Corrosion that occurs preferentially along grain boundaries. f) One element is preferentially removed as a result of corrosion. g) Corrosion resulting from the combined action of chemical attach and mechanical abrasion or wear. h) Corrosion resulting from the combined action of an applied tensile stress and a corrosive environment.

Solution Selective leaching occurs when one element is preferentially removed as a result of corrosion.

Reserve Question 27: Forms of corrosion – stress corrosion Which of the following describes stress corrosion?

a) Oxidation and reduction reactions occur randomly over the surface. b) Two metals/alloys of different compositions are coupled while exposed to an electrolyte. c) Corrosion that results from a difference in concentration of ions or dissolved gases in the electrolyte. d) Localized corrosion that may be initiated at the surface defect. e) Corrosion that occurs preferentially along grain boundaries. f) One element is preferentially removed as a result of corrosion. g) Corrosion resulting from the combined action of chemical attach and mechanical abrasion or wear. h) Corrosion resulting from the combined action of an applied tensile stress and a corrosive environment.

Solution Stress corrosion results from the combined action of an applied tensile stress and a corrosive environment.

Reserve Question 28: Forms of corrosion – uniform corrosion Which of the following describes uniform corrosion?

a) Oxidation and reduction reactions occur randomly over the surface. b) Two metals/alloys of different compositions are coupled while exposed to an electrolyte. c) Corrosion that results from a difference in concentration of ions or dissolved gases in the electrolyte. d) Localized corrosion that may be initiated at the surface defect. e) Corrosion that occurs preferentially along grain boundaries. f) One element is preferentially removed as a result of corrosion. g) Corrosion resulting from the combined action of chemical attach and mechanical abrasion or wear. h) Corrosion resulting from the combined action of an applied tensile stress and a corrosive environment.

Solution Uniform corrosion involves oxidation and reduction reactions that occur randomly over the surface.

Reserve Question 29: Fresh vs seawater Which water environment is more corrosive?

a) Fresh water b) Seawater Solution Seawater, which contains about 3.5% salt, is generally more corrosive than freshwater.

Reserve Question 30: Corrosion prevention – cathodic protection Cathodic protection reduces the corrosion rate of a metal when an electrical connection is made to another metal that is

a) lower in the galvanic series b) higher in the galvanic series Solution Cathodic protection reduces the corrosion rate of a metal by making an electrical connection to another metal that is lower in the galvanic series.

Reserve Question 31: Corrosion prevention – temperature In most instances, one way to reduce the corrosion rate is to

a) Increase the fluid temperature. b) Decrease the fluid temperature.

Solution Decreasing the fluid temperature, in most cases, reduces the corrosion rate.

Reserve Question 32: Galvanizing Galvanizing involves applying a layer of what metal to the surface of steel?

a) b) c) d) e)

zinc iron copper nickel aluminum

Solution Galvanizing involves applying a layer of zinc to the surface of steel.

Reserve Question 33: Pilling – Bedworth ratio I Given the following metal-Pilling-Bedworth ratio combinations, which metals are expected to form protective coatings?

a) b) c) d) e) f) g)

Metal

P-B Ratio

Aluminum

1.28

Chromium

1.99

Copper

1.68

Lithium

0.57

Molybdenum

3.40

Silicon

2.27

Sodium

0.57

aluminum chromium copper lithium molybdenum silicon sodium

Solution A metal normally forms a protective coating if its Pilling-Bedworth ratio is between 1 and 2; thus aluminum, chromium, and copper are expected to form protective coatings.

Reserve Question 34: Pilling – Bedworth ratio II Given the following metal-Pilling-Bedworth ratio combinations, which metals are not expected to form protective coatings?

Metal

P-B Ratio

Aluminum

1.28

Chromium

1.99

Copper

1.68

Lithium

0.57

Molybdenum

3.40

Silicon

2.27

Sodium

0.57

a) b) c) d) e) f) g)

aluminum chromium copper lithium molybdenum silicon sodium

Solution A metal normally does not form a protective coating if its Pilling-Bedworth ratio is less than 1 or greater than 2; thus, lithium, molybdenum, silicon, and sodium are not expected to form protective coatings.

Reserve Question 35: Polymer degradation Polymer degradation may occur from exposure to or as a result of which of the following?

a) b) c) d) e)

Electrochemical reactions Physiochemical reactions Swelling Heat Radiation

Solution Polymer degradation may occur from exposure to heat and radiation, and as a result of physiochemical reactions and swelling.

Reserve Question 36: Swelling The swelling of a polymer results in it becoming

a) Softer b) Harder Solution The swelling of a polymer results in it becoming softer due to the diffusion of solute molecules to positions in between the polymer molecules.

Reserve Question 37: Swelling reduction Polymer deterioration by swelling may be reduced by

a) b) c) d) e) f)

Increasing the degree of crosslinking Decreasing he degree of crosslinking Increasing the molecular weight Decreasing the molecular weight Increasing the degree of crystallinity Decreasing the degree of crystallinity

Solution Polymer deterioration by swelling may be reduced by increasing the degree of crosslinking, increasing the molecular weight, and increasing the degree of crystallinity.

Reserve Questions and Problems Solutions Manual CHAPTER 17 THERMAL PROPERTIES

Reserve Question 01: Constant pressure vs constant volume heat capacity For a given material, which heat capacity is greater? a) At constant pressure, Cp. b) At constant volume, Cv. Solution For a given material, the constant pressure heat capacity, Cp, is always greater than the constant volume value.

Reserve Question 02: Debye temperature I Below the Debye temperature, Cv is virtually independent of temperature. a) True b) False Solution False. Above (not below) the Debye temperature, Cv is virtually independent of temperature.

Reserve Question 03: Debye temperature II For most solid materials, the Debye temperature is below room temperature. a) True b) False Solution True. For most solid materials, the Debye temperature is below room temperature.

Reserve Question 04: Electron contributions In which of the following thermal phenomena do free electrons play a role? a) Thermal conductivity b) Thermal expansion

Solution Free electrons play a role in thermal conduction.

Reserve Question 05: Material thermal expansion Which of the following material types typically has the largest values of the coefficient of thermal expansion? a) Polymers b) Metals c) Ceramics Solution Polymers typically have the largest coefficient of thermal expansion values.

Reserve Question 06: Thermal expansion and atomic bonding energy The greater the atomic bonding energy, a) The larger the interatomic spacing change with increasing temperature. b) The smaller the interatomic spacing change with increasing temperature. c) The larger the value of α l d) The smaller the value of α l Solution The greater the atomic bonding energy, the smaller the interatomic spacing change with increasing temperature, and the smaller the value of α l .

Reserve Question 07: Thermal expansion anisotropy When heated, most materials expand equally in all directions. a) True b) False Solution False. When heated, most materials do not expand equally in all directions; this is indicated by the value of the volume coefficient of thermal expansion, α v, which depends on the crystallographic direction of measurement.

Reserve Problem 08 The difference between the specific heats at constant pressure and volume is described by the expression

c p − cv −

α 2v v0T β

where α v is the volume coefficient of thermal expansion, v0 is the specific volume (i.e. volume per unit mass, or the reciprocal of density), β is the compressibility, and T is the absolute temperature.

Compute he values of Cv , at room temperature (293 K) for copper and nickel using the data in Table –12 17.1, assuming that α v = 3al and given that the values for β for Cu and Ni are 8.35 × 10 and 5.51 × –12 –1 10 (Pa) , respectively.

Solution This problem asks that we calculate the values of Cv for copper and nickel at room temperature using Equation 17.10, the data in Table 17.1, given that αv = 3αl, and also values of the compressibility. From Equation 17.10

α 2v v0T cv = c p − β And, from Table 17.1 and the problem statement cp(Cu) = 386 J/kg-K cp(Ni) = 443 J/kg-K –6 –1 –5 –1 αv(Cu) = (3)[17.0 × 10 (°C) ] = 5.10 × 10 (°C) –6 –1 –5 –1 αv(Ni) = (3)[(13.3 × 10 (°C) ] = 3.99 × 10 (°C) –12 –1 β(Cu) = 8.35 × 10 (Pa) –12 –1 β(Ni) = 5.51 × 10 (Pa) 3

The specific volume is just the reciprocal of the density; thus, in units of m /kg ⎛ ⎞ ⎛ 1000 g ⎞ ⎛ 1 m ⎞3 1 ⎟⎜ =⎜ v0 (Cu) = ⎟ = 1.119 × 10 −4 m3 /kg ⎟⎜ ρ ⎜ 8.94 g/cm3 ⎟ ⎜⎝ kg ⎟⎠ ⎜⎝ 100 cm ⎟⎠ ⎝ ⎠ 1

⎛ ⎞ ⎛ 1000 g ⎞ ⎛ 1 m ⎞3 1 ⎜ ⎟⎜ v0 (Ni) = ⎟ = 1.124 × 10 −4 m 3 /kg ⎟⎟ ⎜ 3 ⎜ ⎜ ⎜ 8.90 g/cm ⎟ ⎝ kg ⎠ ⎝ 100 cm ⎟⎠ ⎝ ⎠

Therefore, for copper

cv (Cu) = c p (Cu) −

α 2v (Cu) v0 (Cu)T β (Cu) 2

⎡5.10 × 10 −5 (°C) −1 ⎤ (1.119 × 10 −4 m 3 /kg ) (293 K) ⎦ = 386 J/kg-K − ⎣ 8.35 × 10 −12 (N/m 2 ) −1

= 376 J/kg-K And, also for nickel

c v (Ni) = 443 J/kg-K −

⎡ −5 (°C) −1 ⎤⎥ ⎢3.99 × 10 ⎣ ⎦

2

(1.124 × 10 −4 m3 /kg ) (293 K)

5.51 × 10 −12 (N/m 2 ) −1

= 433 J/kg-K

Reserve Question 09: Material thermal conductivity Which of the following material types typically has the largest values of the thermal conductivity? a) Polymers b) Metals c) Ceramics Solution Metals typically have the largest values of thermal conductivity.

Reserve Question 10: Thermal conductivity of ceramics In the vicinity of room temperature, how do the thermal conductivities of ceramics change as the temperature increases? a) they decrease b) they increase Solution In the vicinity of room temperature, the thermal conductivities of ceramics decrease as the temperature increases.

Reserve Question 11: Thermal conductivity of polycrystalline vs single crystal How do thermal conductivities of single crystals and polycrystalline materials compare? a) ksingle crystal > kpolycrystalline b) ksingle crystal < kpolycrystalline Solution The thermal conductivity of a single crystal is greater than that of the polycrystalline material because both phonons and free electrons are scattered at grain boundaries; thus, the efficiency of thermal transport is lower for the polycrystalline material.

Reserve Question 12: Thermal shock in brittle materials Which of rapid heating or rapid cooling is more likely to inflict thermal shock in brittle materials?

a) Rapid cooling b) Rapid heating Solution Rapid cooling is more likely to inflict thermal shock in brittle materials.

Reserve Question 13: Thermal shock resistance Which of the following properties lead to a high degree of thermal shock resistance? a) b) c) d) e) f) g) h)

High fracture strength Low fracture strength High thermal conductivity Low thermal conductivity High modulus of elasticity Low modulus of elasticity High coefficient of thermal expansion Low coefficient of thermal expansion

Solution A material that has a high fracture strength, a high thermal conductivity, a low modulus of elasticity, and a low coefficient of thermal expansion will be resistant to thermal shock.

Reserve Question 14: Thermal stress The heating of a homogenous and isotropic rod of solid material for which axial motion is restrained, results in which type of thermal stresses? a) Compressive b) Tensile

Solution Heating of a homogenous and isotropic rod of solid material with restraining axial motion results in compressive thermal stresses.

Reserve Problem 15 (a) If a rod of 1025 steel 0.5 m (19.7 in.) long is heated from 20°C to 80°C (68°F to 176°F) while its ends are maintained rigid, determine the type and magnitude of stress that develops. Assume that at 20°C the rod is stress-free. (b) What will be the stress magnitude if a rod 1 m (39.4 in.) long is used?

–

(c) If the rod in part (a) is cooled from 20°C to 10°C (68°F to 14°F), what type and magnitude of stress will result?

Solution (a) We are asked to compute the magnitude of the stress within a 1025 steel rod that is heated while its ends are maintained rigid. To do this we employ Equation 17.8, using a value of 207 GPa for –6 –1 the modulus of elasticity of steel (Table 7.1), and a value of 12.0 × 10 (°C) for αl (Table 17.1).

Therefore

σ = Eαl (T0 − Tf ) = ( 207 × 10 3 M Pa ) ⎡12.0 × 10 − 6 ( °C) − 1 ⎤ (20 °C − 80 °C) ⎣ ⎦

= –150 MPa (–21,800 psi) The stress will be compressive since its sign is negative. (b) The stress will be the same [–150 MPa (–21,800 psi)], since stress is independent of bar length. (c) Upon cooling the indicated amount, the stress becomes

σ = Eαl (T0 − Tf ) = ( 207 × 10 3 MPa ) ⎡12.0 × 10 − 6 ( °C) − 1 ⎤ ⎡⎣ (20 °C − ( − 10 °C) ⎤⎦ ⎣ ⎦

= +74.5 MPa (+10,900 psi) This stress will be tensile since its sign is positive.

Reserve Problem 16 To what temperature would 23.0 kg of some material at 100°C be raised if 255 kJ of heat is supplied?

Assume a cp value of 423 J/kg-K for this material.

(A) 26.2°C (B) 73.8°C (C) 126°C (D) 152°C Solution The change in temperature (ΔT) when ΔQ of heat is supplied to a mass of material m is equal to a revised form of Equation 17.1 as follows:

ΔQ = ΔT = Tf −T0 c pm where T0 and Tf are the initial and final temperatures, respectively. Solving this equation for the final temperature yields the following:

Tf = T0 +

ΔQ c pm

Using values provided in the problem statement—that is T0 = 100°C ΔQ = 255 kJ = 255,000 J cp = 423 J/kg-K m = 23.0 kg We compute the final temperature as follows: T f = 100°C +

which is answer C.

255,000 J = 126°C (423 J/kg-K)(23.0 kg)

Reserve Problem 17 A rod of some material 0.50 m long elongates 0.40 mm on heating from 50°C to 151°C. What is the value of the linear coefficient of thermal expansion for this material?

–6

–1

(A) 5.30 × 10 (°C) –6 –1 (B) 7.92 × 10 (°C) –5 –1 (C) 1.60 × 10 (°C) –6 –1 (D) 1.24 × 10 (°C) Solution The relationship among change in length (Δl), coefficient of thermal expansion (αl), and change in temperature (ΔT) is given by Equation 17.3b—that is

Δl = αl ΔT = αl (Tf − T0 ) l0 where Tf and T0 are the final and initial temperatures, respectively. Thus, the linear coefficient of thermal expansion may be determined using a rearranged form of the above equation as follows:

αl =

Δl l0 (Tf − T0 )

Incorporation of values of Δl, l0, Tf, and T0 provided in the problem statements leads to the following value of αl αl =

(0.40 × 10 − 3 m) = 7.92 × 10 − 6 ( °C) − 1 (0.50 m)(151 °C − 50 °C)

which is answer B. Reserve Problem 18 Which of the following sets of properties leads to a high degree of thermal shock resistance?

(A) High fracture strength High thermal conductivity High modulus of elasticity High coefficient of thermal expansion (B) Low fracture strength Low thermal conductivity Low modulus of elasticity Low coefficient of thermal expansion (C) High fracture strength High thermal conductivity

Low modulus of elasticity Low coefficient of thermal expansion (D) Low fracture strength Low thermal conductivity High modulus of elasticity High coefficient of thermal expansion Solution The correct answer is C. Thermal shock resistance is defined by Equation 17.9; therefore, a material that has a high TSR will have a high fracture strength, a high thermal conductivity, a low modulus of elasticity, and a low coefficient of thermal expansion.

Reserve Problem 19a (question pool) A 10 meter long square bar of 316 stainless steel (edge length of 5 cm, with a modulus of 193 GPa and a yield point of 290 MPa) is bolted securely in place when its installation temperature was around [I]°C. What is the expected thermal stress in the bar when its service temperature reaches [F]°C? Enter a negative indicate compressive stress, if necessary. The thermal expansion coefficient –6 of 316 stainless steel is 16.0 × 10 1/°C. Solution: In this scenario, the bar is installed and secured at a warmer temperature than its ultimate service temperature. Therefore, as the temperature drops the steel would like to contract and occupy less length, but it cannot do so. The bar instead endures compressive thermal stress via Hooke’s Law (provided the thermal strain is elastic).

σ th = Eε th = Eα L ΔT = Eα L (To − Tf ) = (193,000 MPa)(16 × 10 −6 1/° C)([ I ]° C− [F ]° C) Reserve Problem 20a (question pool) Which of the following 1 kg samples is expected to change temperature the least if 100 kJ of heat is perfectly transferred to each of them at a constant pressure of 1 atmosphere. The initial temperature of each specimen is 25°C. (a) Aluminum (b) Copper (c) Gold (d) Borosilicate Glass (e) Polystyrene

Solution: In this scenario, the material with the highest heat capacity will change temperature the least for a given increase in enthalpy. Polystyrene features the highest heat capacity, so it will experience the smallest temperature change, according to: ΔT =

Q mc p

Reserve Problem 21a (question pool) If you were to locally heat identical geometry plates of the materials listed below with the same heat source, which would increase in temperature the fastest? (a) Polystyrene (b) Aluminum (c) Copper (d) Gold (e) Borosilicate Glass Solution: In this scenario, the material with the highest thermal conductivity will conduct heat the fastest.

Therefore this material will absorb heat and increase in temperature the fastest. In this case, copper has the highest thermal conductivity.

Reserve Problem 22 This ceramic outlier has the highest room-temperature thermal conductivity of about 2,000 watts per meter per kelvin, a value that is five time higher than the best thermally conductive metals. (a) Al2O3 (b) CaF2 (c) TiO2 (d) SrTiO3 (e) C (diamond) Solution: Diamond has the highest thermal conductivity of these options.

Reserve Questions and Problems Solutions Manual CHAPTER 18 MAGNETIC PROPERTIES

Reserve Question 01: Ferromagnetics Which two of the following are ferromagnetic materials? a) b) c) d) e) f) g)

Aluminum oxide Copper Aluminum Titanium Iron (α ferrite) Nickel MnO

h) Fe3O4 i) NiFe2O4 Solution Iron (α ferrite) and nickel are ferromagnetic materials.

Reserve Question 02: Antiferromagnetics Which of the following materials display(s) antiferromagnetic behavior? a) b) c) d) e) f) g)

Aluminum oxide Copper Aluminum Titanium Iron (α ferrite) Nickel MnO

h) Fe3O4 i) NiFe2O4 Solution MnO is an antiferromagnetic material.

Reserve Question 03: Ferrimagnetics Which of the following are ferrimagnetic materials? a) b) c) d) e) f) g)

Aluminum oxide Copper Aluminum Titanium Iron (α ferrite) Nickel MnO

h) Fe3O4 i) NiFe2O4 Solution Fe3O4 and NiFe2O4 are ferrimagnetic materials.

Reserve Problem 04 c

a

d

The formula for yttrium iron garnet (Y3Fe5O12) may be written in the form Y3 Fe2 Fe3 O12 , where the 3+ 3+ superscripts a, c, and d represent different sites on which the Y and Fe ions are located. The spin 3+ 3+ magnetic moments for the Y and Fe ions positioned in the a and c sites are oriented parallel to 3+ one another and antiparallel to the Fe ions in d sites. Compute the number of Bohr magnetons 3+ associated with each Y ion, given the following information: (1) each unit cell consists of eight formula (Y3Fe5O12) units; (2) the unit cell is cubic with an edge length of 1.2376 nm; (3) the 4 saturation magnetization for this material is 1.0 × 10 A/m; and (4) assume that there are 5 Bohr 3+ magnetons associated with each Fe ion.

Solution c

a

d

For this problem, we are given that yttrium iron garnet may be written in the form Y3 Fe2 Fe3 O12 3+ 3+ where the superscripts a, c, and d represent different sites on which the Y and Fe ions are located, and that the spin magnetic moments for the ions on a and c sites are oriented parallel to one 3+ another and antiparallel to the Fe ions on the d sites. We are to determine the number of Bohr 3+ magnetons associated with each Y ion given that each unit cell consists of eight formula units, the unit cell is cubic with an edge length of 1.2376 nm, the saturation magnetization for the material is 4 3+ 1.0 × 10 A/m, and that there are 5 Bohr magnetons for each Fe ion.

The first thing to do is to calculate the number of Bohr magnetons per unit cell, which we will denote nB.

Solving for nB using Equation 18.13, we get nB =

=

M s a3

μB

(1.0 × 104 A/m)(1.2376 × 10−9 m)3 9.27 × 10−24 A-m2 /BM

Now, there are 8 formula units per unit cell or

= 2.04 Bohr magnetons/unit cell

2.04 = 0.255 Bohr magnetons per formula unit.

8

Furthermore, for each formula unit there are two Fe

3+

ions on a sites and three Fe

3+

on d sites which

magnetic moments are aligned antiparallel. Since there are 5 Bohr magnetons associated with each Fe

3+

3+

ion, the net magnetic moment contribution per formula unit from the Fe

magnetons. This contribution is antiparallel to the contribution from the Y 3+

are three Y

3+

ions is 5 Bohr

ions, and since there

ions per formula unit, then

No. of Bohr magnetons/Y 3+ =

0.255 BM + 5 BM 3

= 1.75 BM

Reserve Question 05: Curie temperature At the Curie temperature, the saturation magnetization abruptly diminishes. Which of the following magnetic material types will have Curie temperatures? a) b) c) d) e)

Diamagnetics Paramagnetics Ferromagnetics Antiferromagnetics Ferrimagnetics

Solution Ferromagnetic and ferrimagnetic materials will have Curie temperatures.

Reserve Question 06: Neel temperature With increasing temperature antiferromagnetic materials eventually become which of the following? a) b) c) d) e)

Diamagnetics Paramagnetics Ferromagnetics Antiferromagnetics Ferrimagnetics

Solution With increasing temperature antiferromagnetic materials eventually become paramagnetic.

Reserve Question 07: Domains and crystals In a polycrystalline material, each grain will always consist of just a single domain. a) True b) False Solution False. In a polycrystalline material, each grain may consist of more than one domain.

Reserve Problem 08 A coil of wire 0.1 m long and having 15 turns carries a current of 1.0 A. (a) Compute the flux density if the coil is within a vacuum. (b) A bar of an iron–silicon alloy, the B-H behavior for which is shown in Figure 18.29, is positioned within the coil. What is the flux density within this bar? (c) Suppose that a bar of molybdenum is now situated within the coil. What current must be used to produce the same B field in the Mo as was produced in the iron–silicon alloy part (b) using 1.0 A?

Solution (a) This portion of the problem asks that we compute the flux density in a coil of wire 0.1 m long, having 15 turns, and carrying a current of 1.0 A, and that is situated in a vacuum. Combining Equations 18.1 and 18.3, and solving for B yields

B0 = μ0 H = =

μ0 NI l

(1.257 × 10 −6 H/m) (15 turns) (1.0 A) 0.1 m

= 1.89 × 10 −4 tesla

(b) Now we are to compute the flux density with a bar of the iron-silicon alloy, the B-H behavior for which is shown in Figure 18.29. It is necessary to determine the value of H using Equation 18.1 as

H=

NI (15 turns)(1.0 A) = = 150 A-turns/m l 0.1 m

Using the curve in Figure 18.29, B = 1.65 tesla at H = 150 A-turns/m, as demonstrated below.

(c) Finally, we are to assume that a bar of Mo is situated within the coil, and to calculate the current that is necessary to produce the same B field as when the iron-silicon alloy in part (b) was used.

-4 Molybdenum is a paramagnetic material having a χm of 1.19 × 10 (Table 18.2). Combining Equations 18.2, 18.4, and 18.7 we solve for H

H=

B

μ

=

B

μ0 μr

=

B μ0 (1 + χ m )

And when Mo is positioned within the coil, then, from the above equation

H=

1.65 tesla −6

−4

(1.257 × 10 H/m)(1 + 1.19 × 10 )

=1.312 × 106 A-turns/m

Now, the current may be determined using Equation 18.1:

I=

6 Hl (1.312 × 10 A-turns / m) (0.1 m) = = 8750 A N 15 turns

Reserve Problem 09 The following data are for a transformer steel: H (A/m)

B (tesla)

H (A/m)

B (tesla)

0

0

200

1.04

10

0.03

400

1.28

20

0.07

600

1.36

50

0.23

800

1.39

100

0.70

1000

1.41

150

0.92

(a) What are the values of the initial permeability and initial relative permeability? (b) What is the value of the maximum permeability? (c) At about what H field does this maximum permeability occur? (d) To what magnetic susceptibility does this maximum permeability correspond?

Solution The B-H data for the transformer steel provided in the problem statement are plotted below.

(a) The first four data points are plotted below.

The slope of the initial portion of the curve is μi (as shown), is

μi =

(0.15 − 0) tesla ΔB = =3.0 × 10−3 H/m ΔH (50 − 0) A/m

Also, the initial relative permeability, μri, (Equation 18.4) is just −3

3.0 × 10 H/m μ μri = i = =2387 μ0 1.257 × 10 −6 H/m

(b) The maximum permeability is the tangent to the B-H curve having the greatest slope; it is drawn on the plot below, and designated as μ(max).

The value of μ(max) is (modifying Equation 18.2)

μ (max) =

ΔB (1.3 − 0.3) tesla = = 8.70 × 10−3 H/m ΔH (160 − 45) A-m

(c) The H field at which μmax) occurs is approximately 80 A/m [as taken from the plot shown in part (b)]. (d) We are asked for the maximum susceptibility, χ(max). Combining modified forms of Equations 18.7 and 18.4 yields

χ (max) = μr (max) − 1 =

=

μ (max) −1 μ0

8.70 × 10 −3 H/m 1.257 × 10 −6 H/m

− 1 = 6920

Reserve Problem 10 Estimate saturation values of H for single-crystal iron in [100], [110], and [111] directions.

Solution This problem asks for us to estimate saturation values of H for single crystal iron in the [100], [110], and [111] directions. All we need do is read values of H at the points at which saturation is achieved on the [100], [110], and [111] curves for iron shown in Figure 18.17. Saturation in the [100] direction is approximately 5400 A/m. Corresponding values in [110] and [111] directions are approximately 40,000 and 47,000 A/m, respectively.

Reserve Question 11: Hard/soft materials Which type(s) of magnetic materials may be classified as either soft or hard?

a) b) c) d) e)

Diamagnetic Paramagnetic Ferromagnetic Antiferromagnetic Ferrimagnetic

Solution Ferromagnetic and ferrimagnetic materials may be classified as either soft or hard.

Reserve Question 12: Soft magnetic materials Which of the following characteristics are displayed by soft magnetic materials?

a) b) c) d)

A relatively small hysteresis loop.

A relatively large hysteresis loop.

Magnetization and demagnetization may be achieved using relatively low applied fields.

Magnetization and demagnetization require relatively high applied fields.

Solution Soft materials have relatively small relative hysteresis loops; also, magnetization and demagnetization may be achieved using relatively low applied fields.

Reserve Question 13: Hard magnetic materials Which of the following characteristics are displayed by hard magnetic materials?

a) b) c) d)

A relatively small hysteresis loop.

A relatively large hysteresis loop.

Magnetization and demagnetization may be achieved using relatively low applied fields.

Magnetization and demagnetization require relatively high applied fields.

Solution Hard materials have relatively large hysteresis loops; also, magnetization and demagnetization require relatively high applied fields.

Reserve Problem 14 Using Equation 18.14, determine which of the superconducting elements in Table 20.7 are superconducting at 3 K and in a magnetic field of 15,000 A/m.

Solution

We are asked to determine which of the superconducting elements in Table 18.7 are superconducting at 3 K and in a magnetic field of 15,000 A/m. First of all, in order to be superconductive at 3 K within any magnetic field, the critical temperature must be greater than 3 K. Thus, aluminum, titanium, and tungsten may be eliminated upon inspection. Now, for each of lead, mercury, and tin it is necessary, using Equation 18.14, to compute the value of HC(3)—also substituting for HC(0) from Equation 18.3; if HC(3) is greater than 15,000 A/m then the element will be superconductive. Hence, for Pb ⎡ T 2 ⎤ B (0) ⎡ T2 ⎤ HC (2) = HC (0) ⎢1 − 2 ⎥ = C ⎢1 − 2 ⎥ μ0 ⎢⎣ TC ⎥⎦ TC ⎥⎦ ⎢⎣

⎡ (3.0 K)2 ⎤ ⎢ ⎥ = 5.28 × 10 4 A/m = 1− 2 −6 1.257 × 10 H/m ⎢⎣ (7.19 K) ⎥⎦ 0.0803 tesla

Since this value is greater than 15,000 A/m, Pb will be superconductive.

Similarly for Hg

HC (3) =

⎡ (3.0 K)2 ⎤ ⎢1 − ⎥ =1.56 × 10 4 A/m 2 −6 1.257 × 10 H/m ⎢⎣ (4.15 K) ⎥⎦ 0.0411 tesla

Inasmuch as this value is greater than 15,000 A/m, Hg will be superconductive.

As for Sn

⎡ (3.0 K)2 ⎤ ⎢ ⎥ = 8.48 × 103 A/m HC (3) = 1− 2 −6 1.257 × 10 H/m ⎢⎣ (3.72 K) ⎥⎦ 0.0305 tesla

Therefore, Sn is not superconductive.

Reserve Questions and Problems Solutions Manual CHAPTER 19 OPTICAL PROPERTIES

Reserve Problem 01: Wavelength calculations What are (a) the frequency, and (b) the energy (in eV) of a photon of light having a wavelength of 8.8 –7 –15 × 10 m? The value for Planck’s constant is 4.13 × 10 eV-s. Express your answer to part (a) in scientific notation.

Solution This solution provides a guideline to solving the problem, not the numerical answer. (a) The relationship between wavelength and frequency is c = λν 8

where c, the velocity of light, has a value of 3 × 10 m/s. Therefore, the frequency is equal to

ν=

c

λ

(b) The energy of a photon may be calculated using either the wavelength or the frequency as

E = hν = where h, Planck's constant, has a value of 4.13 × 10

–15

hc

λ eV-s.

Reserve Question 02: Visible wavelength What is the order of magnitude wavelength for visible light? a) b) c) d) e) f)

0.5 Angstroms 0.5 nanometers 0.5 micrometers 0.5 millimeters 0.5 meters 0.5 kilometers

Solution The wavelength of visible light is on the order of 0.5 micrometers.

Reserve Question 03: Bulk metal light transmission In the visible spectrum, a thick metal specimen will be a) Transparent b) Translucent c) Opaque Solution In the visible spectrum, a thick metal specimen is opaque.

Reserve Question 04: Insulator light transmission In the visible spectrum, an electrical insulator that is a single crystal and without porosity is normally a) Transparent b) Opaque c) Translucent Solution In the visible spectrum, an electrical insulator that is a single crystal and without porosity is normally transparent.

Reserve Question 05: Light transmission Match the type of light transmission with its description, Transmits light with relative little absorption.

Translucent Transparent Opaque

Transmits light diffusely Opaque Translucent Transparent Is impervious to light transmission.

Transparent Opaque Translucent

Solution A transparent material transmits light with relatively little absorption.

A translucent material transmits light diffusely.

An opaque material is impervious to light transmission.

Reserve Question 06: Semiconductor light transmission In the visible spectrum, a semiconductor that is a single crystal and nonporous may be a) Transparent b) Translucent c) Opaque Solution In the visible spectrum, a semiconductor that is a single crystal and nonporous may be transparent or opaque.

Reserve Question 07: EM radiation and metals To which of the following electromagnetic radiation types are bulk metals opaque? a) b) c) d) e)

radio waves microwaves infrared radiation ultraviolet radiation X-rays

Solution Bulk metals are opaque to the following radiation types: radio waves, microwaves, infrared radiation, and ultraviolet radiation.

Reserve Question 08: Nonmetallic opaqueness Every nonmetallic material becomes opaque to electromagnetic radiation having some wavelength. a) True b) False Solution True. Every nonmetallic material becomes opaque to electromagnetic radiation having some wavelength. For all nonmetallic materials there is some maximum wavelength below which electron

excitations across the band gap will occur. This results in the adsorption of radiation, and, consequently, the material becomes opaque.

Reserve Question 09: Ceramics and light transmission Match the following material types with their light transmission characteristics.

Single crystal electrical insulators •

Opaque

• •

Translucent Transparent

Polycrystalline and nonporous electrical insulators • Translucent • Transparent • Opaque Porous electrical insulators • Transparent • Opaque • Translucent Solution Single crystal electrical insulators are transparent. Polycrystalline and nonporous electrical insulators are translucent. And, porous electrical insulators are opaque to visible light.

Reserve Question 10: Index of refraction For noncubic crystals, the index of refraction is lowest in the crystallographic direction that has the a) Highest atomic packing density b) Lowest atomic packing density Solution For noncubic crystals, the index of refraction is lowest in the crystallographic direction having the lowest atomic packing density.

Reserve Question 11: Polymer crystallinity and light transmission A completely amorphous and nonporous polymer will be a) Transparent b) Translucent c) Opaque

Solution A completely amorphous and nonporous polymer will be transparent.

Reserve Problem 12a (question pool) A collimated beam containing two different frequencies of light travels through vacuum and is incident on a piece of glass. Which of the schematics below depicts the phenomenon of dispersion within the glass in a qualitative correct manner? Select (e) if none of the options are qualitatively correct. a)

b)

c)

d)

Solution: The index of refraction is inversely proportional to the wavelength of a light ray. Therefore, long wavelength rays (low energy photons) are subject to a lower index of refraction and are therefore refracted less. This means their refraction angle is larger compared to rays that have a shorter wavelength (higher energy photons).

Reserve Problem 13a (question pool) A beam of light is shined on a thin (sub-millimeter thick) single crystal wafer of material. The light source is special since it can be tuned to provide any wavelength of visible light on demand. The specimen is illuminated such that the frequency of light is decreased over time while the transmitted intensity of the light is measured. If the sample becomes transparent when the frequency is less than [F] THz, what is the band gap of the material, in eV? Assume that an intrinsic excitation of electrons is responsible for the absorption.

Solution: The band gap energy is equal to the energy of the photon that is exactly energetic enough to excite electrons across the band gap. Thus, in this experiment, when the frequency is decreased the photon energy decreases. At some point, the frequency will be so low that the photon will no longer have sufficient energy to ionize the semiconductor. For this frequency and below, the semiconductor will be transparent. Thus, at the absorption edge: E g = hf = (4.135 × 10 −15 eV s)([ F ] × 1012 1/s)

Reserve Problem 14 Select the word combination that best completes this statement.

When a semiconductor is exposed to a light source, its intrinsic carrier concentration will increase if the [a] of the light is [b] than band gap of the semiconductor. [a]: intensity, energy, wavelength, frequency, voltage, current, resistance [b]: greater, less

Solution: [a] energy [b] greater

Reserve Question 15: Fluorescence and phosphorescence Match the luminescence characteristics with their descriptions.

Reemission of photons occurs in much less than one second after excitation.

• Phosphorescence • Fluorescence Reemission of photons occurs in more than one second after excitation.

• Fluorescence • Phosphorescence Solution Fluorescence involves reemission of photons in much less than one second after excitation; while phosphorescence involves reemission of photons in more than one second after excitation.

Reserve Question 16: Luminescent materials Only pure materials luminesce. a) True b) False Solution False. Luminescent materials contain impurities.

Reserve Questions and Problems Solutions Manual CHAPTER 20 ECONOMIC, ENVIRONMENTA, AND SOCIETAL ISSUES IN MATERIALS SCIENCE AND ENGINEERING

There are no Reserve Questions and Problems for this chapter.