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Solution Manual for Physical Metallurgy Principles, 4th Edition

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    An ’s Solutions Manual to Accompany Physical Metallurgy Principles, 4 th Edition, SI Version Reza Abbaschian, Lara Abbaschian and Robert E. Reed ‐ Hill

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    Table of Contents Chapter 1 The Structure of Metals 1 Chapter 2 Characterization Technique 1 2 Chapter 3 Crystal Binding 1 9 Chapter 4 Introduction to Dislocations 2 5 Chapter 5 Dislocations and Plastic Deformation 39 Chapter 6 Elements of Grain Boundaries 5 4 Chapter 7 Vacancies 6 1 Chapter 8 Annealing 7 1 Chapter 9 Solid Solutions 8 3 Chapter 10 Phases 9 1 Chapter 11 Binary Phase Diagrams 99 Chapter 12 Diffusion in Substitutional Solid Solutions 11 0 Chapter 13 Interstitial Diffusion 12 0 Chapter 14 Solidification of Metals 1 27 Chapter 15 Nucleation and Growth Kinetic 1 38 Chapter 16 Precipitation Hardening 1 50 Chapter 17 Deformation Twinning and Martensite Reactions 1 58 Chapter 18 The Iron ‐ Carbon Alloy System 1 68 Chapter 19 The Hardening of Steel 1 83 Chapter 20 Selected Nonferrous Alloy Systems 193 Chapter 21 Failure Mechanisms 2 0 1

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    1 CHAPTER 1 THE STRUCTURE OF METALS 1.1 Determine the direction indices for (a) line om , (b) line on , and (c) line op in the accompanying drawing of a cubic unit cell. Solution: (a) The direction indices for line om are [111]. (b) The vector components of line on , in units of , are , 1 and 0 along the x, y and z axes respectively. The corresponding direction indices are accordingly [120]. (c) The components of line op are , , and . Thus the indices are [436]. 1.2 Determine the direction indices of lines (a) qr , (b) qs , and (c) qt .

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    2 Solution: (a) The direction indices of line qr are [001]. (b) The direction indices of line qs are [101]. (c) The direction indices of line qt along the x, y and z axes are , , and , yielding the indices 1.3 In this figure, plane pqr intercepts the x, y, and x axes as indicated. What are the Miller indices of this plane? Solution: The intercepts of plane pqr with the three axes are , , and with the reciprocals , , and , so that the Miller indices of the plane are (12 8 9). 1.4 What are the Miller Indices of plane stu ?

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    3 Solution: This plane has intercepts , -1 and , so that the Miller idices are 1.5 Write the Miller indices for plane vwx . Solution: On the assumption that parallel planes can be described by the same set of Miller indices, this plane should have the same indices as plane (C) of Fig. 1.16, which are (111). 1.6 Linear density in a given crystallogrphic direction represents the fraction of a line length that is occupied by atoms. Similarly, planar density is the fraction of a crystallographic plane occupied by atsom. The fraction of the volume occupied in a unit cell, on the other hand, is called the

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    4 atomic packing factor. The latter should not be confused with bulk densirt, which represents weight per unit volume. (a) Calculate the linear density in the [100], [110], and [111] directions in body-ccentered cubic (bcc) and face-centered cubic (fcc) structures. (b) Calculate planar densities in (100), (110), and (111) planes in bcc and fcc structires. (c) Show that atomic packing factors for bcc, fcc, and hexagonal close-packed (hcp) structures are 0.68, 0.74, and 0.74, respectively. Solution: (a) Refering to the hard ball model of BCC structure shown in Fig 1.1c, it can be seen that the atoms touch each other along the diagonal, [111] direction, of the unit cell. Taking R as radius of the atom, its length is equal to 4R. The lattic parameter, is therefore . The edge of the unit cell, [100], is occupied by two half atoms. The linear density in [100] direction is: For the [110] direction, the length of the diagonal of the cube face is , which is again occupied by two half atoms. The linear density along [111] direction of BCC is obviously 1. For FCC unit cell, the atoms touch each other along [110] direction, as can be seen in Fig 1.2B. The lattice parameter is therefore , and the unit cell diagonal is . Therefore:

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    5 (b) The (100) face of the BCC unit cell, with an area of , is occupied by 4 quarter circle areas. Therefore: The [110] plane of BCC has an area of . This plane is occupied by two circles (4 quarter circles + 1). Therefore: For FCC structure, the [100] face is occupied by 2 circles (4 quarter circles + 1). = 0.785 T he *110+ plane of FCC again has 2 circles (4 quarter circles + 2 half circles). It’s planar density is therefore: (c) The volume of a BCC unit cell is The unit cell contains 2 atoms (1 center atom + 8 quarter corner atoms). The packing factor is therefore: The unit cell of FCC structure contains 4 atoms (8 quarter atoms in corners + 6 half atoms at the center of the faces). Since HCP has the same packaging as FCC, its packafing factor is also 0.74. this can also be calculated following Problem 1.7, by calculating the volume of the unit cell and considering that

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    6 each unit cell contains 6 atoms (3 central atoms + 2 half atoms at the center of each base + 12 one sixth atoms on the corners). 1.7 Consider the three central atoms with the one at the center of the top plane in Fig 1.16. These four atoms, which all touch each other, form an equilateral tethedron with edges equal to 2R. Solution: 1.8 Iron has a bcc structure at room temperature. When heated, it transforms from bcc to fcc at 1185/K. The atomic radii of iron atoms at this temperature are 0.126 and 0.129 nm for bcc and fcc, respectively. What is the percentage volume change upon transformation from bcc to fcc? Solution: Each unit cell of BCC iron contains 2 atoms. The volume of the unit cell is The volume occupied by each atom is therefore Substituting 0.126 nm for R results in:

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    7 Volume occupied per atom in BCC = 0.01232 In FCC, the lattice parameter is equal to , and each unit cell contains four atoms. Therefore: Volume occupied per atom in FCC = % volume change = Iron shrinks by 1.4% as it transforms from BCC to FCC at 1185 K. 1.9 This diagram shows the Thompson Tetrahedron, which is a geometrical figure formed by the four cubic {111} planes. It has special significance with regard to plastic deformation in face- centered cubic metals. The corners of the tetrahedron are marked with the letters A, B, C, and D. The four surfaces of the tetrahedron are defined by the triangles ABC, ABD, ACD, and BCD. Assume that the cube in the above figure corresponds to a face-centered cubic unit cell and indentify the planes, corresponding to each of the four surfaces, in terms of their proper Miller indices. Solution: (a) The indices of plane ABD are (111). (b) The indices of plane ABC are (c) The indices of plane ADC are (d) The indices of plane BCD are

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    8 1.10 The figure accompanying this problem is normally used to represent the unit cell of a close- packed hexagonal metal. Determine Miller indices for the two planes defg and dehj , that are outlined in this drawing Solution: (a) Plane dehj is parallel to the axis and may be assumed to intercept it at infinity. It intercepts the axis at + 1, the at – 1 and the c axis at + 1. This conforms to the Miller indices (b) Plane defg makes indentical intercepts with , and . However, it intercepts the c axis at + . The Miller indices for this latter plane are accordingly 1.11 Two other hexagonal close-paced planes are indicated in this sketch. What are their Miller indices?

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    9 Solution: Both plane klmn and klpq have basal plane intercepts; at – 1, at , and at – 1, while their c axis intercepts are + 1 for klpq and + for klmn . Thus, the indices are respectively. 1.12 Determine the hexagonal close-packed lattice directions of the lines rt , ut , and uv in the figure for this problem. To do this, first determine the vector projection of a line in the basal plane and then add it to the c axis projection of the line. Note that the direction indices of the c axis are [0001], and that if [0001] is considered a vector its magnitude will equal the height of the unit cell. A unit distance along a diagonal axis of Type I, such as the distance or , equals one third of the length of in Fig. 1.20. The magnitude of this unit is thus equal to . Combine these two quantities to obtain the direction indices of each of the lines. Solution: (a) Consider line rt . The basal plane projection of this line equals while its c axis projection is [0001]. Adding these two components yeilds:

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    10 (b) In the case of line ut we have: (c) Finally for line uv , the basal plane projectio for this line equals the addition of this vector to the c axis projection is: 1.13 Place a piece of tracing paper over the Wulff net as described above, and draw an index mark on the tracing paper over the north pole of the Wulff net. Then draw on the tracing paper the proper symbols that identify the three <100> cube poles, the six <110> poles, and the four <111> octahedral poles as in Fig. 1.33. On the assumption that the basic circle is the [010] plane and that the north pole is [100], mark on the tracing paper the correct Miller indices of all of the <100>, <110>, and <111> poles. Draw in the great circles corresponding to the planes of the plotted poles (see Fig. 1.33). Finally, indentify these planes with their Miller indices. Solution: 1.14 Place a piece fo tracing paper over the Wulff net and draw on it the index mark at the north pole of the net as well as the basic circle. Mark on this tracing paper all of the poles shown in order to obtain a 100 standard projection. Now rotate this standard projection about the north- south polar axis by 45°, so that the [110] pole moves to the center of the stereographic projection. In this rotation all of the other poles should also be moved through 45° alond the small circles, of the Wulff net, on which they lie. This type of rotation is facillitated by placing a second sheet of tracing paper over the first, and by plotting the rotated data on this sheet. This

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    11 exercise shows one of the basic rotations that can be made with a stereographic projection. The other primary rotation involves a simple rotation of the tracing paper around the pin passing through the centeres of both the tracing paper and the Wulff net. Solution: Part I In the above figure, the 45° rotation of the poles along the small circles of the Wulff net are shown. Part II The resulting orientations of the poles, after the 45° rotation, are plotted in this stereographic projection.

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    12 CHAPTER 2 – CHARACTERIZATION TECHNIQUES 2.1 If a Bragg angle of 41.31° is observed for the first order diffraction from the {110} planes, of body-centered cubic niobium using copper radiation , what is the interplanar spacing of the {110} planes? Solution: Using the interplanar spacing equation to determine gives: However, by the Bragg equation, assuming a first order reflection, we have Solving this equation for , one obtains: Note by the Bragg law determination, the spacing between {110} planes equals that obtained by the interplanar spacing equation. In the body-center cubic lattice, a set of {110} planes also passes through the center atoms of the unit cells. The first computation above accordingly corresponds to the spacing between the {110} planes of a simple cubic lattice. The second gives the spacing in a body-centered cubic crystal. 2.2 With the aid of a sketch similar to that for a simple cubic lattice in Fig. 2.10, demonstrate that the interplanar spacing of the {110} planes in the body-centered cubic lattice also equals . Solution:

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    13 2.3 Using a geometrical argument similar to that in Prob. 2.2, show that the {100} interplanar spacing in the bcc lattice is and not a, so that the planes are actually {200} planes. Solution: 2.4 Consider the Bragg equation, with respect to first-order reflections, from {100} planes of a bcc metal. By how much of a wavelength do the reflected path-lengths differ for two adjacent parallel (100) planes, if the interplanar spacing, d , is taken as a instead of a /2? Does this explain why the {100} plane is not listed as a reflecting plane in Appendix C? Explain. Solution: If it is assumed that constructive interference occurs at an angle of , which satisfies the Bragg equation for a first order reflection from {100} planes with a separation equal to a, then the path length after reflection will equal one wavelength or . The angle of reflection will be given by: For reflection from planes with a separation of a /2 at this angle, the path-length will be: This signifies both a destructive interference, and that a reflection will not occur when . It also accounts for the fact that the {100} plane is not listed as a reflecting plane in Appendix C.

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