Solution Manual for Physical Metallurgy Principles, 4th Edition

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An’sSolutionsManualtoAccompanyPhysicalMetallurgyPrinciples,4thEdition,SIVersionRezaAbbaschian,LaraAbbaschianandRobertE.Reed‐Hill

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Table of ContentsChapter 1The Structure of Metals1Chapter 2Characterization Technique12Chapter 3Crystal Binding19Chapter 4Introduction to Dislocations25Chapter 5Dislocations and Plastic Deformation39Chapter 6Elements of Grain Boundaries54Chapter 7Vacancies61Chapter 8Annealing71Chapter 9Solid Solutions83Chapter 10Phases91Chapter 11Binary Phase Diagrams99Chapter 12Diffusion in Substitutional Solid Solutions110Chapter 13Interstitial Diffusion120Chapter 14Solidification of Metals127Chapter 15Nucleation and Growth Kinetic138Chapter 16Precipitation Hardening150Chapter 17Deformation Twinning and Martensite Reactions158Chapter 18The Iron‐Carbon Alloy System168Chapter 19The Hardening of Steel183Chapter 20Selected Nonferrous Alloy Systems193Chapter 21Failure Mechanisms201

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1CHAPTER 1THE STRUCTURE OF METALS1.1Determine the direction indices for (a) lineom, (b) lineon, and (c) lineopin the accompanyingdrawing of a cubic unit cell.Solution:(a)The direction indices for lineomare [111].(b)The vector components of lineon, in units of, are, 1 and 0 along the x, y and z axesrespectively. The corresponding direction indices are accordingly [120].(c)The components of lineopare,, and. Thus the indices are [436].1.2Determine the direction indices of lines (a)qr, (b)qs, and (c)qt.

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2Solution:(a)The direction indices of lineqrare [001].(b)The direction indices of lineqsare [101].(c)The direction indices of lineqtalong the x, y and z axes are,, and, yielding the indices1.3In this figure, planepqrintercepts the x, y, and x axes as indicated. What are the Miller indices ofthis plane?Solution:The intercepts of planepqrwith the three axes are,, andwith the reciprocals,, and, so that the Miller indices of the plane are (12 8 9).1.4What are the Miller Indices of planestu?

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3Solution:This plane has intercepts, -1 and, so that the Miller idices are1.5Write the Miller indices for planevwx.Solution:On the assumption that parallel planes can be described by the same set of Miller indices, thisplane should have the same indices as plane (C) of Fig. 1.16, which are (111).1.6Linear density in a given crystallogrphic direction represents the fraction of a line length that isoccupied by atoms. Similarly, planar density is the fraction of a crystallographic plane occupiedby atsom. The fraction of the volume occupied in a unit cell, on the other hand, is called the

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4atomic packing factor. The latter should not be confused with bulk densirt, which representsweight 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) structuresare 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 theatoms touch each other along the diagonal, [111] direction, of the unit cell. Taking R asradius of the atom, its length is equal to 4R. The lattic parameter, is therefore. Theedge 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 againoccupied 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 Fig1.2B. 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 quartercircles + 1). Therefore:For FCC structure, the [100] face is occupied by 2 circles (4 quarter circles + 1).= 0.785The *110+ plane of FCC again has 2 circles (4 quarter circles + 2 half circles). It’s planardensity is therefore:(c) The volume of a BCC unit cell isThe unit cell contains 2 atoms (1center 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 thecenter of the faces).Since HCP has the same packaging as FCC, its packafing factor is also 0.74. this can also becalculated following Problem 1.7, by calculating the volume of the unit cell and considering that

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6each unit cell contains 6 atoms (3 central atoms + 2 half atoms at the center of each base + 12one sixth atoms on the corners).1.7Consider the three central atoms with the one at the center of the top plane in Fig 1.16. Thesefour atoms, which all touch each other, form an equilateral tethedron with edges equal to 2R.Solution:1.8Iron has a bcc structure at room temperature. When heated, it transforms from bcc to fcc at1185/K. The atomic radii of iron atoms at this temperature are 0.126 and 0.129 nm for bcc andfcc, 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 isThevolume occupied by each atom is thereforeSubstituting 0.126 nm for Rresults in:

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7Volume occupied per atom in BCC = 0.01232In 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.1.9This diagram shows the Thompson Tetrahedron, which is a geometrical figure formed by thefour 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, andD. 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 andindentify the planes, corresponding to each of the four surfaces, in terms of their proper Millerindices.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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81.10The figure accompanying this problem is normally used to represent the unit cell of a close-packed hexagonal metal. Determine Miller indices for the two planesdefganddehj, that areoutlined in this drawingSolution:(a)Planedehjis parallel to theaxis and may be assumed to intercept it at infinity. It interceptstheaxis at + 1, theat–1 and the c axis at + 1. This conforms to the Miller indices(b)Planedefgmakes indentical intercepts with, and. However, it intercepts the c axis at+. The Miller indices for this latter plane are accordingly1.11Two other hexagonal close-paced planes are indicated in this sketch. What are their Millerindices?

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9Solution:Both planeklmnandklpqhave basal plane intercepts;at–1,at, andat–1, whiletheircaxis intercepts are + 1 forklpqand +forklmn. Thus, the indices arerespectively.1.12Determine the hexagonal close-packed lattice directions of the linesrt,ut, anduvin the figurefor this problem. To do this, first determine the vector projection of a line in the basal plane andthen add it to thecaxis projection of the line. Note that the direction indices of thecaxis are[0001], and that if [0001] is considered a vector its magnitude will equal the height of the unitcell. A unit distance along a diagonal axis of Type I, such as the distanceor, equals one third ofthe length ofin Fig. 1.20. The magnitude of this unit is thus equal to. Combinethese two quantities to obtain the direction indices of each of the lines.Solution:(a)Consider linert. The basal plane projection of this line equalswhile its c axis projectionis [0001]. Adding these two components yeilds:

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10(b)In the case of lineutwe have:(c)Finally for lineuv, the basal plane projectio for this line equalsthe addition of this vectorto thecaxis projection is:1.13Place a piece of tracing paper over the Wulff net as described above, and draw an index mark onthe tracing paper over the north pole of the Wulff net. Then draw on the tracing paper theproper 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] planeand 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 theplotted poles (see Fig. 1.33). Finally, indentify these planes with their Miller indices.Solution:1.14Place a piece fo tracing paper over the Wulff net and draw on it the index mark at the northpole of the net as well as the basic circle. Mark on this tracing paper all of the poles shown inorder 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 stereographicprojection. In this rotation all of the other poles should also be moved through 45° alond thesmall circles, of the Wulff net, on which they lie. This type of rotation is facillitated by placing asecond sheet of tracing paper over the first, and by plotting the rotated data on this sheet. This

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11exercise shows one of the basic rotations that can be made with a stereographic projection. Theother primary rotation involves a simple rotation of the tracing paper around the pin passingthrough the centeres of both the tracing paper and the Wulff net.Solution:Part IIn the above figure, the 45° rotation of the poles along the small circles of the Wulff net areshown.Part IIThe resulting orientations of the poles, after the 45° rotation, are plotted in this stereographicprojection.

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12CHAPTER 2–CHARACTERIZATION TECHNIQUES2.1If a Bragg angle of 41.31° is observed for the first order diffraction from the {110} planes, ofbody-centered cubic niobium using copperradiation, what is theinterplanar spacing of the {110} planes?Solution:Using the interplanar spacing equation to determinegives:However, by the Bragg equation, assuming a first order reflection, we haveSolving this equation for, one obtains:Note by the Bragg law determination, the spacing between {110} planes equals that obtained bythe interplanar spacing equation. In the body-center cubic lattice, a set of {110} planes alsopasses through the center atoms of the unit cells. The first computation above accordinglycorresponds to the spacing between the {110} planes of a simple cubic lattice. The second givesthe spacing in a body-centered cubic crystal.2.2With the aid of a sketch similar to that for a simple cubic lattice in Fig. 2.10, demonstrate thatthe interplanar spacing of the {110} planes in the body-centered cubic lattice also equals.Solution:

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132.3Using a geometrical argument similar to that in Prob. 2.2, show that the {100} interplanarspacing in the bcc lattice isand nota,so that the planes are actually {200} planes.Solution:2.4Consider the Bragg equation, with respect to first-order reflections, from {100} planes of a bccmetal. By how much of a wavelength do the reflected path-lengths differ for two adjacentparallel (100) planes, if the interplanar spacing,d, is taken asainstead ofa/2? Does this explainwhy 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 Braggequation for a first order reflection from {100} planes with a separation equal toa,then thepath length after reflection will equal one wavelength or. The angle of reflection will be givenby:For reflection from planes with a separation ofa/2 at this angle, the path-lengthwill 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 reflectingplane in Appendix C.

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