HW03 Soln - Problem 3.2 J. F. Shackelford, Introduction to...

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Unformatted text preview: Problem 3.2 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 3.2 What would be an equivalent two-dimensional point lattice for the area-centered hexagon? SOLUTION The two-dimensional point lattices are described in Example 3.1 (p. 61 of the text). There are 5 of them, and all are sketched at the top of page 62. The Note at the bottom of p. 61 explains that there may be some degeneracies in these constructions, such as an area-centered square lattice being equivalent to a simple square lattice in another orientation. This problem asks you to find an equivalent to the area-centered hexagon, reproduced here from the sketch labeled (v) on p. 62. As the outlines of the unit cells show, each cell contains 2 lattice points, one (1) fully contained within the cell and six at the perimeter shared six ways, (6 x ⅙ = 1). An equivalent point lattice can be found be constructing one with a unit cell that contains only a single lattice point per unit cell, known as a "primitive" cell. Begin by "erasing" all of the lines between lattice points above and draw new lines in different unit cell configurations. There are several options, such as a rhombus, Problem 3.2 Solution Professor R. Gronsky page 1 of 2 Problem 3.2 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) or a parallelogram. The only requirement is that the unit cell must have parallel opposing sides, so that the unit cell can be translated along its basis vectors to generate the entire lattice with no gaps, and no overlaps. Consequently, triangles are excluded from the definition of a unit cell. Problem 3.2 Solution Professor R. Gronsky page 2 of 2 Problem 3.10 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 3.10 Calculate the APF of 0.74 for hcp metals. SOLUTION Begin with the definition. The atomic packing factor (APF) is defined on p. 62 of the text as a ratio of the volume of the atoms contained within a single unit cell to the volume of that unit cell. A sample APF calculation is presented in Example 3.6 of the text on page 78. The hexagonal close-packed (hcp) structure is shown in Figure 3.6 of the text (p. 64) to have a "simple" hexagonal unit cell, which by definition is a "primitive" cell containing only 1 lattice point, and a 2 atom motif. The caption to the same figure indicates that the total number of atoms contained within the unit cell is 2: 4 corner atoms contributing 1/6 of their volume, 4 more corner atoms contributing 1/12 of their volume and 1 full atom in the interior of the cell. Next, to visualize the atoms contained within the unit cell, draw some helpful sketches. First, draw the "basal plane" of the hcp structure showing the rhombus-shaped base of the unit cell. In this view it can be seen that the atoms at the 120° corners contribute1/3 of their area to the unit cell and the atoms at the 60° corners contribute 1/6 of their area to the unit cell. Because each of these atoms is shared by two unit cells above and below the plane of the drawing, the three di- Problem 3.10 Solution Professor R. Gronsky page 1 of 5 Problem 3.10 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) mensional fractions are 1/3 × 1/2 = 1/6 and 1/6 × 1/2 = 1/12, as cited above. In the full unit cell, this plane appears at both its top and bottom faces, accounting for all 8 atoms at the corners. What about the other atoms in the unit cell? To visualize these, a different sketch of the "midplane" of atoms in the hcp structure is needed, along with the projection of the unit cell. The location of these atoms above the basal plane is also given in Figure 3.7 (b) of the text on p. 65. The outline of the unit cell shown here represents the projection of the faces of the unit cell that cut through the atoms perpendicular to the plane of the drawing. Note that the unit cell contains a fractional contribution of 3 atoms in this layer, but by the symmetry of the figure, it can be seen that together these fractions constitute 1 complete atom. What is "missing" from the largest atomic segment cut by any one face of the unit cell is restored by the fraction added at the opposing face of the unit cell. Therefore this layer contributes the 1 full atom counted in the caption to Figure 3.6. This confirms that the unit cell contains 2 atoms. Consequently the volume of the atoms occupying the unit cell is simply twice the volume of a single atom. Assuming an atomic radius r, the numerator of the APF expression is Vatoms = 2 (4/3 π r3). Next calculate the volume of the unit cell, which is given by the area of the base (shown in the first figure [basal plane] above) times its height. To calculate the area of the base; note its geometry as shown on the following enlargement. Problem 3.10 Solution Professor R. Gronsky page 2 of 5 Problem 3.10 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) h = a sin 60° 60° a Since the edge length a connects the centers of two atoms that touch (see drawing of the basal plane above and Table 3.3 of the text on page 64), a = 2r, consequently the area of the base of the unit cell can be written Abase = [2r]2 sin 60° The height of the unit cell is the lattice constant c of the hcp structure. Because this is not given in the problem, it must be calculated, made easy by employing an assumption. The assumption is that all atoms are touching, as is the case for any "close-packed" structure. Now make another geometrical sketch for reference... xP 30° a Problem 3.10 Solution Professor R. Gronsky page 3 of 5 Problem 3.10 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) This drawing shows that there is an atom in the midplane position located at position P with coordinates ⅔, ⅓, ½ relative to the origin (center atom on the page), so it is located at elevation c/2 above the basal plane. The tetrahedron sketched in dotted lines gives a geometrical construction that can be used to establish c. First note that the edge of the tetrahedron has length a from before. The distance x along the base of the tetrahedron is therefore given by x= a/2 cos 30◦ Viewed from another perspective, the height of the mid-layer atom above the basal plane (solid triangle) is shown with another triangle construction. One side is a because the atoms also touch along this direction. ! " #$% This triangle that includes x, a and c/2 as its sides permits the side lengths to be summed as a2 = x2 + (c/2)2 Problem 3.10 Solution Professor R. Gronsky page 4 of 5 Problem 3.10 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) Substituting, (c/2) = a − 2 2 and solving, ￿ a/2 cos 30◦ ￿2 c/a = 1.633 Again recognizing that a = 2r, and from above, c = 1.633 [2r], consequently, Vunit cell = Abase × c = [2r]2 sin 60° × 1.633 [2r] Vunit cell = 1.633 [2r]3 sin 60° Finally, solving for the atomic packing factor, APF = 2 ([4/3]π r3 ) 1.633[2r3 ] sin 60◦ and simplifying, APF = [8/3]π 1.633[8] sin 60◦ π [3] [1.633] sin 60◦ π 4.242 APF = APF = APF = 0.74 Problem 3.10 Solution Professor R. Gronsky page 5 of 5 Problem 3.27 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 3.27 (a) Sketch , in a cubic unit cell, [100] and [210] directions. (b) Use a trigonometric calculation to determine the angle between these directions. (c) Use Equation 3.3 to determine this angle. SOLUTION (a) The directions [100] and [210] are specified relative to the coordinate axis system as shown on the following sketch. z y √ 3a = rU4+ + rO2− = 0.105 nm + 0.132 nm 4 a Vunit cell = a3 = (0.548 nm)3 = 0.164 nm3 [100] [210] Vions x 4 3a/2 43 = 4 × π rU4+ + 8 × π rO2− 3 3x (b ) y ( a) 16 π 32 π (0.105 nm)3 + (0.132 nm)3 = 0.0965 nm3 3 3 (b) Noting that the [100] direction is parallel to the x-axis, and the [210] direction connects the origin to the midpoint of the opposite edge,Va triangle can be constructed with these two lengths 0.0965 nm3 ions I P F from the top, it appears as shown in sketch (b) above. Solv= = inscribed in a square. When viewed Vunit cell 0.164 nm3 ing for angle δ from the trigonometric formula for the resulting right triangle Vions = δ = tan yields the value δ = 26.6° −1 ￿ a/2 a ￿ = tan −1 ￿￿ 1 2 Problem 3.27 Solution Professor R. Gronsky page 1 of 2 Vions = Problem 3.27 16 π 32 π (0.105 nm)3 + (0.132 nm)3 = 0.0965 nm3 3 3 IP F = = Vions 0 Engineers, 7th J. F. Shackelford, Introduction to Materials Science for .0965 nm3 Edition, Prentice Hall, New Jersey (2009) Vunit cell ￿ 0.164 nm3 ￿ ￿￿ a/2 −1 1 = tan (c) Equation 3.3 in the text is δ = tan derived from the definition of the vector dot product. In this case a 2 the vectors are the lattice direction vectors [100] and [210]. Substituting into Equation 3.3, −1 and solving, the same result is (fortunately) obtained: 2+0+0 2 cos δ = √ ￿ =√ 5 1 (4 + 1 δ = 26.6°. Problem 3.27 Solution Professor R. Gronsky page 2 of 2 Problem 3.40 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 10 3.40 List the members of the family of prismatic planes for the hexagonal unit cell {01¯ } (see Figure 3.27). SOLUTION Begin by looking up the definition of "prismatic." The prism planes of the hexagonal unit cell ate the vertical facets connecting the "basal" planes at top and bottom of the cell. Figure 3.27 of the text (p. 83) shows one member of the (01¯ family of prism planes, a vertical plane of the 10) hexagonal unit cell. There are six such families comprising the {01¯ } planes. In list form, 10 these planes are: (01¯ , (0¯ , (10¯ , (¯ 10) 110) 10) 1010), (1¯ , (¯ 100) 1100) which are highlighted to show their "prism" orientation, and labeled in the following sketch. c(¯ 1010) (0¯ 110) (¯ 1100) a3 a2 (1¯ 100) (01¯ 10) ¯ (1010) a1 Note that in every case the indices (hkil) satisfy the requirement that h + k = –i. Problem 3.40 Solution Professor R. Gronsky page 1 of 1 Problem 3.41 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 3.41 The four-digit notation system (Miller-Bravais indices) introduced for planes in the hexagonal system can also be used for describing crystal directions. In a hexagonal unit cell 20] sketch (a) the [0001] direction, and (b) the [11¯ direction . SOLUTION The sketch below shows the coordinate axis system used for Miller-Bravais notation (left) and the two directions requested in the problem statement (right). c [0001] (a) a3 a2 a1 [1120] (b ) Note that the [0001] direction (a) is the direction of the c axis, and the [11¯ direction (b) is the 20] direction opposite the a3 axis. Problem 3.41 Solution Professor R. Gronsky page 1 of 1 ...
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This note was uploaded on 01/21/2011 for the course E 45 taught by Professor Gronsky during the Fall '08 term at University of California, Berkeley.

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