CHAPTER 4 IMPERFECTIONS IN SOLIDS PROBLEM SOLUTIONS 4.4 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 (R%) 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 R% Cu Crystal Electro- Structure negativity FCC Valence 2+ C -44 H -64 O -53 Ag +13 FCC 0 1+ Al +12 FCC -0.4 3+ Co -2 HCP -0.1 2+ Cr -2 BCC -0.3 3+ Fe -3 BCC -0.1 2+ Ni -3 FCC -0.1 2+ Pd +8 FCC +0.3 2+ Pt +9 FCC +0.3 2+ Zn +4 HCP -0.3 2+ (a) Ni, Pd, and Pt meet all of the criteria and thus form substitutional solid solutions having complete solubility. (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+. 57 (c) C, H, and O form interstitial solid solutions. These elements have atomic radii that are significantly smaller than the atomic radius of Cu. 4.7 In order to compute composition, in atom percent, of a 92.5 wt% Ag-7.5 wt% Cu alloy, we employ Equation (4.6) as C 'Ag = = C Ag ACu C Ag ACu C CuA Ag x 100 (92.5)(63.55 g / mol) x 100 (92.5)(63.55 g/ mol) (7.5)(107.87g / mol) = 87.9 at% ' CCu = = C CuA Ag C Ag ACu CCuAAg x 100 (7.5)(107.87 g / mol) x 100 (92.5)(63.55 g/ mol) (7.5)(107.87g / mol) = 12.1 at% 4.13 This problem calls for a conversion of composition in atom percent to composition in weight percent. The composition in atom percent for Problem 4.11 is 44.8 at% Ag, 46.2 at% Au, and 9.0 at% Cu. Modification of Equation (4.7) to take into account a three-component alloy leads to the following C Ag = = ' A C Ag Ag ' A ' A C 'Ag AAg C Au CCu Au Cu x 100 (44.8)(107.87 g / mol) x 100 (44.8)(107 .87 g / mol) (46.2)(196.97 g / mol) (9.0)(63.55 g / mol) = 33.3 wt% C Au = ' A C Au Au C' A Ag Ag C' A Au Au 58 C' A Cu Cu x 100 = (46.2)(196.97 g / mol) x 100 (44.8)(107 .87 g / mol) (46.2)(196.97 g / mol) (9.0)(63.55 g / mol) = 62.7 wt% CCu = ' A C Cu Cu C' A Ag Ag = C' A Au Au C' A x 100 Cu Cu (9.0)(63.55 g / mol) x 100 (44.8)(107 .87 g / mol) (46.2)(196.97 g / mol) (9.0)(63.55 g / mol) = 4.0 wt% 4.25 (a) The Burgers vector will point in that direction having the highest linear density. From Section 3.11, the linear density for the [110] direction in FCC is 1/2R, the maximum possible; therefore for FCC b = a [110] 2 From Problem 3.46 the linear density for the [111] direction in BCC is also 1/2R, and therefore for BCC b = a [111] 2 For simple cubic, a unit cell of which is shown in Figure 3.19, the atom spheres touch one another along the cube edges (i.e., in [100] directions) and therefore, the atomic packing is greatest in these directions. Therefore, the Burgers vector is b = a [100] 2 (b) For Cu which has an FCC crystal structure, R = 0.1278 nm (Table 3.1) and a = 2R 2 = 0.3615 nm [Equation (3.1)]; therefore 59 b = = 0.3615 nm 2 a 2 2 (1 ) h 2 k 2 (1 ) 2 2 l 2 (0) = 0.2556 nm For Fe which has a BCC crystal structure, R = 0.1241 nm (Table 3.1) and a 4R = 0.2866 3 nm [Equation (3.3)]; hence b = 0.2866 nm 2 2 2 (1) (1) (1) 2 = 0.2482 nm 4.29 (a) The interfacial defect that exists for this stacking sequence is a twin boundary, which occurs at the following position The stacking sequence on one side of this position is mirrored on the other side. (b) The interfacial defect that exists within this FCC stacking sequence is a stacking fault, which occurs over the region indicated For this region, the BCBC stacking sequence is HCP. 4.31 (a) This portion of the problem calls for a determination of the average grain size of the specimen which microstructure is shown in Figure 9.22a. Seven line segments were drawn across the micrograph, each of which was 60 mm long. The average number of grain boundary intersections for these lines was 6.3. Therefore, the average line length intersected is just 60 60 mm = 9.5 mm 6.3 Hence, the average grain diameter, d, is d = ave. line length int er sected 9.5 mm = = 0.106 mm magnification 90 (b) This portion of the problem calls for us to estimate the ASTM grain size number for this same material. The average grain size number, n, is related to the number of grains per square inch, N, at a magnification of 100x according to Equation 4.16. However, the magnification of this micrograph is not 100x, but rather 90x. Consequently, it is necessary to use the following equation: 2 M NM 2 n 1 100 where NM = the number of grains per square inch at magnification M, and n is the ASTM grain size number. (The above equation makes use of the fact that, while magnification is a length parameter, area is expressed in terms of units of length squared. As a consequence, the number of grains per unit area increases with the square of the increase in magnification.) Solving the above expression for n leads to M log NM 2 log 100 n 1 log 2 From Figure 9.22a, NM is measured to be approximately 4, which leads to 90 log 4 2 log 100 n 1 log 2 = 2.7 61