Unformatted text preview: oduction to  b110 2 2 a2 8 = 3 a2 = = 2.67 2 b111  2 3 2
b111 2 =
3 a2 4 b001  = a = 3 = 2.67 96.5 grains grains 2 = 24.3 2 π (2..5 grains in2 96 25/2) in grains N of Now invoke Equation 4.1 (p. 116 = the text) to relate N24.3 , 2 2 = to G 2 N=
π (2.25/2) in in N = 2(G−1)
Substituting and solving,
G= N = 2(G−1) ln N ln 24.3 +1 = + 1 = 5.6 ln 2 ln 2 so the ASTM Grain Size number determined by this analysis is G = 5.6 NOTE: As cautioned in the problem statement, there are many possible solutions to this problem, depending upon the location of the inscribed sampling area. It therefore would be prudent to sample many different areas and calculate an average grain size number associated with your sample(s) when higher precision is required. Problem 4.25 Solution Professor R. Gronsky page 2 of 2 Problem 4.33
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 4.33 Sketch a few adjacent CaO6 octahedra in the pattern shown in Figure 4.24. Indicate both the nearest neighbor Ca2+–Ca2+ distance, R1, and the nextnearest neighbor Ca2+–Ca2+ distance, R2. SOLUTION Referring to Figure 4.24 as instructed in the problem statement and completing the sketch of a "few" adjacent CaO6 octahedra, the following figure can be generated. R2 R1 The Ca2+ cations are centered in each of the edgesharing octahedra. As requested in the problem statement, the nearest neighbor distance, which is the distance of closest approach of the cations, is labeled as R1 in the figure, and the secondneighbor distance, R2. These are the appropriate distances for this specific plane through the three dimensional structure, which was explicitly requested in the problem statement. Problem 4.33 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 Berkeley.
 Fall '08
 GRONSKY

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