Imperfections in Solids Problem Solutions Part 8

Grain Size Determination 4.45 (a) Using the intercept method determine the mean intercept length, in millimeters, of the specimen whose microstructure is shown in Figure 4.15b; use at least seven straight-line segments. (b) Estimate the ASTM grain size number for this material. Solution (a) Below is shown the photomicrograph of Figure 4.15 b, on which seven straight line segments, each of which is 60 mm long has been constructed; these lines are labeled “1” through “7”. Since the length of each line is 60 mm and there are 7 lines, the total line length ( LT in Equation 4.16) is 420 mm. Tabulated below is the number of grain-boundary intersections for each line: Line Number No. Grains Intersected 1 11 2 10 3 9 4 8.5 5 7 6 10 7 8 Total 63.5 Because LT = 420 mm, P = 63.5 grain-boundary intersections, and the magnification M = 100×, the mean intercept length is computed using Equation 4.16 as follows: = 420 mm LT = PM (63.5 intersections)(100 × ) = 0.066 mm (b) The ASTM grain size number G is computed by substitution of this value for into Equation 4.19a as follows: G = −6.6457log − 3.298 = (−6.6457)[log(0.066 mm)]− 3.298 = 4.55 4.46 (a) Employing the intercept technique, determine the mean intercept length for the steel specimen whose microstructure is shown in Figure 9.25a; use at least seven straight-line segments. (b) Estimate the ASTM grain size number for this material. Solution (a) Below is shown the photomicrograph of Figure 9.25 a, on which seven straight line segments, each of which is 60 mm long has been constructed; these lines are labeled “1” through “7”. Since the length of each line is 60 mm and there are 7 lines, the total line length ( LT in Equation 4.16) is 420 mm. Tabulated below is the number of grain-boundary intersections for each line: Line Number No. Grains Intersected 1 7 2 7 3 7 4 8 5 10 6 6 7 8 Total 53 Because LT = 420 mm, P = 56 grain-boundary intersections, and the magnification M = 90×, the mean intercept length is computed using Equation 4.16 as follows: = LT 420 mm = PM (53 intersections)(90× ) = 0.088 mm (b) The ASTM grain size number G is computed by substitution of this value for into Equation 4.19a as follows: G = −6.6457log − 3.298 = (−6.6457)[log(0.088 mm)]− 3.298 = 3.72 4.47 For an ASTM grain size of 6, approximately how many grains would there be per square inch under each of the following conditions? (a) At a magnification of 100× (b) Without any magnification? Solution (a) This part of problem asks that we compute the number of grains per square inch for an ASTM grain size of 6 at a magnification of 100 ×. All we need do is solve for the parameter n in Equation 4.17, inasmuch as G = 6. Thus n = 2 G−1 6 −1 =2 = 32 grains/in.2 (b) Now it is necessary to compute the value of n for no magnification. In order to solve this problem it is necessary to use Equation 4.18: 2  M = 2G−1  100 ÷  nM  where nM = the number of grains per square inch at magnification M, and G is the ASTM grain size number. Without any magnification, M in the above equation is 1, and therefore, 2  1  = 2 6−1 = 32 n1   100 ÷ And, solving for n1, n1 = 320,000 grains/in.2. 4.48 Determine the ASTM grain size number if 30 grains per square inch are measured at a magnification of 250× . Solution In order to solve this problem we make use of Equation 4.18: 2 M  nM  = 2G−1 100 ÷  where nM = the number of grains per square inch at magnification M, and G is the ASTM grain size number. Solving the above equation for G, and realizing that nM = 30, while M = 250. Some algebraic manipulation of Equation 4.18 is necessary. Taking logarithms of both sides of this equation leads to 2  M   log nM  ÷  = log 2G−1 100     ( ) Or M  = (G −1)log2 log nM + 2 log  100 ÷ And solving this equation for G M  log nM + 2 log   100 ÷ log 2 = G −1 or M  log nM + 2 log   100÷ G= +1 log 2 From the problem statement: nM =30 M = 250 And, therefore  250 log 30 + 2 log   100 ÷  G= + 1 = 8.6 log 2 4.49 Determine the ASTM grain size number if 25 grains per square inch are measured at a magnification of 75× . Solution In order to solve this problem we make use of Equation 4.18—viz. 2 M  nM  = 2G−1 100 ÷  where nM = the number of grains per square inch at magnification M, and G is the ASTM grain size number. By algebraic manipulation of Equation 4.18 (as outlined in Problem 4.48) we make G the dependent variable, which leads to the following expression: M  log nM + 2 log   100÷  G= +1 log 2 For nM = 25 and M = 75, the value of G is  75  log 25 + 2 log   100÷ + 1 = 4.8 G= log 2 4.50 The following is a schematic micrograph that represents the microstructure of some hypothetical metal. Determine the following: (a) Mean intercept length (b) ASTM grain size number, G Solution (a) Below is shown this photomicrograph on which six straight line segments, each of which is 50 mm long has been constructed; these lines are labeled “1” through “6”. Since the length of each line is 50 mm and there are 6 lines, the total line length ( LT in Equation 4.16) is 300 mm. Tabulated below is the number of grain-boundary intersections for each line: Line Number No. Grains Intersected 1 7 2 8 3 8 4 8 5 8 6 8 Total 47 Because LT = 300 mm, P = 47 grain-boundary intersections, and the magnification M = 200 ×, the mean intercept length is computed using Equation 4.16 as follows: = LT 300 mm = PM (47 intersections)(200× ) = 0.032 mm (b) The ASTM grain size number G is computed by substitution of this value for into Equation 4.19a as follows: G = −6.6457log − 3.298 = (−6.6457)[log(0.032 mm)]− 3.298 = 6.64 4.51 The following is a schematic micrograph that represents the microstructure of some hypothetical metal. Determine the following: (a) Mean intercept length (b) ASTM grain size number, G Solution (a) Below is shown this photomicrograph on which six straight line segments, each of which is 50 mm long has been constructed; these lines are labeled “1” through “6”. Since the length of each line is 50 mm and there are 6 lines, the total line length ( LT in Equation 4.16) is 300 mm. Tabulated below is the number of grain-boundary intersections for each line: Line Number No. Grains Intersected 1 8 2 9 3 10 4 8 5 9 6 8 Total 52 Because LT = 300 mm, P = 52 grain-boundary intersections, and the magnification M = 150 ×, the mean intercept length is computed using Equation 4.16 as follows: = LT 300 mm = PM (52 intersections)(150× ) = 0.038 mm (b) The ASTM grain size number G is computed by substitution of this value for into Equation 4.19a as follows: G = −6.6457log − 3.298 = (−6.6457)[log(0.038 mm)]− 3.298 6.14