(1)
BCC = ?BCCM ?crss(2)
FCC = ?FCCM?crss
BCCBCC ?crss = ? FCC ?crss?FCC
BCC5.5?crss x 9.63 x 10-2 = 0.963 MPa ?BCC = FCC ? ?FCC = 0.55?crss
25.
Here crystallographically equivalent positions join ions at cube corners (bv = ao), face diagonals (bv = 2 ao), cube diagonals (bv = 3 ao)
32 atms / m22 aThe most densely packed plane is the (110) in which we have
The shortest vector that will reproduce all elements of the structure is ao. Thus b = a<100>
COMMENT: We note that this is not sufficient for general deformation (e.g. a tensile axis of the type <100> produces zero shear on the 1<100> Burgers vectors. We expect then a<110> Burgers vectors as well. 26.
GIVEN: ? = 1.7 MPa [100] tensile axis (111)[101] slip systems
REQUIRED: ?crss, and crystal structure. Also find flaw in problem statement.
SOLUTION: Since the slip system is of the type {111}<110> the structure is FCC. The problem is misstated since the Burgers vector must lie on the slip plane and [101] does not lie on (111). The slip direction would more appropriately be [101_]. Thus the slip system is (111)[101_] as shown below.
? = [100] [101] ? = [100] [111] ?crss = ? cos? cos? cos? = [100]?[101][100] [111] cos? = 1 x 21 x 3 cos? = 11 ; cos? = 23 11 x ? 0.69 MPa23?crss = 1.7 x 27.
? = edge dislocation
b = Burgers vector
28.
x = start of Burgers circuit y = end of Burgers circuit
FIND: Show energy/area = force/length, that is, surface energy is surface tension in liquids.
DATA: The units of energy are J = W/s or N-m. The units of force are N. SOLUTION: Energy/area = J/m2 =N-m/m2 = N/m = force/length
29. GIVEN: Two grain sizes, 10?m and 40?m
REQUIRED: A) ASTM GS# for both processes, B) Grain boundary area.
SOLUTION: Assume that the grains are in the form of cubes for ease of calculation. The ASTM GS# is defined through the equation: n = 2N-1 where n = # grains/in2 at 100X. N=ASTM GS#
To solve the problem we first convert the grain size to in. where D = length of cube edge in ?m.
?cm?1inD [?m] x 10-4? x ?2.54 cm ?m?? _ D x 3.937 x 10-5 = Din -5-4D10? ? 10 X 3.937 x 10 = 3.937 x 10 in -5-4D40? ? 40 x 3.937 x 10 = 15.75 x 10 in -42-210?D100X = 3.937 x 10 x 10 = 3.937 x 10 in -2-32100XA10? = (3.937 x 10) = 1.550 x 10 in 2-2100X x 10-3 in2 A40? = (15.75 x 10) = 24.812
At 100X linear magnification, the sides of the smaller grains will be:
The area of each grain at 100X will be
Similarly the area of the 40?m grains at 100X is
For the 10?m dia grain, the # of grains per in2 (at box) is
12100X = 645.16grains/in at 100X n10? = 1.550 x 10-312
= = 40.31Similarlyn100Xgrains/in at 100X 40?-324.81 x 10For the 10?m grain size:
645.16 = 2N-1 log10 645.15 = (N - 1) log10 2 log(645.16) + 1 = N _ ASTM#10.3 log 2 For the 40?gs ANSW log(40.31) + 1 = N = ASTM#6.3log2
B. In computing the total g.s. area we will assume 1 in3 of materials. Since there are 6 faces cube and the area of each face is shared by 2 cubes, each cube has an area of 3x
?1?Area of face. G.B. Area =?3? x 3d2 = 3 / d
?d?GB Area (10? gs) = 3/3.937 x 10-4 = 7620in2/in3 GB Area (40? gs) = 3/15.75 x 10-4 = 1905in2/in3 GIVEN:
?ys = 200MPa at GS#4 = 300MPa at GS#6
REQUIRED: ?ys at GS#9
SOLUTION: Recall ?ys = ?o + kd-1/2 (1) for low carbon steel. If d = grain size (assume cubes) load = grain diameter at 100X
30.
