= 14 2 cos? = 1 - 2 + 0 = -1 cos? = 1 14 2 ? = 7?cR = -800 x x 14 2 cos? cos?6 7 14 = -4,572 psi (compression)3? = -800 10.
Burgers vectors lie in the closest packed directions since the distance between equivalent crystallographic positions is shortest in the close-packed directions. This means that the energy associated with the dislocation will be minimum for such dislocations since the energy is proportional to the square of the Burgers vector.
Close packed planes are slip planes since these are the smoothest planes (on an atomic level) and would then be expected to have the lowest critical resolved shear stress. GIVEN: Dislocation lies on (11_1) parallel to intersection of (11_1) and (111) with Burgers vector parallel to [1_1_0]. Structure is FCC.
REQUIRED: A) Burgers vector of dislocation and, B) Character of dislocation.
SOLUTION: A) Since the structure is FCC, the Burgers vector is parallel to <110> and
ahas magnitude _1_0] the scalar multiplier must be . For a Burgers vector parallel to [1
2a/2. Thus b_ = a/2 [1_1_0]. B) We must determine the line direction of the dislocation. From the diagram we see that the BV and line direction are at 60o which means the dislocation is mixed.
GIVEN: Dislocation reaction below:
REQUIRED: Show it is vectorially correct and energetically proper.
aaSOLUTION: [111] + [111] = a [100]
22The sum of the x, y & z components on the LHS must be equal to the corresponding component on the right hand side.
x component (LHS) = x component (RHS)
11. 12.
13.
aa ? (1) + ? (1) = a yes 22??aa (1) + ? (-1)= o yes 22???aa (1) + ? (-1)= o yes 22?2223a23aaa_ b1 _ = + + = = a 44442 3 2_ b1 _ = a2 4 3 2_ b2 _ = a2 4 _ b3 _ = a2 2
y component (LHS) = y component (RHS) z component (LHS) = z component (RHS)
Energy: The reaction is energetically favorable if | b1 | 2 + | b2 | 2 > | b3| 3
33Thus the reaction is favorable since a2 + a2 > a2
44GIVEN: Dislocation in FCC
14.
ab = [101] 2
Parallel to [1_01] i.e. t_ = [1_01]
REQUIRED: Character and slip plane
SOLUTION: Character is found by angle between t_ and b_. Notet_ ? b_ = -1 + 0 + 1 = 0. Thust__b_. Sincet__b_the dislocation is pure edge. To find the slip plane we note that the cross produce ofb_ & t_gives a vector that is normal to the plane in whichb_ & t_lie. This vector so formed has the same indices as the plane since we have a fundamentally cubic structure.
We see from the diagram that these vectors lie on (010).
Thus, we have the plane (01_0) which is the same as the (010) plane. This does not move by glide since planes of the kind {100} are not slip planes for the FCC structure.
FCC metals are more ductile than BCC or HCP because: 1) there is no easy mechanism for nucleation of microcracks in FCC as there is for BCC and HCP; 2) the stresses for plastic deformation are lower in FCC due to the (generally) smoother planes. This means that the microcracks that form in BCC & HCP will have high stresses tending to make them propagate.
For a simple cubic system, the lowest energy Burgers vectors are of the type <001> since this is the shortest distance connecting equivalent atomic positions. This means that the energy is lowest since the strain energy is proportional to the square of the Burgers vector. GIVEN:
At. wt. 0 = 16
At. wt. Mg = 24.32 Same structure as NaCl ? = 3.65 g/cm3
REQUIRED: Find length of Burgers Vector in MgO
SOLUTION: The structure of MgO is shown schematically below along with the shortest Burgers vector. To solve the problem we first note that we require the lattice parameter ao. We can take a sub-section of the unit cell (cross-hatched cube) whose edge aiso units long. 2b = ao 215.
16. 17.
We can calculate the total mass of this cube and the volume and calculate the density. Since the mass is known and the density is known, the volume may be calculated from
which ao may be extracted. Wo= + WMg++We note that there are 40= ions and 4Mg++ ions located at the corners. ? = 3However, an ion at a corner is shared by 8 such cubes. Thus, we have ? ?ao????2?1At. wt O1168-23= = ? = x = x gm Wo1023226.02 x 106.02NAV = 1.33 x 10-23 gm 1At. wt Mg124.32-23 = x = 2.02 x gmWMg++ = x 1023226.02 x NAV10 O= and ? Mg++ ions in our cube.
Thus
(1.33 + 2.02) x 10-238 x 3.35 x 10-233.65 = = 33 / 8-23aaooCheck 8 x 3.35 x 103 ao = 3.65
ao = rMg2+ + ro2- -82ao = 0.419 x 10 cm
+ 0.132nm) ao = 2(0.078nm= 0.419 nm
= 0.42nm aob = = 0.296 nm 2
18.
GIVEN: Critical resolved shear stress (0.34MPa), slip system (111)[1_10], and tensile axis [101]
REQUIRED: Applied stress at which crystal begins to deform and crystal structure. SOLUTION: (A)The situation is shown below
?crss = ? cos? ? cos?
? = angle between tensile axis and slip direction
? = angle between tensile axis and normal to slip plane
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