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Unformatted text preview: 4. GIVEN: Ibl =0.288nm in A3
REQUIRED: Find lattice parameter SOLUTION: Recall the Ag is FCC. For FCC structures the Burgers vector is V: a face b=J3a.= a.
2 J3 ao=ﬁb = 2 x0288 a0 = 0. 407nm diagonal as shown. We see that The (111) plane is shown in a unit cell with all atoms shown. Atoms touch along face
diagonals. The (1 1 1) plane is the most closely packed, and the vectors shown connect equivalent atomic position. Thus b = g [710] etc. Then in general b = g < 110 > B. For NaCl We see that the shortest vector connecting equivalent positions is g—[lw] as shown. This direction lies in both the {100} and {110} planes and both are possible slip planes.
However {110} are the planes most frequently observed as the slip planes. This is
because repulsive interionic forces are minimized on these planes during dislocation
motion. Thus we expect 1/2<110> Burgers vectors and {110} slip planes. 13. GIVEN: Dislocation reaction below: REQUIRED: Show it is vectorially correct and energetically proper.
SOLUTION: % [111] + g—[Tii] =a [100] The sum of the x, y & 2 components on the LHS must be equal to the corresponding
component on the right hand side. _?
a a _
30(1)+3I(1)—ayes x component (LHS) = x component (RHS) .7
%(1)+%o(1)=oyes y component (LHS) = y component (RHS) 7
a a
—1+— 1 ca
2()2r)=ys 2 component (LHS) = 2 component (RHS)
Energy: The reaction is energetically favorable if I b. l 2 +  132' 2 > I b3] 3 2 2 2 2
z g+$+1= Fi=£a
4 4 4 4 2 Thus the reaction is favorable since 3a” + 3:12 > a2 19. GIVEN: tam = 55.2 MPa, (1 l l)ﬂ01] slip system, [112} tensile axis
REQUIRED: Find the highest normal stress that can be applied before dislocation
motion in the [10 1] direction.
SOLUTION: The situation is shown below. Essentially the problem reduces to ﬁnding
the value of the tensile stress when the critical resolved shear is reached. Tm=OOOSBCOS¢
a: [112] L101] ¢=[112] {111} [1121 [E01] = J3 J? 0050 [1121 [1111: J3 J3 cos¢ 2 [1121 [201] _ = [1121 [111]
c059 ———————J1_2 , cos¢ —————m J— r $552=a;0——£—=—~ga
“'" ‘ 2J3 3J5 3J3 B. Would have My the same stress for a BCC metal (at & 0 would be interchanged). GIVEN: o = 1.7 MPa [100] tensile axis (111)[101] slip systems REQUIRED: 1m, 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. 0 : [1001/10i] ¢ = [100] [111] rm, 7: a c050 cosq) [100/o[10i] [100] [111]
COng————'—— COS :————‘
[X's/E ¢ Ixﬁ
1 1
c030=#; cos =—
43 ¢ ﬁ
rm—1.7x—l—x—1—20.69MPa 2 3 ...
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 Summer '08
 Tannebaum

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