HW_Soln_Ch5

# HW_Soln_Ch5 - 4 GIVEN Ibl =0.288nm in A3 REQUIRED Find...

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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()2-r)=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=OOOSB-COS¢ 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——£—=—~g-a “'" ‘ 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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