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Unformatted text preview: 8.1 The dislocation density is just the total dislocation length per unit volume of material (in this case
per cubic millimeters). Thus, the total length in 1000 mm3 of material having a density of 104 mm_2 is just
(104 mm‘2)(1oco mm3) = 107 mm = 104 m = 5.2 mi
. . . . . 1t} —2 .
Similarly, for a dislocation den5ity of 10 mm , the total length is (1010 mm‘2i(1000 mm3)=1013 mm :1910 m =62 31110"6 mi 8.? Below is shown the atomic packing for a BCC {110} type plane. The arrows indicate two different <11 1) type directions. 8.19 (a) Perhaps the easiest way to solve for Go and Ry in Equation (7.5) is to pick two values each of UV and {”2 from Figure 7.15, and then solve two simultaneous equations, which may be set up. For example d‘“2 (mmi‘l’r2 cry (MP3)
4 75
12 17'5
The two equations are thus 75 = on + 4kv 175 : Go + 12% These yield the values of ky = 12.5 MPa(mm)“2 [1810 psiimmim] co = 25 MPa (3530 psi) (b) When d = 1.1} x103 mm, :1'”2 = 31.5 mm'm, and, using Equation (7.5), —112
: +d
”v '30 ky —’1'2 = (25 MPa) + [12.5 MParmm11’2](31.e mm : = 420 MPa (01,000 psi) 8.24 (a) We are asked to show, for a tensile test, that _ E
%CW—(E+1)x100 From Equation (7.6) A A A —A A
%cvv=[ 0 d]x100=[1——d]xt00
0 0 Which is also equal to 
[+3]“th
(1 since Ath0 = lolld, the conservation of volume stipulation in the problem. Now, from the deﬁnition of engineering strain [Equation (6.2)] .'d'oe_1 e s 1
‘ I ‘I ' l—
o o d Substitution in the %CW expression above gives I
_ _Q _ __1 = E
stow—[1 Id]x1ﬂe_[1 8+1:Ixme [8+1]x1co (b) From Figure 6.12, a stress of 400 MPa (58,000 psi) corresponds to a strain of 0.13. Using
the above expression
0.13
X1CU= M 11100 211.5%CW _ e
%CW—[£+1 8.28 (a) We want to compute the ductility of a brass that has a yield strength of 275 MPa (40,000 psi). In order to solve this problem, it is necessary to consult Figures 7.19(a) and (c). From
Figure 7.19(a), a yield strength of 275 l'lea for brass corresponds to 10%CW. A brass that has
been cold—worked 10% will have a ductility of about 44%EL [Figure 7.19(c)].
(b) This portion of the problem asks for the Bn'nell hardness of a 1040 steel having a yield
strength of 690 MPa (100,000 psi). From Figure 7.1Q{a), a yield strength of 690 MP3 for a
1040 steel corresponds to about 11%CW. A 1040 steel that has been cold worked 11% will
have a tensile strength of about 790 lv‘lPa [Figure 7.19(b)]. Finally, using Equation (6.20a) _ TS (MP3) _ 790 MPa _
HB ‘ 3.45 ‘ 3.45 ‘ 230 8.34 (a) The driving force for recrystallization is the difference in internal energy between the strained and unstrained material.
(b) The dn'ving force for grain growth is the reduction in grain boundary energy as the total grain boundary area decreases. 3.38 (a) Using the data given and Equation (7.7) and taking n = 2, we may set up two simultaneous
equations with do and K as unknowns; thus (3.9 X102 mm)2 — d3 = (so min)K 2 2 2_
_d0 (6.6 x 10' mm) — (90 min)K Solution of these expressions yields a value for do' the original grain diameter, of d0 = 0.01 mm, and also
_ 5 2 .
K — 4.73 x 10 mm 7min (b) At 150 min, the diameter is computed as dzsld§+kt : 4:10.01 mm)2 + (4.?3 1110—5 mm27min)(150 min) = 0.085 mm 8.39 This problem calls for us to calculate the yield strength of a brass specimen alter it has been
heated to an elevated temperature at which grain growth was allowed to occur; the yield
strength was given at a grain size of 0.008 mm. It is ﬁrst necessary:r to calculate the constant 00 in Equation (7.5) as so = 0y — kyd'l’2 11'2 = 160 MPa — (12.0 MPa—mm )(0008 mm)_“2 = 25.8 MPa (4046 psi) Next, we must determine the average grain size after the heat treatment. From Figure 7.25 at 600°C after 1000 s (16.7 min) the average grain size of a brass maten'al is about 0.020 mm.
Therefore, calculating 5y at this new grain size using Equation (7.5) we get _ —1!2
cry — 1:50 + kyd = 25.3 MPa + (12.1} MPa—mm1’2)(0.020 mmr“2 = 111 MPa (16,300 psi) ...
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