For Cu, which has an FCC crystal structure, R = nm (Table 3.1) and a = 2R 2 =
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1 7.9 Equations 7.1a and 7.1b, expressions for Burgers vectors for FCC and BCC crstal structures, are of the form a b = uvw where a is the unit cell edge length. Also, since the magnitudes of these Burgers vectors ma be determined from the following equation: a 1/ b = ( u + v + w ) (7.1) determine values of b for copper and molbdenum. You ma want to consult Table 3.1. For Cu, which has an FCC crstal structure, R =.178 nm (Table 3.1) and a = R =.3614 nm (Equation 3.1); also, from Equation 7.1a, the Burgers vector for FCC metals is b = a 11 Therefore, the values for u, v, and w in Equation 7.1 are 1, 1, and, respectivel. Hence, the magnitude of the Burgers vector for Cu is b = a u + v + w.3614 nm (1) (1) ().555 nm = + + = For Mo which has a BCC crstal structure, R =.1363 nm (Table 3.1) and nm (Equation-3.3); also, from Equation 7.1b, the Burgers vector for BCC metals is 4 R a = = b = a 111 Therefore, the values for u, v, and w in Equation 7.1 are 1, 1, and 1, respectivel. Hence, the magnitude of the Burgers vector for Mo is b.3147 nm = (1) + (1) + (1) =.75 nm Excerpts from this work ma be reproduced b instructors for distribution on a not-for-profit basis for testing or instructional purposes onl to students enrolled in courses for which the textbook has been adopted. An other reproduction or translation of this work beond that permitted b Sections 17 or 18 of the 1976 United States Copright Act without the permission of the copright owner is unlawful.
2 7.1 Consider a metal single crstal oriented such that the normal to the slip plane and the slip direction are at angles of 43.1 and 47.9, respectivel, with the tensile axis. If the critical resolved shear stress is MPa, will an applied stress of 5 MPa cause the single crstal to ield? If not, what stress will be necessar? This problem calls for us to determine whether or not a metal single crstal having a specific orientation and of given critical resolved shear stress will ield. We are given that φ = 43.1, λ = 47.9, and that the values of the critical resolved shear stress and applied tensile stress are MPa and 5 MPa respectivel. From Equation 7. τ = σ cos φ cos λ = (5 MPa)(cos 43.1 )(cos 47.9 ) = 4.5 MPa R Since the resolved shear stress (4.5 MPa) is greater than the critical resolved shear stress ( MPa), the single crstal will ield. Excerpts from this work ma be reproduced b instructors for distribution on a not-for-profit basis for testing or instructional purposes onl to students enrolled in courses for which the textbook has been adopted. An other reproduction or translation of this work beond that permitted b Sections 17 or 18 of the 1976 United States Copright Act without the permission of the copright owner is unlawful.
3 7.17 Consider a single crstal of some hpothetical metal that has the FCC crstal structure and is oriented such that a tensile stress is applied along a [1] direction. If slip occurs on a (111) plane and in a [11] direction, compute the stress at which the crstal ields if its critical resolved shear stress is 3. MPa. λ = cos cos uu + vv + ww ( u1 + v1 + w1)( u + v + w) ( 1)( 1) + ()() + ()(1) 1 = ( 1) + + ( 1) = cos = φ ( 1)(1) + ()(1) + ()(1) 1 = cos 1 1 = cos = [ ] ( 1) τ crss 3.Mpa σ = cosφcos λ = = 13.6Mpa
4 7.3 (a) From the plot of ield strength versus (grain diameter) 1/ for a 7 Cu 3 Zn cartridge brass, Figure 7.15, determine values for the constants σ and k in Equation 7.7. (b) Now predict the ield strength of this allo when the average grain diameter is. 1 3 mm. (a) Perhaps the easiest wa to solve for σ and k in Equation 7.7 is to pick two values each of σ and d -1/ from Figure 7.15, and then solve two simultaneous equations, which ma be created. For example d 1/ (mm) 1/ σ (MPa) The two equations are thus σ 75 = + 4 k σ 175 = + 1 k of these equations ield the values of k 1/ = 1.5 MPa (mm) σ = 5 MPa (b) When d =. 1 3 mm, d 1/ =.36 mm 1/, and, using Equation 7.7, σ = σ + kd 1/ ( ) 1/ 1/ = (5 MPa) MPa (mm).36 mm = 35 MPa Excerpts from this work ma be reproduced b instructors for distribution on a not-for-profit basis for testing or instructional purposes onl to students enrolled in courses for which the textbook has been adopted. An other reproduction or translation of this work beond that permitted b Sections 17 or 18 of the 1976 United States Copright Act without the permission of the copright owner is unlawful.
5 Strain Hardening 7.7 (a) Show, for a tensile test, that %CW = if there is no change in specimen volume during the deformation process (i.e., A l = A d l d ). (b) Using the result of part (a), compute the percent cold work experienced b naval brass (the stress-strain behavior of which is shown in Figure 6.1) when a stress of 4 MPa is applied. (a) From Equation 7.8 A A A d d %CW = 1 = 1 1 A A Which is also equal to 1 l 1 l d since A d /A = l /l d, the conservation of volume stipulation given in the problem statement. Now, from the definition of engineering strain (Equation 6.) l l l l l d d = = 1 Or, l l = d Substitution for l /l d into the %CW expression above gives l 1 %CW = 1 1 = 1 1 = 1 ld (b) From Figure 6.1, a stress of 4 MPa corresponds to a strain of.13. Using the above expression.13 %CW = 1 = 1 = 11.5%CW Excerpts from this work ma be reproduced b instructors for distribution on a not-for-profit basis for testing or instructional purposes onl to students enrolled in courses for which the textbook has been adopted. An other reproduction or translation of this work beond that permitted b Sections 17 or 18 of the 1976 United States Copright Act without the permission of the copright owner is unlawful.
6 7.9 Two previousl undeformed specimens of the same metal are to be plasticall deformed b reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular; during deformation the circular cross section is to remain circular, and the rectangular is to remain as such. Their original and deformed dimensions are as follows: Circular (diameter, mm) Rectangular (mm) Original dimensions Deformed dimensions Which of these specimens will be the hardest after plastic deformation, and wh? The hardest specimen will be the one that has experienced the greatest degree of cold work. Therefore, all we need do is to compute the %CW for each specimen using Equation 7.8. For the circular one %CW = A A d 1 A 15. mm 11.4 mm π π = 1 = 43.8%CW 15. mm π 15. mm 11.4 mm π π = 1 = 43.8%CW 15. mm π For the rectangular one (15 mm)(175 mm) (75 mm)( mm) %CW = 1 = 31.4%CW (15 mm)(175 mm) Therefore, the deformed circular specimen will be harder. Excerpts from this work ma be reproduced b instructors for distribution on a not-for-profit basis for testing or instructional purposes onl to students enrolled in courses for which the textbook has been adopted. An other reproduction or translation of this work beond that permitted b Sections 17 or 18 of the 1976 United States Copright Act without the permission of the copright owner is unlawful.
7 7.38 The average grain diameter for a brass material was measured as a function of time at 65 C, which is tabulated below at two different times: Time Grain Diameter (min) (mm) (a) What was the original grain diameter? (b) What grain diameter would ou predict after 15 min at 65 C? (a) Using the data given and Equation 7.9 (taking n = ), we ma set up two simultaneous equations with d and K as unknowns; thus ( mm ) d = (3 min) K ( mm ) d = (9 min) K of these expressions ields a value for d, the original grain diameter, of and a value for K of mm /min d =.1 mm, (b) At 15 min, the diameter d is computed using a rearranged form of Equation 7.9 as d = d + Kt = (.1 mm) + ( mm /min)(15 min) =.85 mm Excerpts from this work ma be reproduced b instructors for distribution on a not-for-profit basis for testing or instructional purposes onl to students enrolled in courses for which the textbook has been adopted. An other reproduction or translation of this work beond that permitted b Sections 17 or 18 of the 1976 United States Copright Act without the permission of the copright owner is unlawful.
8 7.41 An uncold-worked brass specimen of average grai n size.9 mm has a ield strength of 16 MPa. Estimate the ield strength of this allo after it has been heated to 6 C for 1 s, if it is known that the value of k is 1. MPa mm 1/. as In order to solve this problem, it is first necessar to calculate the constant σ in Equation 7.7 σ = σ kd 1/ ( ) 1/ 1/ = 16 MPa 1. MPa mm (.9 mm) = 33.5 MPa Next, we must determine the average grain size af ter the heat treatment. From Figure 7.5 at 6 C (873 K) after 1 s (16.7 min) the average grain size of a brass material is about. mm. Therefore, calculating σ at this new grain size using Equation 7.7 we get σ = σ + kd 1/ ( ) 1/ 1/ = 33.5 MPa + 1. MPa mm (. mm) =6.3 MPa Excerpts from this work ma be reproduced b instructors for distribution on a not- for-profit basis for testing or instructiona l purposes onl to students enrolled in courses for which the textbook has been adopted. An other reproduction or translation of this work beond that permitted b Sections 17 or 18 of the 1976 United States Copright Act without the permission of the copright owner is unlawful.
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