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HW-CH8 - 284 Gupta-’8 l...

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Unformatted text preview: 284 ' Gupta-’8 l DebnnafionandStrengHIeningM-edlanisms QUESTIONS AND PROBLEMS Additional problems and questions for this chapter may be found on both Student and instructor Companion Sites at. m m'ley. com/college/calllster. -—--—~——-————-———~—___....____..—_,.__—_____—________—__ Basic Concepts Characlenlsfl'c.‘ of Dislocations 8.1 To provide some perspective on the dimen- sions of atomic defects, consider a metal spec- imen that has a dislocation density of 105 mm-2 .Suppose that all the dislocations' 1n 1000 mm3 (1 cm3) were somehow removed and linked end to end. How far (in miles) would this chain extend? Now suppose that the den- sity is increased to 109 mm‘2 by cold working. What would be the chain length of dislocations in 1000 mm-3 of material? 8.2 Consider two edge dislocations of opposite sign and having slip planes that are separated by several atomic distances as indicated in the diagram. Briefly describe the defect that results when these two dislocations become aligned with each other. _J_" HT Slip Systems 8.3 (11) Compare planar densities (Section 3.15 and Problem W146 [which appears on the book’s Web site]) for the (100), (110), and (111) planes for FCC. 01) Compare planar densities (Problem 3.44) for the (100). (110), and (111) planes for BCC. 8.4 One slip system for the BCC crystal structure is {110}(111). In a manner similar to Figure 8.6!), sketch a {110}-type plane for the BCC structure, representing atom positions with cir- cles. Now, using arrows, indicate tWo different (111} slip directions within this plane. 8.5 Equations 8.1a and 8.1 b, expressions for Burg- ers vectors for FCC and BCC crystal struc- tures, are of the form I) = gummy) where a is the unit cell edge length. Since the magnitudes of these Burgers vectors may be ‘ determined from the following equation: lbl = $03 + 112 + 1112f” (3.13) determine values of lbj for copper and iron. You may want to consult Equations 3.1 and 3.3 as well as Table 3.1. Slip in Single Crystals 8.6 Consider a metal single crystal oriented so that the normal to the slip plane and the slip direc- tion are at angles of 60° and 35°, respectively, with the tensile axis. If the critical resolved shear stress is 6.2 MPa (900 psi), will an ap- 7 plied stress of 12 MPa (1750 psi) cause the sin- '- gle crystal to yield? If not, what stress will be I; necessary? 8.? Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction If slip occurs on a (111) plane and 1a a [101] direction, and 1s initiated at an applied tensile stress of 13. 9 MPa (2020 psi), compute the critical resolved shear stress. 8.8 (a) A single crystal of a metal that has the BCC crystal structure is oriented so theta tensile stress is applied in the [100] direc— .. tion. If the magnitude of this stress is 4.0' MPa compute the resolved shear stress in = the [111] direction on each of the (110),. (011) and (101) planes. (b) On the basis of these resolved shear stress values, which slip system(s)' is (are) most favorably oriented? 8.9 The critical resolved shear stress for copper is 0.48 MPa (70 psi). Determine the maximum possible yield strength for a single crystal of Cu pulled 1n tension. Strengthening by Grain Size Reduction 8.10 Briefly explain why small-angle grain bound- aries are not as effective in interfering with the _' slip process as are high-angle grain boundaries. ' 8.11 Briefly explain why I-lCP metals are typically _ more brittle than FCC and BCC metals. 8.12 (a) From the plot of yield strength versus; (grain diaineterr“2 for a 70 Cu—30 Zn-_ l g l cartridge brass (Figure 8.15). determine values for the constants no and k,, in Equa— tion 8.7. (11) Now predict the yield strength of this alloy when the average grain diameter is 2.0 x 10‘3 mm. 8.13 If it is assumed that the plot in Figure 8.15 is for noncold-worked brass, determine the grain size of the alloy in Figure 8.19; assume its composition is the same as the alloy in Figure 8.15. Strain Hardening 8.14 'IKvo previously undeformed cylindrical spec- imens of an alloy are to be strain hardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are 15 mm and 12 mm, respectively. The sec- ond specimen, with an initial radius of 11 mm, must have the same deformed hardness as the first specimen: compute the second specimen‘s radius after deformation. 8.15 A cylindrical specimen of cold—worked cop- per has a ductility (%EL) of 15%. If its cold- worked radius is 6.4 mm (0.25 in.), what was its radius before deformation? 8.16 Experimentally, it has been observed for single crystals of a number of metals that the critical resolved shear stress rum is a function of the dislocation density p0 as Tm: = rG-t-AJpD where to and A are constants. For copper. the critical resolved shear stress is 0.69 MPa (100 psi) at a dislocation density of 104 min—2. If it is known that the value of to for copper is 0.069 MPa (10 psi), compute the rm at a dislocation density of 10‘3 mm‘z. Recovery Recrystaflfzafion Grain Growth 8.17 Estimate the fraction of recrystallization from the photomicrograph in Figure 8.21a. 8.18 The average grain diameter for a brass ma- terial was measured as a function of time at 650° C. which is tabulated below at two differ- ent times: Questions and Problems 0 285 Time (min) Grain Diameter (your) 40 5.6 x 10"2 100 8.0 x 10'2 (a) What was the original grain diameter? (b) What grain diameter would you predict af- ter 200 min at 650°C? 8.19 Grain growth is strongly dependent on tem- perature (i.e., rate of grain growth increases with increasing temperature), yet temperature is not explicitly given as a part of Equation 8.9. (a) into which of the parameters in this ex- pression would you expect temperature to be included? (b) 0n the basis of your intuition. cite an ex- plicit expression for this temperature de- pendence. 8.20 An uncold-worked brass specimen of average grain size 0.01 mm has a yield strength of 150 MPa (21,750 psi). Estimate the yield strength of this alloy after it has been heated to 500°C for 1000 3, if it is known that the value of on is 25 MP3 (3625 psi). Defamation ofSerru'cryrtaMne Polymers (Deformation of Hortense“) 8.21 In your own words, describe the mechanisms by which semicrystalline polymers (2) elasti- cally deform and (h) plastically deform. and (c) by which elastomers elastically deform. Factors That influence the Mechanical Properties ofSemicryrtalEne Polymer: Deformation of Elastornen 3.22 Briefly explain how each of the following in- fluences the tensile or yield strength of a semicrystalline polymer and why: (a) Molecular weight 0)) Degree of crystallinity (c) Deformation by drawing (d) Annealing of an undeformed material 8.23 The tensile strength and number-average molecular weight for two poly(rnethyl meth- acrylate) materials are as follows: 286 ' Chapter 8 i Defamation and Strengthening Mechanisms Tensile Nam bar-A vemgc Molecular Strength (MP0) Weight {g/mol) 50 30,000 150 50,000 Estimate the tensile strength at a number- average molecular weight of 40,000 g/mol. 8.24 For each of the following pairs of polymers, do the following: (1) state whether or not it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tenv sile modulus and then cite the reason(s) for your choice; and (3) if it is not possible to de- cide, then state why. (a) Branched and atactic poly(vinyl chlo- ride) with a weight~average molecular weight of 100,000 g/moi; linear and isotac- tic poly(vinyl chloride) having a weight- average molecular weight of 75,000 g/mol (1)) Random styrene-butadiene copolymer with 5% of possible sites crosslinked; block styrene-butadiene copolymer with 10% of possible sites crosslinked (c) Branched polyethylene with a number- average molecular weight of 100,000 g/mol; atactic polypropylene with a number-average molecular weight of 150.000 g/mol 8.25 For each of the following pairs of polymers, plot and label schematic stress—strain curves on the same graph (i.e., make separate plots for parts a, b, and c). (a) Polyisoprene having a number-average molecular Weight of 100,000 glmol and 10% of available sites crosslinked; poly~ DESIGN PROBLEMS isoprene having a number-average molec- ular weight of 100,000 g/mol and 20% of available sites crosslinlted (b) Syndiotactic polypropylene having a Weight-average molecular weight of 100,000 g/mol; atactic polypropylene hav- ing a weight-average molecular weight of 75,000 glmol (c) Branched polyethylene having a number- average molecular weight of 90,000 g/Inol; heavily crosslinked polyethylene having a number-average molecular weight of 90,000 g/mol 826 Which of the following would you expect to be elastomers and which thermosetting polymers at room temperature? Justify each choice. (a) Linear and highly crystalline polyethylene (b) Heavily crosslinked polyisoprene having a glass-transition temperature of 50°C (122° F) (c) Linear and partially amorphous poly(vinyl chloride) 8.21!r Fifteen kilogram of polychloroprene is vul- canized with 5.2 kg sulfur. What fraction of ' the possible crosslink sites is bonded to sulfur - crosslinks, assuming that, on the average, 5.5 sulfur atoms participate in each crosslink? 8.28 The vulcanization of polyisoprene is accom- plished with sulfur atoms according to Equa- tion 8.12. If 45.3 wt% sulfur is combined with polyisoprene, how many crosslinks will he as- sociated with each isoprene repeat unit if it is assumed that, on the average, five sulfur atoms participate in each crosslink? 8.29 Demonstrate, in a manner similar to Equation 8.12. how vulcanization may occur in a chloro~ prene rubber. M Strain Hardening Recrystallization 8.Dl Determine whether or not it is possible to cold work steel so as to give a minimum Brinell hardness of 240, and at the same time have a ductility of at least 15 %EL. Justify your deci- sion. 8.D2 A cylindrical specimen of cold~worked steel has a Brinell hardness of 240. (a) Estimate its ductility in percent elonga- tion. (b) If the specimen remained cylindrical dur- ing deformation and its original radius 1 was 10 mm (0.40 in), determine its radius .7 after deformation. 8.03 A cylindrical rod of 1040 steel originally 11.4 mm (0.45 in.) in diameter is to be cold worked by drawing; the circular cross section will be maintained during defamation. A cold- worked tensile strength in excess of 825 MPa (120,000 psi) and a ductility of at least 12% EL are desired. Furthermore. the final diameter must be 8.9 mm (0.35 in.). Explain how this may be accomplished. Design Problems ' 281 8.D4 A cylindrical brass rod having a minimum ten- sile strength of 450 MPa (65,000 psi), a ductil- ity of at least 13% EL, and a final diameter of 12.7 mm (0.50 in.) is desired. Some 19.0 mm (0.75 in.) diameter brass stock that has been cold worked 35% is available. Describe the procedure you would follow to obtain this ma- terial. Assume that brass experiences crack- ing at 65 % CW. ...
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