材料科学与工程基础英文版试题
《材料科学与工程基础》英文影印版习题及思考题及答案
《材料科学与工程基础》英文习题及思考题及答案第二章习题和思考题Questions and Problems2.6 Allowed values for the quantum numbers ofelectrons are as follows:The relationships between n and the shell designationsare noted in Table 2.1.Relative tothe subshells,l ＝0 corresponds to an s subshelll ＝1 corresponds to a p subshelll ＝2 corresponds to a d subshelll ＝3 corresponds to an f subshellFor the K shell, the four quantum numbersfor each of the two electrons in the 1s state, inthe order of nlmlms , are 100(1/2 ) and 100(-1/2 ).Write the four quantum numbers for allof the electrons inthe L and M shells, and notewhich correspond to the s, p, and d subshells.2.7 Give the electron configurations for the followingions: Fe2+, Fe3+, Cu+, Ba2+,Br-, andS2-.2.17 (a) Briefly cite the main differences betweenionic, covalent, and metallicbonding.(b) State the Pauli exclusion principle.2.18 Offer an explanation as to why covalently bonded materials are generally lessdense than ionically or metallically bonded ones.2.19 Compute the percents ionic character of the interatomic bonds for the followingcompounds: TiO2 , ZnTe, CsCl, InSb, and MgCl2 .2.21 Using Table 2.2, determine the number of covalent bonds that are possible foratoms of the following elements: germanium, phosphorus, selenium, and chlorine.2.24 On the basis of the hydrogen bond, explain the anomalous behavior of waterwhen it freezes. That is, why is there volume expansion upon solidification?3.1 What is the difference between atomic structure and crystal structure?3.2 What is the difference between a crystal structure and a crystal system?3.4Show for the body-centered cubic crystal structure that the unit cell edge lengtha and the atomic radius R are related through a =4R/√3.3.6 Show that the atomic packing factor for BCC is 0.68. .3.27* Show that the minimum cation-to-anion radius ratio for a coordinationnumber of 6 is 0.414. Hint: Use the NaCl crystal structure (Figure 3.5), and assume that anions and cations are just touching along cube edges and across face diagonals.3.48 Draw an orthorhombic unit cell, and within that cell a [121] direction and a(210) plane.3.50 Here are unit cells for two hypothetical metals:(a)What are the indices for the directions indicated by the two vectors in sketch(a)?(b) What are the indices for the two planes drawn in sketch (b)?3.51* Within a cubic unit cell, sketch the following directions:.3.53 Determine the indices for the directions shown in the following cubic unit cell:3.57 Determine the Miller indices for the planesshown in the following unit cell:3.58Determine the Miller indices for the planes shown in the following unit cell: 3.61* Sketch within a cubic unit cell the following planes:3.62 Sketch the atomic packing of (a) the (100)plane for the FCC crystal structure, and (b) the (111) plane for the BCC crystal structure (similar to Figures 3.24b and 3.25b).3.77 Explain why the properties of polycrystalline materials are most oftenisotropic.5.1 Calculate the fraction of atom sites that are vacant for lead at its meltingtemperature of 327_C. Assume an energy for vacancy formation of 0.55eV/atom.5.7 If cupric oxide (CuO) is exposed to reducing atmospheres at elevatedtemperatures, some of the Cu2_ ions will become Cu_.(a) Under these conditions, name one crystalline defect that you would expect toform in order to maintain charge neutrality.(b) How many Cu_ ions are required for the creation of each defect?5.8 Below, atomic radius, crystal structure, electronegativity, and the most commonvalence are tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated.Which of these elements would you expect to form the following with copper:(a) A substitutional solid solution having complete solubility?(b) A substitutional solid solution of incomplete solubility?(c) An interstitial solid solution?5.9 For both FCC and BCC crystal structures, there are two different types ofinterstitial sites. In each case, one site is larger than the other, which site isnormally occupied by impurity atoms. For FCC, this larger one is located at the center of each edge of the unit cell; it is termed an octahedral interstitial site. On the other hand, with BCC the larger site type is found at 0, __, __ positions—that is, lying on _100_ faces, and situated midway between two unit cell edges on this face and one-quarter of the distance between the other two unit cell edges; it is termed a tetrahedral interstitial site. For both FCC and BCC crystalstructures, compute the radius r of an impurity atom that will just fit into one of these sites in terms of the atomic radius R of the host atom.5.10 (a) Suppose that Li2O is added as an impurity to CaO. If the Li_ substitutes forCa2_, what kind of vacancies would you expect to form? How many of thesevacancies are created for every Li_ added?(b) Suppose that CaCl2 is added as an impurity to CaO. If the Cl_ substitutes forO2_, what kind of vacancies would you expect to form? How many of thevacancies are created for every Cl_ added?5.28 Copper and platinum both have the FCC crystal structure and Cu forms asubstitutional solid solution for concentrations up to approximately 6 wt% Cu at room temperature. Compute the unit cell edge length for a 95 wt% Pt-5 wt% Cu alloy.5.29 Cite the relative Burgers vector–dislocation line orientations for edge, screw, andmixed dislocations.6.1 Briefly explain the difference between selfdiffusion and interdiffusion.6.3 (a) Compare interstitial and vacancy atomic mechanisms for diffusion.(b) Cite two reasons why interstitial diffusion is normally more rapid thanvacancy diffusion.6.4 Briefly explain the concept of steady state as it applies to diffusion.6.5 (a) Briefly explain the concept of a driving force.(b) What is the driving force for steadystate diffusion?6.6Compute the number of kilograms of hydrogen that pass per hour through a5-mm thick sheet of palladium having an area of 0.20 m2at 500℃. Assume a diffusion coefficient of 1.0×10- 8 m2/s, that the concentrations at the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.6.7 A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at 1200℃and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 6×10-11m2/s, and the diffusion flux is found to be 1.2×10- 7kg/m2-s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m3. How far into the sheet from this high-pressure side will the concentration be 2.0 kg/m3?Assume a linear concentration profile.6.24. Carbon is allowed to diffuse through a steel plate 15 mm thick. Theconcentrations of carbon at the two faces are 0.65 and 0.30 kg C/m3 Fe, whichare maintained constant. If the preexponential and activation energy are 6.2 _10_7 m2/s and 80,000 J/mol, respectively, compute the temperature at which the diffusion flux is 1.43 _ 10_9 kg/m2-s.6.25 The steady-state diffusion flux through a metal plate is 5.4_10_10 kg/m2-s at atemperature of 727_C (1000 K) and when the concentration gradient is _350kg/m4. Calculate the diffusion flux at 1027_C (1300 K) for the sameconcentration gradient and assuming an activation energy for diffusion of125,000 J/mol.10.2 What thermodynamic condition must be met for a state of equilibrium to exist? 10.4 What is the difference between the states of phase equilibrium and metastability?10.5 Cite the phases that are present and the phase compositions for the followingalloys:(a) 90 wt% Zn–10 wt% Cu at 400℃(b) 75 wt% Sn–25wt%Pb at 175℃(c) 55 wt% Ag–45 wt% Cu at 900℃(d) 30 wt% Pb–70 wt% Mg at 425℃(e) 2.12 kg Zn and 1.88 kg Cu at 500℃(f ) 37 lbm Pb and 6.5 lbm Mg at 400℃(g) 8.2 mol Ni and 4.3 mol Cu at 1250℃.(h) 4.5 mol Sn and 0.45 mol Pb at 200℃10.6 For an alloy of composition 74 wt% Zn–26 wt% Cu, cite the phases presentand their compositions at the following temperatures: 850℃, 750℃, 680℃, 600℃, and 500℃.10.7 Determine the relative amounts (in terms of mass fractions) of the phases forthe alloys and temperatures given inProblem 10.5.10.9 Determine the relative amounts (interms of volume fractions) of the phases forthe alloys and temperatures given inProblem 10.5a, b, and c. Below are given theapproximate densities of the various metalsat the alloy temperatures:10.18 Is it possible to have a copper–silveralloy that, at equilibrium, consists of a _ phase of composition 92 wt% Ag–8wt% Cu, and also a liquid phase of composition 76 wt% Ag–24 wt% Cu? If so, what will be the approximate temperature of the alloy? If this is not possible,explain why.10.20 A copper–nickel alloy of composition 70 wt% Ni–30 wt% Cu is slowly heatedfrom a temperature of 1300_C .(a) At what temperature does the first liquid phase form?(b) What is the composition of this liquid phase?(c) At what temperature does complete melting of the alloy occur?(d) What is the composition of the last solid remaining prior to completemelting?10.28 .Is it possible to have a copper–silver alloy of composition 50 wt% Ag–50 wt%Cu, which, at equilibrium, consists of _ and _ phases having mass fractions W_ _0.60 and W_ _ 0.40? If so, what will be the approximate temperature of the alloy?If such an alloy is not possible, explain why.10.30 At 700_C , what is the maximum solubility (a) of Cu in Ag? (b) Of Ag in Cu?第三章习题和思考题3.3If the atomic radius of aluminum is 0.143nm, calculate the volume of its unitcell in cubic meters.3.8 Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomicweight of 55.85 g/mol. Compute and compare its density with the experimental value found inside the front cover.3.9 Calculate the radius of an iridium atom given that Ir has an FCC crystal structure,a density of 22.4 g/cm3, and an atomic weight of 192.2 g/mol.3.13 Using atomic weight, crystal structure, and atomic radius data tabulated insidethe front cover, compute the theoretical densities of lead, chromium, copper, and cobalt, and then compare these values with the measured densities listed in this same table. The c/a ratio for cobalt is 1.623.3.15 Below are listed the atomic weight, density, and atomic radius for threehypothetical alloys. For each determine whether its crystal structure is FCC,BCC, or simple cubic and then justify your determination. A simple cubic unitcell is shown in Figure 3.40.3.21 This is a unit cell for a hypotheticalmetal:(a) To which crystal system doesthis unit cell belong?(b) What would this crystal structure be called?(c) Calculate the density of the material, given that its atomic weight is 141g/mol.3.25 For a ceramic compound, what are the two characteristics of the component ionsthat determine the crystal structure?3.29 On the basis of ionic charge and ionic radii, predict the crystal structures for thefollowing materials: (a) CsI, (b) NiO, (c) KI, and (d) NiS. Justify your selections.3.35 Magnesium oxide has the rock salt crystal structure and a density of 3.58 g/cm3.(a) Determine the unit cell edge length. (b) How does this result compare withthe edge length as determined from the radii in Table 3.4, assuming that theMg2_ and O2_ ions just touch each other along the edges?3.36 Compute the theoretical density of diamond given that the CUC distance andbond angle are 0.154 nm and 109.5°, respectively. How does this value compare with the measured density?3.38 Cadmium sulfide (CdS) has a cubic unit cell, and from x-ray diffraction data it isknown that the cell edge length is 0.582 nm. If the measured density is 4.82 g/cm3 , how many Cd 2+ and S 2—ions are there per unit cell?3.41 A hypothetical AX type of ceramic material is known to have a density of 2.65g/cm 3 and a unit cell of cubic symmetry with a cell edge length of 0.43 nm. The atomic weights of the A and X elements are 86.6 and 40.3 g/mol, respectively.On the basis of this information, which of the following crystal structures is (are) possible for this material: rock salt, cesium chloride, or zinc blende? Justify your choice(s).3.42 The unit cell for Mg Fe2O3 (MgO-Fe2O3) has cubic symmetry with a unit celledge length of 0.836 nm. If the density of this material is 4.52 g/cm 3 , compute its atomic packing factor. For this computation, you will need to use ionic radii listed in Table 3.4.3.44 Compute the atomic packing factor for the diamond cubic crystal structure(Figure 3.16). Assume that bonding atoms touch one another, that the angle between adjacent bonds is 109.5°, and that each atom internal to the unit cell is positioned a/4 of the distance away from the two nearest cell faces (a is the unit cell edge length).3.45 Compute the atomic packing factor for cesium chloride using the ionic radii inTable 3.4 and assuming that the ions touch along the cube diagonals.3.46 In terms of bonding, explain why silicate materials have relatively low densities.3.47 Determine the angle between covalent bonds in an SiO44—tetrahedron.3.63 For each of the following crystal structures, represent the indicated plane in themanner of Figures 3.24 and 3.25, showing both anions and cations: (a) (100)plane for the rock salt crystal structure, (b) (110) plane for the cesium chloride crystal structure, (c) (111) plane for the zinc blende crystal structure, and (d) (110) plane for the perovskite crystal structure.3.66 The zinc blende crystal structure is one that may be generated from close-packedplanes of anions.(a) Will the stacking sequence for this structure be FCC or HCP? Why?(b) Will cations fill tetrahedral or octahedral positions? Why?(c) What fraction of the positions will be occupied?3.81* The metal iridium has an FCC crystal structure. If the angle of diffraction forthe (220) set of planes occurs at 69.22°(first-order reflection) when monochromatic x-radiation having a wavelength of 0.1542 nm is used, compute(a) the interplanar spacing for this set of planes, and (b) the atomic radius for aniridium atom.4.10 What is the difference between configuration and conformation in relation topolymer chains? vinyl chloride).4.22 (a) Determine the ratio of butadiene to styrene mers in a copolymer having aweight-average molecular weight of 350,000 g/mol and weight-average degree of polymerization of 4425.(b) Which type(s) of copolymer(s) will this copolymer be, considering thefollowing possibilities: random, alternating, graft, and block? Why?4.23 Crosslinked copolymers consisting of 60 wt% ethylene and 40 wt% propylenemay have elastic properties similar to those for natural rubber. For a copolymer of this composition, determine the fraction of both mer types.4.25 (a) Compare the crystalline state in metals and polymers.(b) Compare thenoncrystalline state as it applies to polymers and ceramic glasses.4.26 Explain briefly why the tendency of a polymer to crystallize decreases withincreasing molecular weight.4.27* For each of the following pairs of polymers, do the following: (1) state whetheror not it is possible to determine if one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.(a) Linear and syndiotactic polyvinyl chloride; linear and isotactic polystyrene.(b) Network phenol-formaldehyde; linear and heavily crosslinked ci s-isoprene.(c) Linear polyethylene; lightly branched isotactic polypropylene.(d) Alternating poly(styrene-ethylene) copolymer; randompoly(vinylchloride-tetrafluoroethylene) copolymer.4.28 Compute the density of totally crystalline polyethylene. The orthorhombic unitcell for polyethylene is shown in Figure 4.10; also, the equivalent of two ethylene mer units is contained within each unit cell.5.11 What point defects are possible for MgO as an impurity in Al2O3? How manyMg 2+ ions must be added to form each of these defects?5.13 What is the composition, in weight percent, of an alloy that consists of 6 at% Pband 94 at% Sn?5.14 Calculate the composition, in weight per-cent, of an alloy that contains 218.0 kgtitanium, 14.6 kg of aluminum, and 9.7 kg of vanadium.5.23 Gold forms a substitutional solid solution with silver. Compute the number ofgold atoms per cubic centimeter for a silver-gold alloy that contains 10 wt% Au and 90 wt% Ag. The densities of pure gold and silver are 19.32 and 10.49 g/cm3 , respectively.8.53 In terms of molecular structure, explain why phenol-formaldehyde (Bakelite)will not be an elastomer.10.50 Compute the mass fractions of αferrite and cementite in pearlite. assumingthat pressure is held constant.10.52 (a) What is the distinction between hypoeutectoid and hypereutectoid steels?(b) In a hypoeutectoid steel, both eutectoid and proeutectoid ferrite exist. Explainthe difference between them. What will be the carbon concentration in each?10.56 Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 727_C(a) What is the proeutectoid phase?(b) How many kilograms each of total ferrite and cementite form?(c) How many kilograms each of pearlite and the proeutectoid phase form?(d) Schematically sketch and label the resulting microstructure.10.60 The mass fractions of total ferrite and total cementite in an iron–carbon alloyare 0.88 and 0.12, respectively. Is this a hypoeutectoid or hypereutectoid alloy?Why?10.64 Is it possible to have an iron–carbon alloy for which the mass fractions of totalferrite and proeutectoid cementite are 0.846 and 0.049, respectively? Why orwhy not?第四章习题和思考题7.3 A specimen of aluminum having a rectangular cross section 10 mm _ 12.7 mmis pulled in tension with 35,500 N force, producing only elastic deformation. 7.5 A steel bar 100 mm long and having a square cross section 20 mm on an edge ispulled in tension with a load of 89,000 N , and experiences an elongation of 0.10 mm . Assuming that the deformation is entirely elastic, calculate the elasticmodulus of the steel.7.7 For a bronze alloy, the stress at which plastic deformation begins is 275 MPa ,and the modulus of elasticity is 115 Gpa .(a) What is the maximum load that may be applied to a specimen with across-sectional area of 325mm, without plastic deformation?(b) If the original specimen length is 115 mm , what is the maximum length towhich it may be stretched without causing plastic deformation?7.8 A cylindrical rod of copper (E _ 110 GPa, Stress (MPa) ) having a yield strengthof 240Mpa is to be subjected to a load of 6660 N. If the length of the rod is 380 mm, what must be the diameter to allow an elongation of 0.50 mm?7.9 Consider a cylindrical specimen of a steel alloy (Figure 7.33) 10mm in diameterand 75 mm long that is pulled in tension. Determine its elongation when a load of 23,500 N is applied.7.16 A cylindrical specimen of some alloy 8 mm in diameter is stressed elasticallyin tension. A force of 15,700 N produces a reduction in specimen diameter of 5 _ 10_3 mm. Compute Poisson’s ratio for this material if its modulus of elasticity is 140 GPa .7.17 A cylindrical specimen of a hypothetical metal alloy is stressed in compression.If its original and final diameters are 20.000 and 20.025 mm, respectively, and its final length is 74.96 mm, compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 105 Gpa and 39.7 GPa,respectively.7.19 A brass alloy is known to have a yield strength of 275 MPa, a tensile strength of380 MPa, and an elastic modulus of 103 GPa . A cylindrical specimen of thisalloy 12.7 mm in diameter and 250 mm long is stressed in tension and found to elongate 7.6 mm . On the basis of the information given, is it possible tocompute the magnitude of the load that is necessary to produce this change inlength? If so, calculate the load. If not, explain why.7.20A cylindrical metal specimen 15.0mmin diameter and 150mm long is to besubjected to a tensile stress of 50 Mpa; at this stress level the resulting deformation will be totally elastic.(a)If the elongation must be less than 0.072mm,which of the metals in Tabla7.1are suitable candidates? Why ?(b)If, in addition, the maximum permissible diameter decrease is 2.3×10-3mm,which of the metals in Table 7.1may be used ? Why?7.22 Cite the primary differences between elastic, anelastic, and plastic deformationbehaviors.7.23 diameter of 10.0 mm is to be deformed using a tensile load of 27,500 N. It mustnot experience either plastic deformation or a diameter reduction of more than7.5×10-3 mm. Of the materials listed as follows, which are possible candidates?Justify your choice(s).7.24 A cylindrical rod 380 mm long, having a diameter of 10.0 mm, is to besubjected to a tensile load. If the rod is to experience neither plastic deformationnor an elongation of more than 0.9 mm when the applied load is 24,500 N,which of the four metals or alloys listed below are possible candidates?7.25 Figure 7.33 shows the tensile engineering stress–strain behavior for a steel alloy.(a) What is the modulus of elasticity?(b) What is the proportional limit?(c) What is the yield strength at a strain offset of 0.002?(d) What is the tensile strength?7.27 A load of 44,500 N is applied to a cylindrical specimen of steel (displaying thestress–strain behavior shown in Figure 7.33) that has a cross-sectional diameter of 10 mm .(a) Will the specimen experience elastic or plastic deformation? Why?(b) If the original specimen length is 500 mm), how much will it increase inlength when t his load is applied?7.29 A cylindrical specimen of aluminumhaving a diameter of 12.8 mm and a gaugelength of 50.800 mm is pulled in tension. Usethe load–elongation characteristics tabulatedbelow to complete problems a through f.(a)Plot the data as engineering stressversusengineering strain.(b) Compute the modulus of elasticity.(c) Determine the yield strength at astrainoffset of 0.002.(d) Determine the tensile strength of thisalloy.(e) What is the approximate ductility, in percent elongation?(f ) Compute the modulus of resilience.7.35 (a) Make a schematic plot showing the tensile true stress–strain behavior for atypical metal alloy.(b) Superimpose on this plot a schematic curve for the compressive truestress–strain behavior for the same alloy. Explain any difference between thiscurve and the one in part a.(c) Now superimpose a schematic curve for the compressive engineeringstress–strain behavior for this same alloy, and explain any difference between this curve and the one in part b.7.39 A tensile test is performed on a metal specimen, and it is found that a true plasticstrain of 0.20 is produced when a true stress of 575 MPa is applied; for the same metal, the value of K in Equation 7.19 is 860 MPa. Calculate the true strain that results from the application of a true stress of 600 Mpa.7.40 For some metal alloy, a true stress of 415 MPa produces a plastic true strain of0.475. How much will a specimen of this material elongate when a true stress of325 MPa is applied if the original length is 300 mm ? Assume a value of 0.25 for the strain-hardening exponent n.7.43 Find the toughness (or energy to cause fracture) for a metal that experiences bothelastic and plastic deformation. Assume Equation 7.5 for elastic deformation,that the modulus of elasticity is 172 GPa , and that elastic deformation terminates at a strain of 0.01. For plastic deformation, assume that the relationship between stress and strain is described by Equation 7.19, in which the values for K and n are 6900 Mpa and 0.30, respectively. Furthermore, plastic deformation occurs between strain values of 0.01 and 0.75, at which point fracture occurs.7.47 A steel specimen having a rectangular cross section of dimensions 19 mm×3.2mm (0.75in×0.125in.) has the stress–strain behavior shown in Figure 7.33. If this specimen is subjected to a tensile force of 33,400 N (7,500lbf ), then(a) Determine the elastic and plastic strain values.(b) If its original length is 460 mm (18 in.), what will be its final length after theload in part a is applied and then released?7.50 A three-point bending test was performed on an aluminum oxide specimenhaving a circular cross section of radius 3.5 mm; the specimen fractured at a load of 950 N when the distance between the support points was 50 mm . Another test is to be performed on a specimen of this same material, but one that has a square cross section of 12 mm length on each edge. At what load would you expect this specimen to fracture if the support point separation is 40 mm ?7.51 (a) A three-point transverse bending test is conducted on a cylindrical specimenof aluminum oxide having a reported flexural strength of 390 MPa . If the speci- men radius is 2.5 mm and the support point separation distance is 30 mm ,predict whether or not you would expect the specimen to fracture when a load of 620 N is applied. Justify your prediction.(b) Would you be 100% certain of the prediction in part a? Why or why not?7.57 When citing the ductility as percent elongation for semicrystalline polymers, it isnot necessary to specify the specimen gauge length, as is the case with metals.Why is this so?7.66 Using the data represented in Figure 7.31, specify equations relating tensilestrength and Brinell hardness for brass and nodular cast iron, similar toEquations 7.25a and 7.25b for steels.8.4 For each of edge, screw, and mixed dislocations, cite the relationship between thedirection of the applied shear stress and the direction of dislocation line motion.8.5 (a) Define a slip system.(b) Do all metals have the same slip system? Why or why not?8.7. One slip system for theBCCcrystal structure is _110__111_. In a manner similarto Figure 8.6b sketch a _110_-type plane for the BCC structure, representingatom positions with circles. Now, using arrows, indicate two different _111_ slip directions within this plane.8.15* List four major differences between deformation by twinning and deformationby slip relative to mechanism, conditions of occurrence, and final result.8.18 Describe in your own words the three strengthening mechanisms discussed inthis chapter (i.e., grain size reduction, solid solution strengthening, and strainhardening). Be sure to explain how dislocations are involved in each of thestrengthening techniques.8.19 (a) From the plot of yield strength versus (grain diameter)_1/2 for a 70 Cu–30 Zncartridge brass, Figure 8.15, determine values for the constants _0 and ky inEquation 8.5.(b) Now predict the yield strength of this alloy when the average grain diameteris 1.0 _ 10_3 mm.8.20. The lower yield point for an iron that has an average grain diameter of 5 _ 10_2mm is 135 MPa . At a grain diameter of 8 _ 10_3 mm, the yield point increases to 260MPa. At what grain diameter will the lower yield point be 205 Mpa ?8.24 (a) Show, for a tensile test, thatif there is no change in specimen volume during the deformation process (i.e., A0 l0 _Ad ld).(b) Using the result of part a, compute the percent cold work experienced bynaval brass (the stress–strain behavior of which is shown in Figure 7.12) when a stress of 400 MPa is applied.8.25 Two previously undeformed cylindrical specimens of an alloy are to be strainhardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are 16 mm and11 mm, respectively. The second specimen, with an initial radius of 12 mm, musthave the same deformed hardness as the first specimen; compute the secondspecimen’s radius after deformation.8.26 Two previously undeformed specimens of the same metal are to be plasticallydeformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular is to remain as such. Their original and deformeddimensions are as follows:Which of these specimens will be the hardest after plastic deformation, and why?8.27 A cylindrical specimen of cold-worked copper has a ductility (%EL) of 25%. Ifits coldworked radius is 10 mm (0.40 in.), what was its radius beforedeformation?8.28 (a) What is the approximate ductility (%EL) of a brass that has a yield strengthof 275 MPa ?(b) What is the approximate Brinell hardness of a 1040 steel having a yieldstrength of 690 MPa?8.41 In your own words, describe the mechanisms by which semicrystalline polymers(a) elasticallydeform and (b) plastically deform, and (c) by which elastomerselastically deform.8.42 Briefly explain how each of the following influences the tensile modulus of asemicrystallinepolymer and why:(a) molecular weight;(b) degree of crystallinity;(c) deformation by drawing;(d) annealing of an undeformed material;(e) annealing of a drawn material.8.43* Briefly explain how each of the following influences the tensile or yieldstrength of a semicrystalline polymer and why:(a) molecular weight;。
材料科学专业英语复习题.docx
复习题1、On the basis of crystal structure, using the following equation and values of parameters, compute the theoretical density for sodium chloride (NaCI). How does this compare with its measured density? Measured density of NaCI is 2.16 g/cm3.Theoretical density formula:n' = the number of formula units1 within the unit cellSX C = the sum of the atomic weights of all cations in the formula unitSX A = the sum of the atomic weights of all anions in the formula unitV c = the unit cell volumeN A = Avogadro's number, 6.023 X 1023 formula units/mol4Na = 22.99 g/molA C} = 35.45 g/molr Na+ =0.102nm r c f= 0.18 InmONa+ O Cl-Solution:According to the structure of NaCI, each unit cell consist of 4 NaCI formula units. Therefore,n =4.Furthermore, since the unit cell of NaCI is cubic, therefore, V c=a3, a being the edge length of the unit cell. Again with referred to the structure of NaCI,a=2r Na++2r c fThus,V c=(2r Na++2r C i_)3Using the theoretical density formula,_ 〃'(4卬+ 4q)P = (2『3 + 2,C1 )3'A_ 4(22.99 + 35.45)~ [2(0.102 X 10~7) + 2(0.181 X 10-7)]3(6.023 X 1023)=2.14 g/cm3This value agrees fairly well with the experimental value.2、Calculate the number of vacancies per cubic meter in gold at 900°C. The energy for vacancy formation is 1.602xl0'19 J/atom. Furthermore, the density and atomic weight for Au are 18.63 g/cm3 (at 900°C) and 196.9 g/mol, respectively.N〃 = B (-新N v: number of equilibrium vacanciesN: total number of atomic sitesQ v: formation energy of a vacancyk: Boltzmann constant (1.38xlO'23J/K)T: temperature in KelvinSolution:In order to calculate the number of vacancies per cubic meter in gold at 900°C, it has to be known the total number of atomic sites per cubmic meter in gold first. N can be computed as,N=(18.63xl06g/m3/196.9 g/mol)x6.02xl023=5.7xl028Substituting the calculated N and other parameters into the equation, we have- ( 1.602 x 10T9 \N v = 5.7 x 1028 exp —~~——~——~~—— = 2.87 X 1024v 1.38 x 10-23 x (900 + 273)J3、Explain the relative orientation between Burgers vector and dislocation line for edge dislocation, screw dislocation, and mixed dislocation.Solution:For edge dislocation, Burgers vector is perpendicular to the dislocation line.For screw dislocation, Burgers vector is parallel to the dislocation line.For mixed dislocation, the angle between Burgers vector and dislocation is higher than 0° but lower than 90°.4、For both FCC and BCC crystal structures, there are two different types of interstitial sites. In each case, one site is larger than the other; and is normally occupied by impurity atoms. For FCC, this larger one is located at the center of each edge of the unit cell; it is termed an octahedral interstitial site. On the other hand, with BCC the larger site type is found at positions 0 % %; it is termed a tetrahedral interstitial site. For both FCC and BCC crystal structures, compute the radius r of an impurity atom that will just fit into one of these sites in terms of the atomic radius R of the host atom.Solution:Assuming the lattice parameter of both FCC and BCC crystal structures is a.Then, for FCC structure, a 2 + a 2 = (4R)2For octahedral interstitial position, 1， L 、r 0 = -(2V2R - 2R )= 0.414RFor tetragonal interstitial position, I a o a~ a~~「T =北产 + 0 + (J = °・225RFor BCC structure, a 2 + a 2 + a 2 = (4R)2For octahedral interstitial position, 1/4V3 \r 0 - 2(~R-2R ) = 0・155RFor tetragonal interstitial position,2V3 , V3 o (—R)2 + (—R)2 - R = 0.291R O sites T sitesBCCA copper-nickel alloy of composition 70 wt% Ni-30 wt% Cu is slowly heated from a temperature of 1300°C (2370°F).(a) At what temperature does the first liquid phase form?(b) What is the composition of this liquid phase?(c) At what temperature does complete melting of the alloy occur?(d) What is the composition of the last solid remaining prior to complete melting?Composition (at% Ni)(Cu) Composition (wt% Ni) (Ni)Is it possible to have a copper-nickel alloy that, at equilibrium, consists of an a phase of composition 37 wt% Ni-63 wt% Cu, and also a liquid phase of composition 20 wt% Ni-80 wt% Cu? If so, what will be the approximate temperature of the alloy? If this is not possible, explain why.Composition (at% Ni)(Cu) Composition (wt% Ni) (Ni)Solution：No, it is not possible to have such an alloy. From the phase diagram, it can be seen that if the composition of a phase is 37wt.%Ni-63wt.%Cu, at equilibrium, the corresponding composition of liquid phase should be 22wt.%Ni-78wt.%Cu.7、(a) Using the NaCl-H?。
材料科学与工程专业英语1-18单元课后翻译答案
Unit 1 Translation.1."材料科学〞涉与到研究材料的结构与性能的关系.相反,材料工程是根据材料的结构与性质的关系来涉与或操控材料的结构以求制造出一系列可预定的性质.2.实际上,所有固体材料的重要性质可以分为六类：机械、电学、热学、磁学、光学、腐蚀性.3.除了结构与性质,材料科学与工程还有其他两个重要的组成部分,即加工与性能.4.工程师或科学家越熟悉材料的各种性质、结构、性能之间的关系以与材料的加工技术,根据以上的原则,他或她就会越自信与熟练地对材料进行更明智的选择.5.只有在少数情况下,材料才具有最优或最理想的综合性质.因此,有时候有必要为某一性质而牺牲另一性能.6.Interdisciplinary dielectric constantSolid material<s> heat capacityMechanical property electromagnetic radiationMaterial processing elastic modulus7.It was not until relatively recent times that scientists came to understand the relationships between the structural elements of materials and their properties.8. Materials engineering is to solve the problem during the manufacturing and application of materials.9.10.Mechanical properties relate deformation to an applied load or force.Unit 21.金属是电和热很好的导体,在可见光下不透明；擦亮的金属表面有金属光泽.2.陶瓷是典型的导热导电的绝缘体,并且比金属和聚合物具有更高的耐热温度和耐恶劣环境性能.3.用于高科技领域的材料有时也被称为先进材料.4.压电陶瓷在电场作用下膨胀和收缩；反之,当它们膨胀和收缩时,他们也能产生一个电场.5.随着能够观察单个原子或者分子的扫描探针显微镜的出现,操控和移动原子和分子以形成新结构成为可能,因此,我们能通过一些简单的原子水平的构建就可以设计出新的材料.6.advanced materials ceramic materialshigh-performance materials clay mineralsalloy implantglass fibre carbon nanotube7.Metallic materials have large numbers of nonlocalized electrons and many properties of metals are directlyattributable to these electrons.8.Many of polymeric materials are organic compounds with very large molecular structures.9.Semiconductors hace electrical properties that are intermediate between the electrical conductors<viz. metalsand metal alloys> and insulators<viz. ceramics and polymers>.10.Biomaterials must not produce toxic substances and must be compatible with body tissues.Unit 31．金属的行为〔性质〕不同于陶瓷的行为〔性质〕,陶瓷的行为〔性质〕不同于聚合物的行为〔性质〕. 2．原子结构主要影响化学性质、物理性质、热学性质、电学性能、磁性能、光学性能.微结构和宏观结构虽也能影响这些性能但是他们主要影响力学性能和化学反应速率.3．金属的强度表明原子是通过强的键结合在一起的.4．元素的原子序数表明该元素的原子核内带正电的质子数.而原子的原子量则表明该原子核中质子数与中子数.5．Microstructure macrostructureChemical reaction atomic weightBalanced electrical charge positively charged proton6. 100 atoms form thousands of different substances ranging from the air we breathe to the metal used to supporttall buildings,7．The facts suggests that metallic atoms are held together bu strong bonds.8. Microstructure which includes features that cannot be seen with the naked eye,but using a microscope.Macrostructure includes features that can be seen with the naked eye.9. The atomic weight is the sum of proton number and neutron number in the nucleus.Unit 41．当密度小于水的密度时,物体将漂浮在水面上,当密度大于水的密度时,物体会沉降.相似的,当比重小于1时,物体将漂浮在水面上,当比重大于1时,物体会沉降.2．由于相互排斥而往相反的方向移动,导致磁通量密度比真空中小,这种材料为反磁性材料.3．使磁通量密度提高1倍以上小于或等于10倍的材料叫顺磁性材料,使磁通量密度提高10倍以上的材料叫铁磁性材料.4．某些铁磁性材料,特别是一些粉末状或夹层铁,钢或镍合金的相对导磁率可高达1000000.反磁性材料的相对导磁率小于1,但是到目前还没有哪种材料的相对导磁率远小于1.5．当顺磁性或铁磁性的芯插入线圈时,其磁感应系数等于相对磁导率乘以空芯时的磁感应系数. 6．specific gravity boiling point magnetic inductioncoefficient of thermal conductivity glass transition temperaturenon-ferrous metals linear coefficient of thermal expansionmass per unit of volume7. Properties that describe how a substance changes into a completely different substance are called chemicalproperties.8. Phase is a physical property of matter and matter can exist in four phases: solid, liquid, gas and plasma.9. At some temperature below the melting point, polymers start to lose their crystalline structure but the moleculesremain linked in chains, which tesults in a soft and pliable material.10. In engineering applications, permeability is often expressed in relative, rather than in absolute.Unit 51. 金属的力学性能决定了材料的使用X围与期望的服役寿命.2. 因此,一般多测几次以得到力学性能,报导的数值一般是平均值或者计算的统计最小值.3．材料的承载方式极大地影响了材料的力学性能,也决定了材料失效形式,以与在失效前是否有预警. 4．然而,受力弯曲时会产生一个应力分布,应力大小与到轴线的垂直距离有关.5．材料受到低于临界压力即屈服强度的力时,材料才会发生弹性形变.6．Test specimen static loading force normal axisEngineering strain critical stress yield strength stress areaStress- strain curve7. Temperatures below room temperature generally cause an increase in strength properties of metallic alloys;while ductility, fracture toughness, and elongation usually decrease.8. From the respective of what is happening within a material, stress is the internal distribution of forces withina body that balance and react to the loads applied to it.9. Engineering strain is defined as the amount of deformation in the direction of the applied force divided bythe initial length of the material.10. A material with high strength and high ductility will hace more toughness than a material with low strengthand high ductility.Unit 61. 随着影响我们星球上人类生存条件的社会问题的即将出现,材料科学与工程界有责任和机遇通过解决未来世界的需求—在能量、交通、住房、饮食、回收和健康方面的需求来改变世界.2. 不发达国家的人口增长率远高于1.4%的世界平均人口增长率.3. 全球能源使用的预算在2025以前将以每年1.7%速度增长,这比世界人口增长率快多了.4. 此外,发达地区的人均能量使用量是不发达地区人均能量使用量的九倍以上.5. gross domestic product materials science and engineering market economySocietal issues economic index sanitationGross national product popilation growth rate6. Some things that have been constant over time are human innovation and creativity, the engineer’s ability toaddress societal needs, and the entrepreneurial spirit of engineering.7. We have witnessed the re-shaping of our lives through revolutions that hace taken place in medicine,telecommunications, and transportantion industries.8. Eighteen percent of the world’s population lacks access to safe drinking water and nearly 40% has no access tosanitation.9. Materials and society are interlinked, and it is only rational that we should see a close relation between the MSEresearch agenda and societal issues that affect the human condition on the globe.Unit 71.从化学角度来说,金属是一类容易失电子以形成正离子的元素,它与其他金属原子形成金属键.2.金属键的无方向性被认为是金属具有延展性的主要原因.3.存在着共价键的晶体只有在原子之间的键断裂的情况下变形,从而导致晶体破裂.4.合金特别是为满足更高应用要求的合金比如喷气发动机,可能含有十种以上的元素.5.delocalized electron electronic structurealkaline earth metal chemical cellnuclear charge electric conductivity6.Metals are sometimes described as a lattice of positive ions surrounded by a cloud of delocalized electrons.7.Metals in general have superior electric and thermal conductivity, high luster and density, and the ability to bedeformed under stress without cleaving.8.An alloy is a mixture of two or more elements in solid solution in which the major component is a metal.bining different ration of metals as alloys modifies the properties of pure metals to produce desirablecharacteristics.Unit 81.超耐热合金的发展非常依赖于化学与加工的创新,并主要受到航空和能源工业的推动.2.抗蠕变性能主要取决于晶体结构内位错速度的放缓.3.超耐热合金在加工方面的发展使超耐热合金的操作温度大幅度提高.4.单晶高温合金是运用改进的定向凝固技术而形成单晶的,因此在材料中并无晶界.5.faced-centred cubic crystal structure turbine entry temperaturemetallic materials phase stabilitynuclear reactor synthesis of nanoparticle6.Superalloys typically hace an sustenitic faced-centered cubic crystal structure.7.Superalloys are used where there is a need for high temperature strength and corrosion/oxidation resistance.8.Superalloys are widely used in aircraft ,submarines, nuclear reactors and military electric motors.9.At high temperatures the gaseous aluminum chloride<or fluoride> is transfereed to the surface of part anddiffuseds inside.Unit 91.腐蚀过程从本质上说是一个电化学的过程,有着与电池相同的本质特征.2. 从矿物中提炼金属所需能源的问题与后续的腐蚀和能量释放直接相关.3. 当电子与中和的正离子〔如电解液中的氢离子〕发生反应时,阴极处电子得以平衡.4．Protective film circuitFree electron electron transferMetal cation anode reaction5.Some metals, such as gold and silver, can be found in the earth in their natural, metallic state, and they havelittle tendency to corrode.6.Oxidation is the process of stripping electrons from an atom and reduction occurs when an electron is addedto an atom.7.If the surface becomes wet, corrosion may take place through ionic exchange in the surface water layerbetween the anode and cathode.8.Corrosion is commonly classification based on the appearance of the corroded material.Unit 101. 我们要观察〔研究〕这些性能,看它们与我们所期望的陶瓷的组成有多匹配.2. 在高于玻璃化转变温度的高温下,玻璃不再具有脆性行为,而表现为粘稠液体.3. 它们显示出优异的力学性能、抗腐蚀/氧化性能,或电学、光学抑或是磁学性能.4. 一般认为,先进是最近100年才发展起来的,而传统的基于粘土的陶瓷早已在25000多年前就被使用了.5. the glass transition temperature ionic-covalent bondStress distribution coefficient of thermal expansionGlass optical fibre materials science and engineeringSolid-oxide fuel cells electron microscopy6. Diamond, which is classified as a ceramic, has the highest thermal conductivity of any known material.7. Ceramic are stronger in compression than in tension, whereas metals have comparable tensile and compressivestrength.8. Ceramics generally have low toughness, although combining them in composites can dramatically improvethis property.9. The functions of ceramic products are dependent on their chemical composition and microstructure, whichdetermines their properties.Unit 111. 材料科学与工程领域经常是根据四大方面—合成与加工,结构与组成,性质与性能之间的相互联系来定义的.2．我们不仅要考虑具有完美晶格和理想结构〔的情况〕,我们也要考虑材料中不可避免的结构缺陷的存在,甚至是无定形的.3. 通过热压可使孔径减小从而得到高密度产品.4. 在运输时,厂方要提供关于产品危害方面的信息.5. crystalline ceramics grain boundaryAlkaline earth oxide oxide additiveTriple point saturation magnetizationTelevision tube the color scale6.To understand the behavior and properties of any material, it is essential to understand its structure.7.The grain size is determined by the size of the initial powder particles and the way in which they wereconsolidated.8.Transparent or translucent ceramics require that we limit the scattering of light by pores and second-phaseparticles.9.Alumina ceramics are used as electrical insulators because of theie high electrical resistivity and lowdielectric constant.Unit 121. 材料的选择是任何组分设计至关重要的环节,尤其在植入体和其它医疗器械方面是特别重要的.2. 我们能进行承载应用的三种主要材料是金属、聚合物和陶瓷.3. 高密度、高纯度的氧化铝被大量的用于植入物,特别是在需要承载压力的髋关节修复和牙移植中.4. 在陶瓷或陶瓷复合材料中,氧化锆的磨损率远远高于氧化锆铝的磨损率.5. controlled reaction stress shieldingTotal hip prosthese strain-to-failure ratioMechanical stress flexural strengthMartensitic transformation6.Biomaterial is a non-viable material used in a medical device intended to interact with biological systems.7.These repairs become necessary when the existing part becomes diseased, damaged, or just simply wears out.8.Because of its low density, cancellous bone has a lower E and higher strain-to-failure ratio than cortical bone.9.Eliminating stress shielding, by reducing E, is one of the primary motivations for the development ofbioceramic composites.10.There are questions concerning the long-term effect of radiation emission from zirconia ceramics.Unit 131. 聚合物的俗名叫塑料,这个词指的是一大类具有许多性质和用途的天然材料和合成材料.2. 聚合物合成是一个把叫做单体的小分子通过共价键的结合形成链的过程.3. 支化聚合物分子是由一条带有一个或多个侧基或支链的主链组成.一些特殊的支化聚合物有星型聚合物、梳状聚合物和刷状聚合物.4. 某些生物聚合物是由一系列不同的胆识结构却相关的单体组成的,例如聚核苷酸是由核苷组成的.5. persistence length cross-linkPolar monomer nucleic acidPolymerization polyelectrolyte6.Most commercially important polymers today are entirely synthetic and produced in high volume, onappropriately scaled organic synthetic techniques.7.Some biological polymers are composed of a variety of different but structurally related monomers, suchas polynucleotides composed of nucleotide sbunits.8. A polymer molecule with a high degree of crosslinking is referred to as a polymer network.9.In polymers, however, the molecular mass may be expressed in terms of degree of polymerization,essentially the number of monomer Units which comprise the polymer.Unit 141. 大量合成的聚合物具有碳-碳骨架,这是因为碳原子具有与其它原子形成更强更稳定的键的优异性能.2. 它们在一定X围内软化,这与完好晶体相具有非常明确的熔点不同.3. 分子量取决于其合成时的条件,因此分子量可能分布很宽或分布很窄.4. Goodyear 很偶然的发现了在橡胶中加入硫磺并加热这个混合物能使橡胶变硬,对氧化和化学进攻能力的敏感性降低.5. thermosetting plastic cross-sectional areaPolymerization reaction double bondChemical composition carboxylic acidMelting point degradation by oxidation6.Polymer with different chemical composition has different physical and chemical property.7. A thermosetting plastic is shaped through irreversible chemical processes and therefore cannot be reshapedreadily.8.Natural rubber is not a useful polymer because it is too soft and too chemically reactive.9.Various substances may be added to polymers to provide protection against the effects of sunlight or againstdegradation by oxidation.Unit 151. 逐渐增强的环境意识促使包装薄膜与其加工既要方便使用又要具有环境友好的特点.2. 显而易见,实现这些性能对控制和改进机械性能和阻隔性能是非常重要的.3. 在羧酸、醇、醛、酮的含氧生物降解过程中,由水和热引发的过氧化反应可以使之降解成低摩尔质量的物质,这就是碳氢聚合物力学性能降低的主要原因.4. 用持久耐用的聚合物做短期使用的包装材料并不合理,另外也是因为包装材料被食物污染后再进行物理回收是不切合实际的.5. natural gas packaging materialsAroma compound bioplastic materialChemical structure the life cycle of biomass6. Bacteria , fungi, enzymes start the bioassimilation giving rise to biomass and CO2 that finally form the humus.7. The bioplastic aim is to imitate the life cycle of biomass, which includes conservation of fossil resources, waterand CO2 production.8. During the oxo-degradation of carboxylic acid, molecules of alcohols, aldehydes and ketones degradable withlow molar mass are produced by peroxidation initiated by heat or light .9. While most of the commercialized biopolymer materials are biodegradable, these are not fully compostable inreal composting conditions, which vary with temperature and relative humidity.Unit 161. 比如,多相金属在微观尺度上是复合材料.但一般意义上的复合材料是指通过键的作用使两种或多种不同的材料结合在一起的材料..2. 最常见的是,复合材料有一个连续的叫基体的本体相,还有一个分散的非连续的叫增强相的相.3. 先进材料采用了树脂与纤维的复合材料,一般为碳/石墨,凯芙拉或玻璃纤维与环氧树脂的复合材料.纤维具有高的硬度,而聚合物树脂基体能保持复合材料的结构.4. 如果复合材料设计和制备合理的话,复合材料就既具有增强相的强度又具有基体的韧性从而得到了性能的理想组合,这是任何一种组分单独存在时所具备的性能.5. composite material reinforcement materialFiberglass matrix materialStrengthening mechanism conventional material6. A composite is commonly defined as a combination of two or more distinct materials, each of which retainsits own distinctive properties, to create a new material with properties that cannot be achieved by any of the components acting alone.7.Carbon-epoxy composties are two thirds the weight of aluminum, and two and a half times as stiff.Composites are resistant to fatigue damage and harsh enviroments, and are repairable.8.According to the conception of composite , reinforced plastics, metal-matrix composites, ceramic-matrixcomposites and concrete are composites.9.In fiber-reinforced composites, the fiber is the primary load-bearing component. Fiberglass and carbon fibercomposites are examples of fiber-reinforced composites.Unit 171. 震荡、撞击或者重复的周期性应力能导致两层的界面处发生薄层分离,这种情况叫剥离.2.3. 事实上,工业材料既要质轻又要牢固的要求是推动复合材料发展的主要动力.4. 提到飞机,值得铭记的是复合材料不像金属〔如铝〕那样在压力的作用下会完全解体.5. orthotropic thermosetThermoplastic Young’s ModulusMechanical property constants extreme enviroment6.In contrast, isotropic material < for example, aluminuim or steel>, in standard wrought forms, typicallyhave the same stiffness regardless of the directional orientation of the applied forces and /or moments.7.The greatest advantage of composite materials is strength and stiffness combined with lightness.8.This makes them ideal for use in products thar are exposed to extreme enviroments such as boats,chemical-handling equipments and spacecrafts.posites will never totally replace tranditional materials like steel, but in many cased they are just whatwe need.Unit 181. 具有相分离的聚合物共混材料经常出现纳米尺度的相.2 在过去几十年里研究的基于溶胶-凝胶化学的有机-无机纳米复合材料已基本淡出纳米复合材料的研究.3. 理解粒子的性质随着尺寸降低到纳米级别而发生改变,这对于优化所得到的纳米复合材料很重要.4. 廉价石墨的生产尚未实现,石墨的广泛使用呕待石墨合成技术的突破.5. electro-optical property bactericidal propertyBlock copolymer interfacial phenomenaExfoliated graphene morphology control6.The field of nanotechnology is one of the most popular areas for current research and development inbasically all technical discillines.7.Nanoscale is considered where the dimensions of the particle, platelet or fiber modification are in the rangeof 1～100nm.8.These improvements are key to future aircraft and wind energy turbine applications.9.Nanostructured surfaces have been noted to yield superhydrophobic character and exceptional adhesion.。
材料科学与工程基础双语期中复习
材料科学与工程基础双语期中复习-CAL-FENGHAI-(2020YEAR-YICAI)_JINGBIAN1. Valence band 价带；价电子能级展宽成的带；2. 位错线的运动方向总是垂直垂直于位错线。
正确3. 费米能级是，在T=0K时，金属原子中电子被填充的最高能级，以下能级全满，以上能级全空。
正确4. 以原子半径R为单位，fcc晶体的点阵常数a 是 2（2）1/2R5. fcc金属晶体具有较高的强度、硬度和熔点。
错误6. 晶体物质的共同特点是都具有金属键。
错误7. 在界面两侧性质发生突然变化的是两个不同的相，否则是同一相。
正确8. 石墨、富勒烯、巴基管和金刚石都是碳的同素异形体，均有相同的晶体结构。
错误9. 钢是含碳量低于2%的碳铁合金。
正确10. 绝大多数的金属均以金属键的方式结合，它的基本特点是电子共有化。
正确11. APF是致密度；晶胞内原子总体积与晶胞体积之比；原子堆积因子。
12. 价带未填满是导体。
13. 每一个bcc晶胞，八面体间隙 6个，四面体间隙 12 个。
14. 金属原子的配位数越大，邻近的原子数越多，相互作用越强，原子半径越小。
错误15. 扩散系数一般表示为D=D。
Exp(-Q/RT)，显然扩散激活能与扩散激系数成正比，Ｑ值越大，Ｄ值越大。
错误16. 在四个量子数中，ms是确定体系角动量在磁场方向的分量。
错误17. 立方晶胞中，晶面指数（111）和晶向指数【111】是垂直的。
18. 氯化钠是面心立方结构，其晶胞分子数是 4 。
19. 由液态转变为固态的过程称为凝固，亦称结晶。
错误20. 同一周期中原子共价半径随电子数的增加而增加。
错误21. bcc结构的配位数是 8 。
22. 元素的电负性是指元素的原子在化合物中把电子引向自己的能力。
正确23. 满带与空带重叠是半导体。
24. covalent bond 是指共价键。
25. 尖晶石晶体属于立方晶系，每个单位晶胞由相同体系的4个A块和4个B块所构成，共有32个八面体间隙和64个四面体间隙。
材料工程基础期末试题
Exam Questions ofFundamentals of Materials Engineering1. What are materials? Give the basic classifications of materials based on chemical makeup and atomic structure. (15 points)2. Define Materials Science and Engineering. Why do we study materials science and engineering? (15 points)3. Contrast compression, injection, and transfer molding techniques that are used to form plastic materials. (15 points)4. Translate the following paragraph into Chinese（20 points）:A special group of dispersion-strengthened nanocomposite materials containing particles 10 to 250nm in diameter is classified as particulate composites. These dispersoids, usually a metallic oxide, are introduced into the matrix by means other than traditional phase transformations. Even though the small particles are not coherent with the matrix, they block the movement of dislocations ad produce a pronounced strengthening effect. At room temperature, the dispersion-strengthened composites may be weaker than traditional age-hardened alloys, which contain a coherent precipitate. However, because the composites do not catastrophically soften by overaging, overtempering, grain growth, or coarsening of the dispersed phase, the strength of the composite decreases only gradually with increasing temperature. Furthermore, their creep resistance is superior to that of metals and alloys.5. Glass, aluminum, and various plastic materials are utilized for containers. Make a list of the advantages and disadvantages of using each of these three material types; include such factors as cost, recyclability, and energy consumption. (15 points)6. What do you think about the environment impact of materials production? (20 points)。
材料科学与工程基础专业课模拟试题(A)
中北大学材料科学与工程基础专业课模拟试题一、选择（10分，每题0.5分。
）1. The stimulus of electrical properties is . A: load B: force C:electric field D: magnetic field2. consist of a polymer resin as the matrix, with fibers as the reinforcement medium. A: Polymer-matrix composites B: Metal-matrix composites C: Ceramic-matrix composites D: Carbon –carbon composites3. relates to the arrangement of its internal components. A: property B: structure C:processing D: application4. are very thin single crystals that have extremely large length-to-diameter ratios. A: Fibers B: Wires C: Particles D: Whiskers5. Atomic packing factor of HCP structure is .A: 0.68 B: 0.34 C: 0.74 D: 0.56 6. has the sodium chloride crystal structure.A: CsBr B:CsI C: ZrO 2D: MgO7. Most glass-forming operations are carried out within the working range —between the and softening temperatures.A: melting; working B: softening; annealing C: working; softening D: annealing; strain8. Hydrocarbons are composed of and . The intramolecular bonds are covalent.A: hydrogen; carbon B: oxygen; carbonC: hydrogen; oxygen D: carbon; nitrogen9. consists of passing a piece of metal between two rolls.A: Forging B: Extrusion C:Casting D:Rolling10. A load produces an elongation and positive linear strain.A: Compressive B: Shear C: Torsional D: Tensile11. is based on the weight fraction of molecules within the various size ranges.A: Number-average molecular weight B:Weight-average molecular weightC: Z-average molecular weight D: Viscosity-average molecular weight12. For many metals and other alloys, Poisson’s ratio is in .A: 0.1~0.2 B: 0.25~0.35 C: 0.45~0.55 D: 0.6~0.713. can be given either as percent elongation or area reduction.A: Ductility B: Resilience C: Toughness D: Tensile strength 14. may be thought of as stiffness.A: Young's modulus B: Tensile strengthC: Shear modulus D: Compressive strength15.The corresponds to the temperature at which the viscosity is 4×106Pa·s.A: melting point B:working point C: softening point D:annealing point 16. The coordination number of sodium chloride crystal structure is .A: 2 B: 4 C: 6 D: 817.Addition polymerization is used in the synthesis of .A: phenolformaldehyde B: polyvinyl chlorideC: the thermosetting polyesters D: polycarbonates18.For FCC structure, a total of atoms are assigned to a given unit cell.A: four B: two C: six D: eight19.The thermal behavior of solids can be represented in terms of .A: heat capacity B:electrical conductivityC:dielectric constant D:elastic modulus20. is the point of the initial departure from linearity of the stress-strain curve.A: Yielding strength B: Tensile strength C: Proportional limit D: Shear strength二、名词解释（20分，每题4分）1.Mer2.Yield strength:3.vitrificationposite5. Processing三、简答题（30分，1，2，3必做题，每题7分；4，5，6选做其中之一，9分）4. Make comparisons of thermoplastic and thermosetting polymers (a) on the basis of mechanical characteristics upon heating, and (b) according to possible molecular structures.5. What is the difference between graphite structure and diamond structure？四、计算题（40分，每题10分）1. Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol. Compute (1) the volume of a unit cell and (2) its density.2. A specimen of aluminum having a rectangular cross section 10 mm×12.7 mm and70mm long is pulled in tension with 35,500 N forces, producing only elastic deformation. Calculate the resulting strain and elongation.3. For a bronze alloy, the stress at which plastic deformation begins is 275 MPa, and the modulus of elasticity is 115 GPa.(a) What is the maximum load that may be applied to a specimen with a cross-sectional area of 325mm2 without plastic deformation?(b) If the original specimen length is 115 mm, what is the maximum length to which it may be stretched without causing plastic deformation?4. The number-average molecular weight of a polypropylene is 1,000,000 g/mol, Compute the number-average degree of polymerization. And the number-average degree of polymerization is 1,000,000 , compute the number-average molecular weight of a polypropylene.。
材料科学与工程专业英语第三版-翻译以及答案
UNIT 1一、材料根深蒂固于我们生活的程度可能远远的超过了我们的想象，交通、装修、制衣、通信、娱乐（recreation）和食品生产，事实上（virtually），我们生活中的方方面面或多或少受到了材料的影响。
历史上，社会的发展和进步和生产材料的能力以及操纵材料来实现他们的需求密切（intimately）相关，事实上，早期的文明就是通过材料发展的能力来命名的（石器时代、青铜时代、铁器时代）。
二、早期的人类仅仅使用（access）了非常有限数量的材料，比如自然的石头、木头、粘土（clay）、兽皮等等。
随着时间的发展，通过使用技术来生产获得的材料比自然的材料具有更加优秀的性能。
这些性材料包括了陶瓷（pottery）以及各种各样的金属，而且他们还发现通过添加其他物质和改变加热温度可以改变材料的性能。
此时，材料的应用（utilization）完全就是一个选择的过程，也就是说，在一系列有限的材料中，根据材料的优点来选择最合适的材料，直到最近的时间内，科学家才理解了材料的基本结构以及它们的性能的关系。
在过去的100年间对这些知识的获得，使对材料性质的研究变得非常时髦起来。
因此，为了满足我们现代而且复杂的社会，成千上万具有不同性质的材料被研发出来，包括了金属、塑料、玻璃和纤维。
三、由于很多新的技术的发展，使我们获得了合适的材料并且使得我们的存在变得更为舒适。
对一种材料性质的理解的进步往往是技术的发展的先兆，例如：如果没有合适并且没有不昂贵的钢材，或者没有其他可以替代（substitute）的东西，汽车就不可能被生产，在现代、复杂的（sophisticated）电子设备依赖于半导体（semiconducting）材料四、有时，将材料科学与工程划分为材料科学和材料工程这两个副学科（subdiscipline）是非常有用的，严格的来说，材料科学是研究材料的性能以及结构的关系，与此相反，材料工程则是基于材料结构和性能的关系，来设计和生产具有预定性能的材料，基于预期的性能。
材料科学与工程基础英文版试题
材料科学与工程基础”考试试题–英文原版教材班（注：第1、2、3题为必做题；第4、5、6、7题为选择题，必须二选一。
共100分）1. Glossary (2 points for each)1) crystal structure: The arrangement of the atoms in a materialinto a repeatable lattice.2) basis (or motif): A group of atoms associated with a lattice.3) packing fractor: The fraction of space in a unit cell occupiedby atoms.4) slip system: The combination of the slip plane and the slipdirection.5) critical size: The minimum size that must be formed by atomsclustering together in the liquid before the solid particle is stable and begins to grow.6) homogeneous nucleation: Formation of a critically sized solidfrom the liquid by the clustering together of a large number of atoms at a high undercooling (without an external interface).7) coherent precipitate:A precipitate whose crystal structure andatomic arrangement have a continuous relationship with matrix from which precipitate is formed.8) precipitation hardening: A strengthening mechanism thatrelies on a sequence of solid state phase transformations in a dispersion of ultrafine precipitates of a 2nd phase. This is same as age hardening. It is a form of dispersion strengthening.9) diffusion coefficient: A temperature-dependent coefficientrelated to the rate at which atom, ion, or other species diffusion.The DC depends on temperature, the composition and microstructure of the host material and also concentration of the diffusion species.10) uphill diffusion: A diffusion process in which species movefrom regions of lower concentration to that of higher concentration.2. Determine the indices for the planes in the cubic unit cell shown in Figure 1. (5 points)Fig. 1Solution : A(-364), B(-340), C(346). 3. Determine the crystal structure for the following: (a) a metal with a 0 = 4.9489 Å, r = 1.75 Å and one atom per lattice point; (b) a metal with a 0 = 0.42906 nm, r = 0.1858 nm and one atom per lattice point. (10 points) Solution : (a)fcc; (b) bcc. 4-1. What is the characteristic of brinell hardness test, rockwell hardness test and V ickers hardness test? What are the effects of strain rate and temperature on the mechanical properties of metallic materials? (15 points) 4-2. What are the effects of cold-work on metallic materials? How to eliminate those effects? And what is micro-mechanism for the eliminating cold-work effects? (15 points) 5-1. Based on the Pb-Sn-Bi ternary diagram as shown in Fig. 2, try to (1)Show the vertical section of 40wt.%Sn; (5 points) (2) Describe the solidification process of the alloy 2# with very low cooling speed (including phase and microstructure changes); (5 points) (3)Plot the isothermal section at 150o C. (5 points)5-2. A 1mm sheet of FCC iron is used to contain N 2 in a heated exchanger at 1200o C. The concentration of N at one surface is 0.04 atomic percent and the concentration at the second surface is 0.005 atomic percent. At 1000 oC, if same N concentration is demanded atFig. 2the second surface and the flux of N becomes to half of that at 1200o C, then what is the thickness of sheet? (15 points)6-1. Supposed that a certain liquid metal is undercooled until homogeneous nucleation occurs. (15 points)(1)How to calculate the critical radius of the nucleus required?Please give the deduction process.(2)For the Metal Ni, the Freezing Temperature is 1453︒C, the LatentHeat of Fusion is 2756 J/cm3, and the Solid-liquid Interfacial Energy is 255⨯10-7J/cm2. Please calculate the critical radius at 1353︒C. (Assume that the liquid Ni is not solidified.)6-2. Fig.3 is a portion of the Mg-Al phase diagram. (15 points) (1)If the solidification is too rapid, please describe the solidificationprocess of Mg-10wt%Al alloy.(2)Please describe the equilibrium solidification process ofMg-20wt%Al alloy, and calculate the amount of each phase at 300︒C.Fig. 37-1. Figure 4 shows us the Al-Cu binary diagram and some microstructures found in a cooling process for an Al-4%Cu alloy. Please answer following questions according to this figure. (20 points)Fig. 4 (1)What are precipitate, matrix and microconstituent? Please point them out in the in the figure and explain. (2)Why is need-like precipitate not good for dispersion strengthening? The typical microstructure shown in the figure is good or not? why? (3)Please tell us how to obtain the ideal microstructure shown in this figure. (4)Can dispersion strengthened materials be used at high temperature? Please give the reasons (comparing with cold working strengthening) 7-2. Please answer following questions according to the time-temperature-transformation (TTT) diagram as shown in Fig. 5. (20 points) (1)What steel is this TTT diagram for? And what means P, B, and M in the figure? (2)Why dose the TTT diagram exhibits a ‘C’ shape? (3)Point out what microconstituent will be obtained after austenite is cooled according to the curves I, II, III and IV . (4)What is microstructural difference between the curve I and the curve II? (5)How to obtain the steel with the structure of (a) P+B (b) P +M+A (residual) (c) P+B+M+A (residual) (d) F ull tempered martensite If you can, please draw the relative cooling curve or the flow chart of heat treatment.Fig. 5 III III IV。
材料科学与工程基础(英文)_南京航空航天大学中国大学mooc课后章节答案期末考试题库2023年
材料科学与工程基础（英文）_南京航空航天大学中国大学mooc课后章节答案期末考试题库2023年1.The driving force for steady-state diffusion is the __________.答案:concentration gradient2.Diffusion coefficient is with the increasing diffusion temperature.答案:exponentially increased;3.Due to , alloys are usually than pure metals of the solvent.答案:solid solution strengthening, stronger;4.The finer the grains, the larger the , and .答案:strength, hardness, toughness;5.With plastic deformation,the increase of dislocationdensity will result in .答案:higher strength;6.In general, Brinell Hardness test is to measure thematerial’s hardness.答案:relatively softer7.Yield strength is corresponding to the occurrenceof deformation.答案:noticeable plastic8.Strain Hardening is also named as .答案:work hardening9.Vacancy diffusion is usually interstitial one.答案:slower than10.Edge and screw dislocations differ in what way?答案:angle between Burgers vector and line direction.11. A ____ may form when impurity atoms are added to a solid, in which case theoriginal crystal structure is retained and no new phases are formed.答案:solid solution12.One explanation for why graphite powder acts so well as a “solid lubricant”is .答案:carbon atoms in graphite are covalently bonded within planar layers but have weaker secondary bonds between layers13.Substitutional atom (impurity) is an example of ______.答案:point defect14.Interstitial solid solution belongs to .答案:finite solid solution;15.The atomic packing factor for FCC is .答案:0.7416.The coordination number of BCC crystal structure is .答案:817.The crystal structure of Cu is ?答案:FCC18.How many atoms does the face centered cubic unit cell contain?答案:Four19.If the electron configuration of Fe is 1s2 2s2 2p6 3s2 3p6 3d6 4s2, then theelectron configurations for the Fe3+ is 1s2 2s2 2p6 3s2 _____.答案:3p6 3d520.Bonds in most metals are referred to as ______.答案:Non-directional21.Covalent bonding occurs as a result of _________ sharing.答案:electron22.Which of the following is NOT an example of primary bonding?答案:Van der Waals23.Atomic weight (A) of an element corresponds to the weighted average of theatomic masses of the atom’s naturally occurring ___________.答案:isotopes24.The point on a phase diagram where the maximum number of allowablephases are in equilibrium is .答案:eutectic point25.Sterling silver (92.5%Ag/7.5%Cu) is an example of ___________.答案:Solid solution26.Engineering stress-strain curve and true stress-strain curve are equal up to .答案:Yeild point27.Among thefollowingtypical transformations of austenite in steels,____________transformation is diffusionless.答案:martensitic28.The heat-treatable aluminum alloy can be strengthened by .答案:Both of above29.In the as-quenched state, martensite is very hard and so brittle that a heattreatment known as must be accomplished sequently.答案:tempering30.During heat treatment of steel, austenite transforms into martensite by .答案:quenching31.Which of the following plane has the highest planar density for fcc.答案:(111)32.Which of the following describes recrystallization?答案:Diffusion dependent with no change in phase composition33.Heating the cold-worked metal progresses in three stages: .答案:recovery, recrystallization, grain growth;34.Strength is increased by making dislocation motion .答案:difficult35.The boundary above which only liquid phase exist is called _________.答案:liquidus36.We have an annealed carbon steel which has hardness of 150HBS. Supposewe know the hardness of Pearlite is 200HBS and the hardness of Ferrite is 80HBS, determine the carbon amount of this steel.答案:0.45%37.The maximum solubility of C in γ-austenite - solid solution is .答案:2.1438.In a plain steel that contains 0.2 percentage carbon, we should expect: .答案:a 25% pearlite and 75% pro-eutectoid ferrite39. A copper-nickel alloy is high-temperature heat treated; the diffusion of Cuinto Ni and Ni into Cu regions is referred to as _____________________.答案:Inter-diffusion40.The phase diagram of Sn-Pb alloy is called .答案:Eutectic phase diagram。
材料科学与工程专业英语第三版-翻译以及答案
UNIT 1一、材料根深蒂固于我们生活的程度可能远远的超过了我们的想象，交通、装修、制衣、通信、娱乐（recreation）和食品生产，事实上（virtually），我们生活中的方方面面或多或少受到了材料的影响。
历史上，社会的发展和进步和生产材料的能力以及操纵材料来实现他们的需求密切（intimately）相关，事实上，早期的文明就是通过材料发展的能力来命名的（石器时代、青铜时代、铁器时代）。
二、早期的人类仅仅使用（access）了非常有限数量的材料，比如自然的石头、木头、粘土（clay）、兽皮等等。
随着时间的发展，通过使用技术来生产获得的材料比自然的材料具有更加优秀的性能。
这些性材料包括了陶瓷（pottery）以及各种各样的金属，而且他们还发现通过添加其他物质和改变加热温度可以改变材料的性能。
此时，材料的应用（utilization）完全就是一个选择的过程，也就是说，在一系列有限的材料中，根据材料的优点来选择最合适的材料，直到最近的时间内，科学家才理解了材料的基本结构以及它们的性能的关系。
在过去的100年间对这些知识的获得，使对材料性质的研究变得非常时髦起来。
因此，为了满足我们现代而且复杂的社会，成千上万具有不同性质的材料被研发出来，包括了金属、塑料、玻璃和纤维。
三、由于很多新的技术的发展，使我们获得了合适的材料并且使得我们的存在变得更为舒适。
对一种材料性质的理解的进步往往是技术的发展的先兆，例如：如果没有合适并且没有不昂贵的钢材，或者没有其他可以替代（substitute）的东西，汽车就不可能被生产，在现代、复杂的（sophisticated）电子设备依赖于半导体（semiconducting）材料四、有时，将材料科学与工程划分为材料科学和材料工程这两个副学科（subdiscipline）是非常有用的，严格的来说，材料科学是研究材料的性能以及结构的关系，与此相反，材料工程则是基于材料结构和性能的关系，来设计和生产具有预定性能的材料，基于预期的性能。
材料科技英语试题及答案
材料科技英语试题及答案一、选择题（每题2分，共20分）1. What is the most common type of material used in the construction of bridges?A. SteelB. ConcreteC. WoodD. Plastic2. The process of hardening metal by heating and cooling is known as:A. AnnealingB. TemperingC. QuenchingD. Forging3. Which of the following is not a property of materials?A. DensityB. ElasticityC. ColorD. Thermal conductivity4. The term "nanomaterials" refers to materials with at least one dimension in the size range of:A. 1-100 nanometersB. 1-100 micrometersC. 1-100 millimetersD. 1-100 centimeters5. What is the primary function of a catalyst in a chemical reaction?A. To increase the temperatureB. To provide energyC. To speed up the reaction without being consumedD. To slow down the reaction6. The strength of a material is often measured by its:A. DuctilityB. Tensile strengthC. MalleabilityD. Hardness7. Which of the following is a type of composite material?A. GlassB. CeramicC. AlloyD. Fiberglass8. The SI unit for measuring thermal expansion is:A. CelsiusB. KelvinC. JouleD. Degree Celsius per meter Kelvin9. What is the main difference between amorphous and crystalline materials?A. ColorB. ShapeC. Atomic arrangementD. Density10. The term "strain" in materials science refers to:A. The amount of deformation per unit lengthB. The force applied to a materialC. The change in shape of a materialD. The resistance to deformation二、填空题（每题2分，共20分）11. The process of changing the physical or chemical properties of a material is known as ________.12. The ________ of a material is its ability to resist deformation.13. The ________ is a material that can withstand high temperatures without significant loss of strength.14. The ________ of a material is the measure of its ability to conduct heat.15. A ________ is a material that can be easily deformed without breaking.16. The ________ of a material is its resistance to wear or abrasion.17. The ________ is the process of joining two pieces of metal by heating them to a molten state.18. The ________ is the process of removing excess material to create a desired shape.19. The ________ of a material is its ability to return to its original shape after deformation.20. The ________ is the study of the behavior of materials under various conditions.三、简答题（每题10分，共30分）21. Explain the difference between ductile and brittle materials.22. Describe the process of annealing and its effects on materials.23. Discuss the importance of material selection in the design of aeronautical components.四、论述题（每题15分，共30分）24. Discuss the role of materials science in the development of new technologies.25. Analyze the environmental impact of material production and disposal, and suggest ways to minimize these effects.答案：一、1-5: B C C A C6-10: B D D A C A二、11. Alteration12. Rigidity13. Refractory14. Thermal conductivity15. Ductile material16. Hardness17. Fusion18. Machining19. Elasticity20. Material science三、21. 略22. 略23. 略四、24. 略25. 略。
《材料科学与工程基础》英文影印版习题及思考题及答案
《材料科学与工程基础》英文习题及思考题及答案第二章习题和思考题Questions and Problems2.6 Allowed values for the quantum numbers ofelectrons are as follows:The relationships between n and the shell designationsare noted in Table 2.1.Relative tothe subshells,l ＝0 corresponds to an s subshelll ＝1 corresponds to a p subshelll ＝2 corresponds to a d subshelll ＝3 corresponds to an f subshellFor the K shell, the four quantum numbersfor each of the two electrons in the 1s state, inthe order of nlmlms , are 100(1/2 ) and 100(-1/2 ).Write the four quantum numbers for allof the electrons inthe L and M shells, and notewhich correspond to the s, p, and d subshells.2.7 Give the electron configurations for the followingions: Fe2+, Fe3+, Cu+, Ba2+,Br-, andS2-.2.17 (a) Briefly cite the main differences betweenionic, covalent, and metallicbonding.(b) State the Pauli exclusion principle.2.18 Offer an explanation as to why covalently bonded materials are generally lessdense than ionically or metallically bonded ones.2.19 Compute the percents ionic character of the interatomic bonds for the followingcompounds: TiO2 , ZnTe, CsCl, InSb, and MgCl2 .2.21 Using Table 2.2, determine the number of covalent bonds that are possible foratoms of the following elements: germanium, phosphorus, selenium, and chlorine.2.24 On the basis of the hydrogen bond, explain the anomalous behavior of waterwhen it freezes. That is, why is there volume expansion upon solidification?3.1 What is the difference between atomic structure and crystal structure?3.2 What is the difference between a crystal structure and a crystal system?3.4Show for the body-centered cubic crystal structure that the unit cell edge lengtha and the atomic radius R are related through a =4R/√3.3.6 Show that the atomic packing factor for BCC is 0.68. .3.27* Show that the minimum cation-to-anion radius ratio for a coordinationnumber of 6 is 0.414. Hint: Use the NaCl crystal structure (Figure 3.5), and assume that anions and cations are just touching along cube edges and across face diagonals.3.48 Draw an orthorhombic unit cell, and within that cell a [121] direction and a(210) plane.3.50 Here are unit cells for two hypothetical metals:(a)What are the indices for the directions indicated by the two vectors in sketch(a)?(b) What are the indices for the two planes drawn in sketch (b)?3.51* Within a cubic unit cell, sketch the following directions:.3.53 Determine the indices for the directions shown in the following cubic unit cell:3.57 Determine the Miller indices for the planesshown in the following unit cell:3.58Determine the Miller indices for the planes shown in the following unit cell: 3.61* Sketch within a cubic unit cell the following planes:3.62 Sketch the atomic packing of (a) the (100)plane for the FCC crystal structure, and (b) the (111) plane for the BCC crystal structure (similar to Figures 3.24b and 3.25b).3.77 Explain why the properties of polycrystalline materials are most oftenisotropic.5.1 Calculate the fraction of atom sites that are vacant for lead at its meltingtemperature of 327_C. Assume an energy for vacancy formation of 0.55eV/atom.5.7 If cupric oxide (CuO) is exposed to reducing atmospheres at elevatedtemperatures, some of the Cu2_ ions will become Cu_.(a) Under these conditions, name one crystalline defect that you would expect toform in order to maintain charge neutrality.(b) How many Cu_ ions are required for the creation of each defect?5.8 Below, atomic radius, crystal structure, electronegativity, and the most commonvalence are tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated.Which of these elements would you expect to form the following with copper:(a) A substitutional solid solution having complete solubility?(b) A substitutional solid solution of incomplete solubility?(c) An interstitial solid solution?5.9 For both FCC and BCC crystal structures, there are two different types ofinterstitial sites. In each case, one site is larger than the other, which site isnormally occupied by impurity atoms. For FCC, this larger one is located at the center of each edge of the unit cell; it is termed an octahedral interstitial site. On the other hand, with BCC the larger site type is found at 0, __, __ positions—that is, lying on _100_ faces, and situated midway between two unit cell edges on this face and one-quarter of the distance between the other two unit cell edges; it is termed a tetrahedral interstitial site. For both FCC and BCC crystalstructures, compute the radius r of an impurity atom that will just fit into one of these sites in terms of the atomic radius R of the host atom.5.10 (a) Suppose that Li2O is added as an impurity to CaO. If the Li_ substitutes forCa2_, what kind of vacancies would you expect to form? How many of thesevacancies are created for every Li_ added?(b) Suppose that CaCl2 is added as an impurity to CaO. If the Cl_ substitutes forO2_, what kind of vacancies would you expect to form? How many of thevacancies are created for every Cl_ added?5.28 Copper and platinum both have the FCC crystal structure and Cu forms asubstitutional solid solution for concentrations up to approximately 6 wt% Cu at room temperature. Compute the unit cell edge length for a 95 wt% Pt-5 wt% Cu alloy.5.29 Cite the relative Burgers vector–dislocation line orientations for edge, screw, andmixed dislocations.6.1 Briefly explain the difference between selfdiffusion and interdiffusion.6.3 (a) Compare interstitial and vacancy atomic mechanisms for diffusion.(b) Cite two reasons why interstitial diffusion is normally more rapid thanvacancy diffusion.6.4 Briefly explain the concept of steady state as it applies to diffusion.6.5 (a) Briefly explain the concept of a driving force.(b) What is the driving force for steadystate diffusion?6.6Compute the number of kilograms of hydrogen that pass per hour through a5-mm thick sheet of palladium having an area of 0.20 m2at 500℃. Assume a diffusion coefficient of 1.0×10- 8 m2/s, that the concentrations at the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.6.7 A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at 1200℃and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 6×10-11m2/s, and the diffusion flux is found to be 1.2×10- 7kg/m2-s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m3. How far into the sheet from this high-pressure side will the concentration be 2.0 kg/m3?Assume a linear concentration profile.6.24. Carbon is allowed to diffuse through a steel plate 15 mm thick. Theconcentrations of carbon at the two faces are 0.65 and 0.30 kg C/m3 Fe, whichare maintained constant. If the preexponential and activation energy are 6.2 _10_7 m2/s and 80,000 J/mol, respectively, compute the temperature at which the diffusion flux is 1.43 _ 10_9 kg/m2-s.6.25 The steady-state diffusion flux through a metal plate is 5.4_10_10 kg/m2-s at atemperature of 727_C (1000 K) and when the concentration gradient is _350kg/m4. Calculate the diffusion flux at 1027_C (1300 K) for the sameconcentration gradient and assuming an activation energy for diffusion of125,000 J/mol.10.2 What thermodynamic condition must be met for a state of equilibrium to exist? 10.4 What is the difference between the states of phase equilibrium and metastability?10.5 Cite the phases that are present and the phase compositions for the followingalloys:(a) 90 wt% Zn–10 wt% Cu at 400℃(b) 75 wt% Sn–25wt%Pb at 175℃(c) 55 wt% Ag–45 wt% Cu at 900℃(d) 30 wt% Pb–70 wt% Mg at 425℃(e) 2.12 kg Zn and 1.88 kg Cu at 500℃(f ) 37 lbm Pb and 6.5 lbm Mg at 400℃(g) 8.2 mol Ni and 4.3 mol Cu at 1250℃.(h) 4.5 mol Sn and 0.45 mol Pb at 200℃10.6 For an alloy of composition 74 wt% Zn–26 wt% Cu, cite the phases presentand their compositions at the following temperatures: 850℃, 750℃, 680℃, 600℃, and 500℃.10.7 Determine the relative amounts (in terms of mass fractions) of the phases forthe alloys and temperatures given inProblem 10.5.10.9 Determine the relative amounts (interms of volume fractions) of the phases forthe alloys and temperatures given inProblem 10.5a, b, and c. Below are given theapproximate densities of the various metalsat the alloy temperatures:10.18 Is it possible to have a copper–silveralloy that, at equilibrium, consists of a _ phase of composition 92 wt% Ag–8wt% Cu, and also a liquid phase of composition 76 wt% Ag–24 wt% Cu? If so, what will be the approximate temperature of the alloy? If this is not possible,explain why.10.20 A copper–nickel alloy of composition 70 wt% Ni–30 wt% Cu is slowly heatedfrom a temperature of 1300_C .(a) At what temperature does the first liquid phase form?(b) What is the composition of this liquid phase?(c) At what temperature does complete melting of the alloy occur?(d) What is the composition of the last solid remaining prior to completemelting?10.28 .Is it possible to have a copper–silver alloy of composition 50 wt% Ag–50 wt%Cu, which, at equilibrium, consists of _ and _ phases having mass fractions W_ _0.60 and W_ _ 0.40? If so, what will be the approximate temperature of the alloy?If such an alloy is not possible, explain why.10.30 At 700_C , what is the maximum solubility (a) of Cu in Ag? (b) Of Ag in Cu?第三章习题和思考题3.3If the atomic radius of aluminum is 0.143nm, calculate the volume of its unitcell in cubic meters.3.8 Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomicweight of 55.85 g/mol. Compute and compare its density with the experimental value found inside the front cover.3.9 Calculate the radius of an iridium atom given that Ir has an FCC crystal structure,a density of 22.4 g/cm3, and an atomic weight of 192.2 g/mol.3.13 Using atomic weight, crystal structure, and atomic radius data tabulated insidethe front cover, compute the theoretical densities of lead, chromium, copper, and cobalt, and then compare these values with the measured densities listed in this same table. The c/a ratio for cobalt is 1.623.3.15 Below are listed the atomic weight, density, and atomic radius for threehypothetical alloys. For each determine whether its crystal structure is FCC,BCC, or simple cubic and then justify your determination. A simple cubic unitcell is shown in Figure 3.40.3.21 This is a unit cell for a hypotheticalmetal:(a) To which crystal system doesthis unit cell belong?(b) What would this crystal structure be called?(c) Calculate the density of the material, given that its atomic weight is 141g/mol.3.25 For a ceramic compound, what are the two characteristics of the component ionsthat determine the crystal structure?3.29 On the basis of ionic charge and ionic radii, predict the crystal structures for thefollowing materials: (a) CsI, (b) NiO, (c) KI, and (d) NiS. Justify your selections.3.35 Magnesium oxide has the rock salt crystal structure and a density of 3.58 g/cm3.(a) Determine the unit cell edge length. (b) How does this result compare withthe edge length as determined from the radii in Table 3.4, assuming that theMg2_ and O2_ ions just touch each other along the edges?3.36 Compute the theoretical density of diamond given that the CUC distance andbond angle are 0.154 nm and 109.5°, respectively. How does this value compare with the measured density?3.38 Cadmium sulfide (CdS) has a cubic unit cell, and from x-ray diffraction data it isknown that the cell edge length is 0.582 nm. If the measured density is 4.82 g/cm3 , how many Cd 2+ and S 2—ions are there per unit cell?3.41 A hypothetical AX type of ceramic material is known to have a density of 2.65g/cm 3 and a unit cell of cubic symmetry with a cell edge length of 0.43 nm. The atomic weights of the A and X elements are 86.6 and 40.3 g/mol, respectively.On the basis of this information, which of the following crystal structures is (are) possible for this material: rock salt, cesium chloride, or zinc blende? Justify your choice(s).3.42 The unit cell for Mg Fe2O3 (MgO-Fe2O3) has cubic symmetry with a unit celledge length of 0.836 nm. If the density of this material is 4.52 g/cm 3 , compute its atomic packing factor. For this computation, you will need to use ionic radii listed in Table 3.4.3.44 Compute the atomic packing factor for the diamond cubic crystal structure(Figure 3.16). Assume that bonding atoms touch one another, that the angle between adjacent bonds is 109.5°, and that each atom internal to the unit cell is positioned a/4 of the distance away from the two nearest cell faces (a is the unit cell edge length).3.45 Compute the atomic packing factor for cesium chloride using the ionic radii inTable 3.4 and assuming that the ions touch along the cube diagonals.3.46 In terms of bonding, explain why silicate materials have relatively low densities.3.47 Determine the angle between covalent bonds in an SiO44—tetrahedron.3.63 For each of the following crystal structures, represent the indicated plane in themanner of Figures 3.24 and 3.25, showing both anions and cations: (a) (100)plane for the rock salt crystal structure, (b) (110) plane for the cesium chloride crystal structure, (c) (111) plane for the zinc blende crystal structure, and (d) (110) plane for the perovskite crystal structure.3.66 The zinc blende crystal structure is one that may be generated from close-packedplanes of anions.(a) Will the stacking sequence for this structure be FCC or HCP? Why?(b) Will cations fill tetrahedral or octahedral positions? Why?(c) What fraction of the positions will be occupied?3.81* The metal iridium has an FCC crystal structure. If the angle of diffraction forthe (220) set of planes occurs at 69.22°(first-order reflection) when monochromatic x-radiation having a wavelength of 0.1542 nm is used, compute(a) the interplanar spacing for this set of planes, and (b) the atomic radius for aniridium atom.4.10 What is the difference between configuration and conformation in relation topolymer chains? vinyl chloride).4.22 (a) Determine the ratio of butadiene to styrene mers in a copolymer having aweight-average molecular weight of 350,000 g/mol and weight-average degree of polymerization of 4425.(b) Which type(s) of copolymer(s) will this copolymer be, considering thefollowing possibilities: random, alternating, graft, and block? Why?4.23 Crosslinked copolymers consisting of 60 wt% ethylene and 40 wt% propylenemay have elastic properties similar to those for natural rubber. For a copolymer of this composition, determine the fraction of both mer types.4.25 (a) Compare the crystalline state in metals and polymers.(b) Compare thenoncrystalline state as it applies to polymers and ceramic glasses.4.26 Explain briefly why the tendency of a polymer to crystallize decreases withincreasing molecular weight.4.27* For each of the following pairs of polymers, do the following: (1) state whetheror not it is possible to determine if one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.(a) Linear and syndiotactic polyvinyl chloride; linear and isotactic polystyrene.(b) Network phenol-formaldehyde; linear and heavily crosslinked ci s-isoprene.(c) Linear polyethylene; lightly branched isotactic polypropylene.(d) Alternating poly(styrene-ethylene) copolymer; randompoly(vinylchloride-tetrafluoroethylene) copolymer.4.28 Compute the density of totally crystalline polyethylene. The orthorhombic unitcell for polyethylene is shown in Figure 4.10; also, the equivalent of two ethylene mer units is contained within each unit cell.5.11 What point defects are possible for MgO as an impurity in Al2O3? How manyMg 2+ ions must be added to form each of these defects?5.13 What is the composition, in weight percent, of an alloy that consists of 6 at% Pband 94 at% Sn?5.14 Calculate the composition, in weight per-cent, of an alloy that contains 218.0 kgtitanium, 14.6 kg of aluminum, and 9.7 kg of vanadium.5.23 Gold forms a substitutional solid solution with silver. Compute the number ofgold atoms per cubic centimeter for a silver-gold alloy that contains 10 wt% Au and 90 wt% Ag. The densities of pure gold and silver are 19.32 and 10.49 g/cm3 , respectively.8.53 In terms of molecular structure, explain why phenol-formaldehyde (Bakelite)will not be an elastomer.10.50 Compute the mass fractions of αferrite and cementite in pearlite. assumingthat pressure is held constant.10.52 (a) What is the distinction between hypoeutectoid and hypereutectoid steels?(b) In a hypoeutectoid steel, both eutectoid and proeutectoid ferrite exist. Explainthe difference between them. What will be the carbon concentration in each?10.56 Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 727_C(a) What is the proeutectoid phase?(b) How many kilograms each of total ferrite and cementite form?(c) How many kilograms each of pearlite and the proeutectoid phase form?(d) Schematically sketch and label the resulting microstructure.10.60 The mass fractions of total ferrite and total cementite in an iron–carbon alloyare 0.88 and 0.12, respectively. Is this a hypoeutectoid or hypereutectoid alloy?Why?10.64 Is it possible to have an iron–carbon alloy for which the mass fractions of totalferrite and proeutectoid cementite are 0.846 and 0.049, respectively? Why orwhy not?第四章习题和思考题7.3 A specimen of aluminum having a rectangular cross section 10 mm _ 12.7 mmis pulled in tension with 35,500 N force, producing only elastic deformation. 7.5 A steel bar 100 mm long and having a square cross section 20 mm on an edge ispulled in tension with a load of 89,000 N , and experiences an elongation of 0.10 mm . Assuming that the deformation is entirely elastic, calculate the elasticmodulus of the steel.7.7 For a bronze alloy, the stress at which plastic deformation begins is 275 MPa ,and the modulus of elasticity is 115 Gpa .(a) What is the maximum load that may be applied to a specimen with across-sectional area of 325mm, without plastic deformation?(b) If the original specimen length is 115 mm , what is the maximum length towhich it may be stretched without causing plastic deformation?7.8 A cylindrical rod of copper (E _ 110 GPa, Stress (MPa) ) having a yield strengthof 240Mpa is to be subjected to a load of 6660 N. If the length of the rod is 380 mm, what must be the diameter to allow an elongation of 0.50 mm?7.9 Consider a cylindrical specimen of a steel alloy (Figure 7.33) 10mm in diameterand 75 mm long that is pulled in tension. Determine its elongation when a load of 23,500 N is applied.7.16 A cylindrical specimen of some alloy 8 mm in diameter is stressed elasticallyin tension. A force of 15,700 N produces a reduction in specimen diameter of 5 _ 10_3 mm. Compute Poisson’s ratio for this material if its modulus of elasticity is 140 GPa .7.17 A cylindrical specimen of a hypothetical metal alloy is stressed in compression.If its original and final diameters are 20.000 and 20.025 mm, respectively, and its final length is 74.96 mm, compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 105 Gpa and 39.7 GPa,respectively.7.19 A brass alloy is known to have a yield strength of 275 MPa, a tensile strength of380 MPa, and an elastic modulus of 103 GPa . A cylindrical specimen of thisalloy 12.7 mm in diameter and 250 mm long is stressed in tension and found to elongate 7.6 mm . On the basis of the information given, is it possible tocompute the magnitude of the load that is necessary to produce this change inlength? If so, calculate the load. If not, explain why.7.20A cylindrical metal specimen 15.0mmin diameter and 150mm long is to besubjected to a tensile stress of 50 Mpa; at this stress level the resulting deformation will be totally elastic.(a)If the elongation must be less than 0.072mm,which of the metals in Tabla7.1are suitable candidates? Why ?(b)If, in addition, the maximum permissible diameter decrease is 2.3×10-3mm,which of the metals in Table 7.1may be used ? Why?7.22 Cite the primary differences between elastic, anelastic, and plastic deformationbehaviors.7.23 diameter of 10.0 mm is to be deformed using a tensile load of 27,500 N. It mustnot experience either plastic deformation or a diameter reduction of more than7.5×10-3 mm. Of the materials listed as follows, which are possible candidates?Justify your choice(s).7.24 A cylindrical rod 380 mm long, having a diameter of 10.0 mm, is to besubjected to a tensile load. If the rod is to experience neither plastic deformationnor an elongation of more than 0.9 mm when the applied load is 24,500 N,which of the four metals or alloys listed below are possible candidates?7.25 Figure 7.33 shows the tensile engineering stress–strain behavior for a steel alloy.(a) What is the modulus of elasticity?(b) What is the proportional limit?(c) What is the yield strength at a strain offset of 0.002?(d) What is the tensile strength?7.27 A load of 44,500 N is applied to a cylindrical specimen of steel (displaying thestress–strain behavior shown in Figure 7.33) that has a cross-sectional diameter of 10 mm .(a) Will the specimen experience elastic or plastic deformation? Why?(b) If the original specimen length is 500 mm), how much will it increase inlength when t his load is applied?7.29 A cylindrical specimen of aluminumhaving a diameter of 12.8 mm and a gaugelength of 50.800 mm is pulled in tension. Usethe load–elongation characteristics tabulatedbelow to complete problems a through f.(a)Plot the data as engineering stressversusengineering strain.(b) Compute the modulus of elasticity.(c) Determine the yield strength at astrainoffset of 0.002.(d) Determine the tensile strength of thisalloy.(e) What is the approximate ductility, in percent elongation?(f ) Compute the modulus of resilience.7.35 (a) Make a schematic plot showing the tensile true stress–strain behavior for atypical metal alloy.(b) Superimpose on this plot a schematic curve for the compressive truestress–strain behavior for the same alloy. Explain any difference between thiscurve and the one in part a.(c) Now superimpose a schematic curve for the compressive engineeringstress–strain behavior for this same alloy, and explain any difference between this curve and the one in part b.7.39 A tensile test is performed on a metal specimen, and it is found that a true plasticstrain of 0.20 is produced when a true stress of 575 MPa is applied; for the same metal, the value of K in Equation 7.19 is 860 MPa. Calculate the true strain that results from the application of a true stress of 600 Mpa.7.40 For some metal alloy, a true stress of 415 MPa produces a plastic true strain of0.475. How much will a specimen of this material elongate when a true stress of325 MPa is applied if the original length is 300 mm ? Assume a value of 0.25 for the strain-hardening exponent n.7.43 Find the toughness (or energy to cause fracture) for a metal that experiences bothelastic and plastic deformation. Assume Equation 7.5 for elastic deformation,that the modulus of elasticity is 172 GPa , and that elastic deformation terminates at a strain of 0.01. For plastic deformation, assume that the relationship between stress and strain is described by Equation 7.19, in which the values for K and n are 6900 Mpa and 0.30, respectively. Furthermore, plastic deformation occurs between strain values of 0.01 and 0.75, at which point fracture occurs.7.47 A steel specimen having a rectangular cross section of dimensions 19 mm×3.2mm (0.75in×0.125in.) has the stress–strain behavior shown in Figure 7.33. If this specimen is subjected to a tensile force of 33,400 N (7,500lbf ), then(a) Determine the elastic and plastic strain values.(b) If its original length is 460 mm (18 in.), what will be its final length after theload in part a is applied and then released?7.50 A three-point bending test was performed on an aluminum oxide specimenhaving a circular cross section of radius 3.5 mm; the specimen fractured at a load of 950 N when the distance between the support points was 50 mm . Another test is to be performed on a specimen of this same material, but one that has a square cross section of 12 mm length on each edge. At what load would you expect this specimen to fracture if the support point separation is 40 mm ?7.51 (a) A three-point transverse bending test is conducted on a cylindrical specimenof aluminum oxide having a reported flexural strength of 390 MPa . If the speci- men radius is 2.5 mm and the support point separation distance is 30 mm ,predict whether or not you would expect the specimen to fracture when a load of 620 N is applied. Justify your prediction.(b) Would you be 100% certain of the prediction in part a? Why or why not?7.57 When citing the ductility as percent elongation for semicrystalline polymers, it isnot necessary to specify the specimen gauge length, as is the case with metals.Why is this so?7.66 Using the data represented in Figure 7.31, specify equations relating tensilestrength and Brinell hardness for brass and nodular cast iron, similar toEquations 7.25a and 7.25b for steels.8.4 For each of edge, screw, and mixed dislocations, cite the relationship between thedirection of the applied shear stress and the direction of dislocation line motion.8.5 (a) Define a slip system.(b) Do all metals have the same slip system? Why or why not?8.7. One slip system for theBCCcrystal structure is _110__111_. In a manner similarto Figure 8.6b sketch a _110_-type plane for the BCC structure, representingatom positions with circles. Now, using arrows, indicate two different _111_ slip directions within this plane.8.15* List four major differences between deformation by twinning and deformationby slip relative to mechanism, conditions of occurrence, and final result.8.18 Describe in your own words the three strengthening mechanisms discussed inthis chapter (i.e., grain size reduction, solid solution strengthening, and strainhardening). Be sure to explain how dislocations are involved in each of thestrengthening techniques.8.19 (a) From the plot of yield strength versus (grain diameter)_1/2 for a 70 Cu–30 Zncartridge brass, Figure 8.15, determine values for the constants _0 and ky inEquation 8.5.(b) Now predict the yield strength of this alloy when the average grain diameteris 1.0 _ 10_3 mm.8.20. The lower yield point for an iron that has an average grain diameter of 5 _ 10_2mm is 135 MPa . At a grain diameter of 8 _ 10_3 mm, the yield point increases to 260MPa. At what grain diameter will the lower yield point be 205 Mpa ?8.24 (a) Show, for a tensile test, thatif there is no change in specimen volume during the deformation process (i.e., A0 l0 _Ad ld).(b) Using the result of part a, compute the percent cold work experienced bynaval brass (the stress–strain behavior of which is shown in Figure 7.12) when a stress of 400 MPa is applied.8.25 Two previously undeformed cylindrical specimens of an alloy are to be strainhardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are 16 mm and11 mm, respectively. The second specimen, with an initial radius of 12 mm, musthave the same deformed hardness as the first specimen; compute the secondspecimen’s radius after deformation.8.26 Two previously undeformed specimens of the same metal are to be plasticallydeformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular is to remain as such. Their original and deformeddimensions are as follows:Which of these specimens will be the hardest after plastic deformation, and why?8.27 A cylindrical specimen of cold-worked copper has a ductility (%EL) of 25%. Ifits coldworked radius is 10 mm (0.40 in.), what was its radius beforedeformation?8.28 (a) What is the approximate ductility (%EL) of a brass that has a yield strengthof 275 MPa ?(b) What is the approximate Brinell hardness of a 1040 steel having a yieldstrength of 690 MPa?8.41 In your own words, describe the mechanisms by which semicrystalline polymers(a) elasticallydeform and (b) plastically deform, and (c) by which elastomerselastically deform.8.42 Briefly explain how each of the following influences the tensile modulus of asemicrystallinepolymer and why:(a) molecular weight;(b) degree of crystallinity;(c) deformation by drawing;(d) annealing of an undeformed material;(e) annealing of a drawn material.8.43* Briefly explain how each of the following influences the tensile or yieldstrength of a semicrystalline polymer and why:(a) molecular weight;。
材料导论习题2有答案(英文版)
材料科学与工程导论1 Polymerization (聚合作用) is the process by which small molecules (分子) consisting of one unit or a few units are chemically joined to create these giant molecules. Those small molecule units are called______A) polymers B) monomers (单体)C) oligomers D)elastomers2 Choose the type of polymer material you might select for the following applications: an automobile bumper (保险杠), a mineral water bottle, and an insulating (绝热的) cable.A) For automobile bumper, the best choice of polymer is Natural Rubber.B) For mineral water bottle, the best choice of polymer is Low-Density（密度）Polyethylene (聚乙烯)C) For insulating cable, the best choice of polymer is polystyrene (聚苯乙烯).D) For mineral water bottle, the best choice of polymer is Polyethylene Terephthalate.3) In general, for a given type of thermoplastic (热塑性塑胶) the tensile (可延展的) strength, creep resistance impact toughness, wear resistance, and melting temperature all increase with______A) degree of polymerization B)density of branching C)tacticity D) crystallinity4) ABS, composed of acrylonitrile, butadiene, and styrene (苯乙烯), is one of the most common polymer materials. Styrene and acrylonitrile form a liner copolymer (异量分子聚合物) (SNA) that serves as a matrix. Styrene and butadiene also form a liner copolymer, BS rubber, which acts as the ______ material.A) fire retardant B) filler C) cross-linking D)plasticizer5) Epoxies （环氧树脂）are thermosetting polymers formed from molecules containing a tight C-O-C ring. During polymerization, the C-O-C rings are opened and the bonds are rearranged to join the molecules. If epoxy is used as an adhesive (粘剂) for a variety of applications, which kind of adhesives does it belong to?A) Chemically Reactive AdhesivesB) Evaporation or Diffusion AdhesivesC) Hot-Melt Adhesives D) Pressure-Sensitive Adhesives6) What area the major advantages associated with plastic compared to ceramics(陶瓷), glasses, and metallic materials?A) lightweight and corrosion(铁锈，腐蚀)-resistantB) high strength and high-temperature resistantC) high-temperature resistant and insulatingD) high stiffness and corrosion-resistant7) Which one is not a thermoset polymer?A) Polyethylene B) Natural Rubber C) Epoxy D) Phenolic8) Depending on the degree of cross-linking, the polyurethanes behave as thermosetting polymers, thermoplastics (热塑性塑胶), or elastomers. These polymers find application as fibers, coatings, and foams for furniture, mattress, and insulation. Why polyurethanes are versatile?A) for their polar repeat unit B) for their linear structureC) for their multi-functional monome r D) for their tactictity structure9) Liquid Crystalline(透明的) Polymers are polymers which behave as ______A) liquid B) thermoplastic C) thernoset D) oriented(取向) rods10) The glass temperature (Tg) is typically about ______ times the absolute melting temperature (Tm).A) 0.2 to 0.3 B) 0.5 to 0.7C) 0.8 to 1.0 D) 1.0 to 2.011) Most thermoplastic exhibit a non-Newtonian and visicoelastic behavior. The stress and train are not linearly related for most parts of the stress-strain curve. The viscoelastic behavior means when an external (外部的) force is applied to a thermoplastic _______ deformation occurs.A) only elastic B) both elastic and plastic (塑性)C) only plastic D) neither elastic nor plastic12) Compare the tensile strength of LDPE, HDPE, PVE, PP, and arrange them in sequence(顺序) from high strength to low tensile strength.A) LOPE>HDPE>PVE>PP B) LDPE>HDPE>PP>PVCC) PVC>PP>HDPE>LDPE D) PVC>HDPE>PP>LDPE13) Degree of polymerization is usually used to characterize ________A) cross-linking network B) thermosetting polymerC) polymer degradation D) thermoplastic polymer14) Silicones(硅) are important elastomer based on chains composed of silicon and oxygen atoms. The silicone rubbers provide high-temperature resistance, permitting use of the elastomer at temperature as high as _____℃. Low molecular weight silicones form liquids and are known as silicon oils.A) 100 B) 150 C) 200 D) 30015) There are a lot of thermoplastic processing methods, typical forming process includes: extrusion, blow molding, injection molding, thermoforming, calendaring and spinning. If we want to produce sping water bottles, which processing way is best choice?A) extrusion B) blow molding C) injection molding D) thermoforming16) The recycling of thermoplastics is relatively easy and practiced widely. Many of the everyday plastic products you encounter (bags, soda bottles, yogurt containers, etc.) have numbers stamped on them. For PRT products, the number is 1. For HDPE and LDPE, the numbers are_______, respectively. Other plastics are marked number 7,A) 3 and 4 B) 4 and 5 C) 2 and 4D) 2 and 317) The temperature above which a polymer burns, chars, or decompose. Which team is appropriately used in describing the remperature.A) Tg B) Td C) Tm D) HDT18) Which forming process is not a best choice for thermosetting polymers.A) Compression molding B)transfer moldingC) reaction injection D) spinning19) Elastomers are thermoplastics or lightly cross-linked thermosets that exhibit greater than _______ elastic deformation.A) 0.2% B ) 2% C) 20% D)200%20) Thermoplastic elastomers combine feature of both thermoplastics and elastomers.At high temperature, these polymers behave as ________ and are plastically into shapes, at how temperature, they behave as _________A) thermoplastics; elastomers B) thermoplastics; thermosetting polymersC) elastomers; thermoplastic D) elastomers; thermosetting polymers21 Define (a) a thermoplastic, (b) thermosetting plastics, (c) elastomers, and (d) thermoplastic elastomers.22) Explain what the following terms mean in polymer materials: decompositions temperate (Td), heat distortion temperature (HDT), glass temperature (Tg), and melting temperature (Tm).23) Typical forming processes for thermoplastic are schematicly shown blew, give each forming process a name and explain what kind of plastic products can be made by corresponding forming process.挤出吹塑延展喷涂…………24) The molecular weight of polymethyl methacrylate (PMMA) is 250,000g/mol. If all of the polymer chains are the same length.a) calculate the degree of polymerization, andb) the number of chains in 1g of the polymers.25) Explain why low-density polyethylene is good to make grocery bags, however, ultra-high molecular weight polyethylene must be used where strength and very high wear resistance is needed.26) How do the glass temperature of polyethylene, polypropylene, and polymethyl methacrylate compare? Explain their difference, based on the structure of the monomer?27) Give a materials example in your everyday life, explain the relationship of their structure and properties.bdada aacdb bcddb cbdda。
材料科学与工程专业英语第三版-翻译以及答案
材料科学与工程专业英语第三版-翻译以及答案Unit 1材料在我们生活中的影响可能远远超出我们的想象。
从交通、装修、制衣、通信、娱乐到食品生产，材料无处不在。
历史上，社会的发展和进步与生产材料的能力密切相关。
早期的文明就是通过材料发展的能力来命名的（石器时代、青铜时代、铁器时代）。
早期的人类仅仅使用了极少量的材料，如自然的石头、木头、粘土和兽皮。
随着时间的发展，通过技术生产的材料比自然材料具有更好的性能，如各种金属和陶瓷。
人们还发现，通过添加其他物质和改变加热温度可以改变材料的性能。
在过去的100年中，科学家对材料的基本结构和性能关系有了更深入的理解。
为满足现代复杂社会的需求，成千上万种不同性质的材料被研发出来，包括金属、塑料、玻璃和纤维。
新技术的发展使我们获得了更适合的材料，使我们的生活更加舒适。
对材料性质的理解进步往往是技术发展的先兆。
例如，如果没有合适且不昂贵的钢材或其他替代品，汽车就不可能生产。
现代复杂的电子设备依赖于半导体材料。
将材料科学和工程划分为两个副学科是非常有用的。
材料科学研究材料性能和结构的关系，而材料工程则基于材料结构和性能的关系，设计和生产具有预定性能的材料。
材料科学家开发或合成新材料，而材料工程师生产新产品或运用现有材料开发生产技术。
大多数材料学毕业生同时接受材料科学和工程的培训。
五、材料的“structure”指的是其内在成分的排列。
在原子水平上，结构包括原子或分子与其他相关的原子或分子的组织，而在更大的结构领域上，其包括大的原子团。
最后，结构单元可以通过肉眼看到的称为宏观结构。
六、“Property”指的是材料对外部刺激的反应。
材料的特征取决于其对外部刺激的反应程度。
通常，材料的性质与其形状及大小无关。
七、所有固体材料的重要性质可以概括为六类：机械、电学、热学、磁学、光学和腐蚀性。
对于每一种性质，其都有一种对特定刺激引起反应的能力。
比如，机械性能与施加压力引起的形变有关，而电性能则与电场有关。
材料科学复习资料(英文版)(doc 6页)
Thermoplastic plastics are generally best for type impactof service conditionsPolycarbonates ( PC ) Provides the substrate for CD-ROMs because they possess excellent optical properties as well as hardness and toughness .Better toughness allow plastics compete strongly with optical glassesAnaerobic adhesives retain their fluid state when exposed to oxygen , but when squeezed into thin joints that block the . oxygen , they set up into hard , strong adhesives.The advantage ( s )of Light weight , load spreading , join sealing would adhesive bonding have over other joining methodsA field that involves the generation and application of knowledge relating to the composition,structure and processing of materials to their properties and use, Materials Science And EngineeringRecycling Plastics is the most effective approach to making plastic more environmentally friendly .Blow molding can be applied to thermoplastics only .Rayon, Polyacrylonitrile are used to Produce carbon fibersAccelerators and activators are (is)used to speed sulfur vulcanizing . Higher stiffness compared with reinforcement is not the requirement for matrix used in composites.The cost in materials selection usually dominates the final choice.Tensile strength is the maximum stress developed in a material during a tensile test._____Toughness________represents the energy per unit volume of a material required to produce fracture under static conditions.A useful tool that is used to study the crystal structures of solids by measuring the angles of electrons glancing off material specimens. XRDTrade name for aramid fiber , which is used for bulletProof vests and advanced composites for aerospace applications . KevlarThe major source of raw materials for synthetic rubber. OilA substance capable of holding materials together by surface attachment Adhesive Diamond is a transparent crystalline carbon.Which is not the advantage of Carbon/Carbon composites ?Easy oxidation at high temperaturePlywood is an example of Laminar composites.The hardness of plastics closely correlates to tensile strength塑料的硬度和抗拉强度密切相关。
材料科学基础英文版习题
What is the difference between Vickers and Knoop hardness tests? Chapter 6, Problem 13 Describe the slip mechanism that enables a metal to be plastically deformed without fracture. Chapter 6, Problem 14
ChapБайду номын сангаасer 6，Problem 15 (a) What are the principal slip planes and slip directions for FCC metals? (b) What are the principal slip planes and slip directions for BCC metals? (c) What are the principal slip planes and slip directions for HCP metals? Chapter 6, Problem 16 Describe the deformation twinning process that occurs in some metals when they are plastically deformed. Chapter 6, Problem 17 What is the difference between the slip and twinning mechanisms of plastic deformation of metals? Chapter 6，Problem 18 What other types of slip planes are important other than the basal planes for HCP metals with low c/a ratios? Chapter 6，Problem 19 What important role does twinning play in the plastic deformation of metals with regard to deformation of metals by slip? Chapter 6, Problem 20 Define the critical resolved shear stress for a pure metal single crystal? What happens to the metal from the macroscale point of view and behavior point of view once critical resolved shear stress is exceeded? Chapter 6, Problem 21 By what mechanism do grain boundaries strengthen metals? Chapter 6, Problem 22 (a) What is solid-solution strengthening? Describe the two main types, (b) What are two
高三英语材料科学练习题40题答案解析版
高三英语材料科学练习题40题答案解析版1. In the field of material science, a new ______ has been discovered which is very light and strong.A. materialB. substanceC. alloyD. matter答案：A。
解析：本题考查材料科学领域的基础词汇。
A选项“material”通常指用于制造或构建某物的材料，在材料科学语境下，新发现的这种又轻又强的东西可以直接称为一种新的材料。
B选项“substance”含义更广泛，指任何一种物质，但不如“material”在材料科学领域那么具体。
C选项“alloy”是合金的意思，这里没有提及是合金相关，所以不符合。
D选项“matter”是物质的总称，比较抽象，没有“material”准确。
2. The ______ used in this experiment are all carefully selected.A. materialsB. substancesC. alloysD. matters答案：A。
解析：根据句子中的“used in this experiment”（用于这个实验中的）可知是具体的材料，“materials”侧重于表示可用于制作或构建的材料，是可数名词复数形式，符合语境。
B选项“substances”虽然也表示物质，但没有“materials”在此处合适。
C选项“alloys”是合金，这里没有表明是合金材料。
D选项“matters”是物质的统称，过于宽泛。
3. Many modern buildings ______ new types of materials to make them more energy - efficient.A. useB. usesC. usingD. used答案：A。
解析：本题考查主谓一致。
材料科学与工程基础英文版第五版课后习题
Homework 11.1 What are materials? List eight commonly encountered engineering materials. Answer1.1: Materials are substances of which something is composed or made. Steels, aluminum alloys, concrete, wood, glass, plastics, ceramics and electronic materials.1.2 What are the main classes of engineering materials?Answer1.2: Metallic, polymeric, ceramic, composite, and electronic materials are the five main classes.1.3 What are some of the important properties of each of the five main classes of engineering materials?Answer1.3:Metallic Materials• many are relatively strong and ductile at room temperature• some have good strength at high temperature• most have relatively high electrical and thermal conductivitiesPolymeric Materials• generally are poor electrical and thermal conductors• most have low to medium strengths• most have low densities• most are relatively easy to process into final shape• some are transparentCeramic Materials• generally have high ha rdness and are mechanically brittle• some have useful high temperature strength• most have poor electrical and thermal conductivitiesComposite Materials• have a wide range of strength from low to very high• some have very high strength-to-weight ratios (e.g. carbon-fiber epoxy materials)• some have medium strength and are able to be cast or formed into a variety of sha (e.g. fiberglass-polyester materials)• some have useable strengths at very low cos t (e.g. wood and concrete)Electronic Materials• able to detect, amplify and transmit electrical signals in a complex manner• are light weight, compact and energy efficient1.8 What are nanomaterials? What are some proposed advantages of using nanomaterials over their conventional counterparts?Answer1.8: Are defined as materials with a characteristic length scale smaller than 100 nm. The length scale could be particle diameter, grain size in a material, layer thicknessin a sensor, etc. These materials have properties different than that at bulk scale or at themolecular scale. These materials have often enhanced properties and characteristics because of their nano-features in comparison to their micro-featured counterparts. The structural, chemical, electronic, and thermal properties (among other characteristics) are often enhanced at the nano-scale.Homework 2Chapter 3, Problem 4What are the three most common metal crystal structures? List five metals that have each of these crystal structures. Chapter 3, Solution 4The three most common crystal structures found in metals are: body-centered cubic (BCC), face-centered cubic (FCC), and hexagonal close-packed (HCP). Examples of metals having these structures include the following. BCC:iron,α-vanadium, tungsten, niobium, and chromium.FCC: copper, aluminum, lead, nickel, and silver. HCP: magnesium, titanium,α-zinc, beryllium, and cadmium.Chapter 3, Problem 5For a BCC unit cell, (a) how many atoms are there inside the unit cell, (b) what is the coordination number for the atoms, (c) what is the relationship between the length of the side a of the BCC unit cell and the radius of its atoms, and (d) APF = 0.68 or 68%Chapter 3, Solution 5(a) A BCC crystal structure has two atoms in each unit cell. (b) A BCC crystal structure has a coordination number of eight . (c) In a BCC unit cell, one complete atom and two atom eighths toucheach other along the cube diagonal. This geometry translates into the relationship 4.R =Chapter 3, Problem 6For an FCC unit cell, (a) how many atoms are there inside the unit cell, (b) What is the coordination number for the atoms, (c) 24R a =, and (d) what is the atomic packing factor?Chapter 3, Solution 6(a) Each unit cell of the FCC crystal structure contains four atoms. (b) The FCC crystal structure has acoordination number of twelve . (d) By definition, the atomic packing factor is given as:volume of atoms in FCC unit cellAtomic packing factor volume of the FCC unit cell=These volumes, associated with the four-atom FCC unit cell, are33416433atoms V R R ππ⎡⎤==⎢⎥⎣⎦and 3unit cellV a =where a represents the lattice constant. Substitutinga =33unit cellV a ==The atomic packing factor then becomes,3316APF (FCC unit cell)3632R R ππ⎛⎫⎛⎫== ⎪ ⎪ ⎪⎝⎭⎝⎭=0.74 Chapter 3, Problem 7For an HCP unit cell (consider the primitive cell), (a) how many atoms are there inside the unit cell, (b) What is the coordination number for the atoms, (c) what is the atomic packing factor, (d) what is the ideal c/a ratio for HCP metals, and (e) repeat a through c considering the “larger” cell.Chapter 3, Solution 7The primitive cell has (a) two atoms/unit cell; (b) The coordination number associated with the HCP crystal structure is twelve . (c)the APF is 0.74 or 74%; (d) The ideal c/a ratio for HCP metals is 1.633; (e) all answers remain the same except for (a) where the new answer is 6.Homework 3 Chapter 3, Problem 25Lithium at 20︒C is BCC and has a lattice constant of 0.35092 nm. Calculate a value for the atomic radius of a lithium atom in nanometers.Chapter 3, Solution 25For the lithium BCC structure, which has a lattice constant of a = 0.35092 nm, the atomic radius is,R ===0.152 nmPalladium is FCC and has an atomic radius of 0.137 nm. Calculate a value for its lattice constant a in nanometers.Chapter 3, Solution 27Letting a represent the FCC unit cell edge length and R the palladium atomic radius,Chapter 3, Problem 31 Draw the following directions in a BCC unit cell and list the position coordinates of the atoms whose centers are intersected by the direction vector: (a ) [100] (b ) [110] (c ) [111]Chapter 3, Solution 31Chapter 3, Solution 324 or R a R ====0.387 nm(1, 0, 0)yxzyxz[111]x = +1y = -1 z = -1 x = +1 y = -1 z = 0(a) (b)[110]x = -½ y = 1(c)x = – ⅓ y = – ⅓(d)A cubic plane has the following axial intercepts: . What are the Miller indicesof this plane?Chapter 3, Solution 46Given the axial intercepts of (⅓, -⅔, ½), the reciprocal intercepts are:Multiplying by 2 to clear the fraction, the Miller indices areChapter 3, Problem 50Determine the Miller indices of the cubic crystal plane that intersects the following positioncoordinates:Chapter 3, Solution 50First locate the three position coordinates as shown. Next, connect points a and b and extend the line to point d . Complete the plane by connecting point d to c and point c to b . Using (1, 0, 1) as the plane origin, x = -1, y = 1 and z = –1. The intercept reciprocals are thusThe Miller indices are121332, , a b c ==-=11313,, 2.2x y z ==-=.(634)1122(, 0, ); (0,0,1); (1,1,1).1111,1, 1.x y z =-==-.(111)a(½, 0, ½ )。
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材料科学与工程基础”考试试题–英文原版教材班（注：第1、2、3题为必做题；第4、5、6、7题为选择题，必须二选一。
共100分）1. Glossary (2 points for each)1) crystal structure: The arrangement of the atoms in a materialinto a repeatable lattice.2) basis (or motif): A group of atoms associated with a lattice.3) packing fractor: The fraction of space in a unit cell occupiedby atoms.4) slip system: The combination of the slip plane and the slipdirection.5) critical size: The minimum size that must be formed by atomsclustering together in the liquid before the solid particle is stable and begins to grow.6) homogeneous nucleation: Formation of a critically sized solidfrom the liquid by the clustering together of a large number of atoms at a high undercooling (without an external interface).7) coherent precipitate:A precipitate whose crystal structure andatomic arrangement have a continuous relationship with matrix from which precipitate is formed.8) precipitation hardening: A strengthening mechanism thatrelies on a sequence of solid state phase transformations in a dispersion of ultrafine precipitates of a 2nd phase. This is same as age hardening. It is a form of dispersion strengthening.9) diffusion coefficient: A temperature-dependent coefficientrelated to the rate at which atom, ion, or other species diffusion.The DC depends on temperature, the composition and microstructure of the host material and also concentration of the diffusion species.10) uphill diffusion: A diffusion process in which species movefrom regions of lower concentration to that of higher concentration.2. Determine the indices for the planes in the cubic unit cell shown in Figure 1. (5 points)Fig. 1Solution : A(-364), B(-340), C(346). 3. Determine the crystal structure for the following: (a) a metal with a 0 = 4.9489 Å, r = 1.75 Å and one atom per lattice point; (b) a metal with a 0 = 0.42906 nm, r = 0.1858 nm and one atom per lattice point. (10 points) Solution : (a)fcc; (b) bcc. 4-1. What is the characteristic of brinell hardness test, rockwell hardness test and V ickers hardness test? What are the effects of strain rate and temperature on the mechanical properties of metallic materials? (15 points) 4-2. What are the effects of cold-work on metallic materials? How to eliminate those effects? And what is micro-mechanism for the eliminating cold-work effects? (15 points) 5-1. Based on the Pb-Sn-Bi ternary diagram as shown in Fig. 2, try to (1)Show the vertical section of 40wt.%Sn; (5 points) (2) Describe the solidification process of the alloy 2# with very low cooling speed (including phase and microstructure changes); (5 points) (3)Plot the isothermal section at 150o C. (5 points)5-2. A 1mm sheet of FCC iron is used to contain N 2 in a heated exchanger at 1200o C. The concentration of N at one surface is 0.04 atomic percent and the concentration at the second surface is 0.005 atomic percent. At 1000 oC, if same N concentration is demanded atFig. 2the second surface and the flux of N becomes to half of that at 1200o C, then what is the thickness of sheet? (15 points)6-1. Supposed that a certain liquid metal is undercooled until homogeneous nucleation occurs. (15 points)(1)How to calculate the critical radius of the nucleus required?Please give the deduction process.(2)For the Metal Ni, the Freezing Temperature is 1453︒C, the LatentHeat of Fusion is 2756 J/cm3, and the Solid-liquid Interfacial Energy is 255⨯10-7J/cm2. Please calculate the critical radius at 1353︒C. (Assume that the liquid Ni is not solidified.)6-2. Fig.3 is a portion of the Mg-Al phase diagram. (15 points) (1)If the solidification is too rapid, please describe the solidificationprocess of Mg-10wt%Al alloy.(2)Please describe the equilibrium solidification process ofMg-20wt%Al alloy, and calculate the amount of each phase at 300︒C.Fig. 37-1. Figure 4 shows us the Al-Cu binary diagram and some microstructures found in a cooling process for an Al-4%Cu alloy. Please answer following questions according to this figure. (20 points)Fig. 4 (1)What are precipitate, matrix and microconstituent? Please point them out in the in the figure and explain. (2)Why is need-like precipitate not good for dispersion strengthening? The typical microstructure shown in the figure is good or not? why? (3)Please tell us how to obtain the ideal microstructure shown in this figure. (4)Can dispersion strengthened materials be used at high temperature? Please give the reasons (comparing with cold working strengthening) 7-2. Please answer following questions according to the time-temperature-transformation (TTT) diagram as shown in Fig. 5. (20 points) (1)What steel is this TTT diagram for? And what means P, B, and M in the figure? (2)Why dose the TTT diagram exhibits a ‘C’ shape? (3)Point out what microconstituent will be obtained after austenite is cooled according to the curves I, II, III and IV . (4)What is microstructural difference between the curve I and the curve II? (5)How to obtain the steel with the structure of (a) P+B (b) P +M+A (residual) (c) P+B+M+A (residual) (d) F ull tempered martensite If you can, please draw the relative cooling curve or the flow chart of heat treatment.Fig. 5 III III IV。