材料科学基础英文版习题

材料科学基础(英)5

10
11
Some rules
The following features are of interest in connection with compositional aspects of ternary systems. (1) If a straight line such as CD is drawn from one corner of the 等比例规则：沿成分三 triangle (i.e. from component C) to 角形的顶点 C引一条与对 intersect the opposite side (i.e. AB), 边相交的线段 then the ratio of CD the，在此 other two components (i.e. A and B) is 线段CD上的成分的合金 constant in两组元的质量分 alloys represented by 中A和B points lying on the line. Considering 数比值相等，但含 C的质 alloy compositions along CD, as the 量分数不同，距离 composition point C点越 becomes 远，C的含量越少。 progressively further away from C the percentage of C present in the alloys decreases, but the relative amounts of B and A remain the same.
Chapter 5 Ternary equilibrium diagrams （三元平衡相图）
5.1 The representing way of ternary phase diagrams There are 2 variables for

高三英语材料科学单选题60题

高三英语材料科学单选题60题1.Metal is a common material in engineering. Which of the following is not a metal?A.ironB.aluminumC.glassD.copper答案：C。

本题考查材料科学中金属的概念。

铁（iron）、铝（aluminum）和铜（（copper）都是常见的金属。

而玻璃（（glass）是一种非金属材料，主要由硅酸盐等组成。

2.Which material is often used for insulation?A.steelB.plasticC.goldD.silver答案：B。

在材料科学中，塑料（（plastic）常被用作绝缘材料。

钢（steel）、金（（gold）和银（（silver）都是导电性能较好的金属材料，不适合用于绝缘。

3.In the field of materials science, which one is a composite material?A.woodB.paperC.concreteD.water答案：C。

混凝土（（concrete）是一种复合材料，由水泥、骨料和水等组成。

木头（（wood）是天然材料。

纸（（paper）主要由纤维素等组成，不是复合材料。

水（water）是一种化合物，不是复合材料。

4.Which material is known for its high hardness?A.rubberB.leatherC.diamondD.cloth答案：C。

在材料科学中，钻石（（diamond）以其高硬度而闻名。

橡胶（rubber）、皮革（leather）和布（cloth）的硬度都较低。

5.Which of the following materials is ductile?A.brickB.glassC.copperD.stone答案：C。

在材料科学中，铜（（copper）是一种具有延展性（（ductile）的金属材料。

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;。

高三英语材料科学单选题60题

高三英语材料科学单选题60题1. In the field of materials science, the "atom" is considered as the basic ______.A. unitB. elementC. particleD. molecule答案：A。

本题考查名词词义辨析。

选项A“unit”有“单位；单元”的意思，“atom”（原子）被视为材料科学中的基本“单位”，符合语境。

选项B“element”指化学元素；选项C“particle”侧重于粒子；选项D“molecule”指分子。

在材料科学中，原子通常被描述为基本单位，其他选项不符合。

2. The study of materials science often involves the analysis of various ______.A. substancesB. compoundsC. mixturesD. materials答案：D。

此题考查名词的含义。

选项A“substances”指物质；选项B“compounds”指化合物；选项C“mixtures”指混合物；选项D“materials”指材料。

材料科学的研究对象主要是各种“材料”，其他选项虽然也与物质相关，但不如“materials”直接和准确。

3. When testing the properties of a new material, the ______ of heat transfer is an important factor.A. processB. methodC. wayD. means答案：A。

本题考查名词的用法。

选项A“process”强调过程；选项B“method”侧重方法；选项C“way”比较常用，指方式、道路；选项D“means”意为手段、工具。

在测试新材料性能时，热传递的“过程”是重要因素，A 选项最符合题意。

20172材料科学基础英文版(4)

Equilibrium boundary concentrations at different temperatures
2019/4/3
Segregation energies at different temperatures
Results and discussion (5)
G G 3 8 k J / m o l P P
Applied stress
2.25Cr1Mo steel
2019/4/3
Outline (1)
• Equilibrium grain boundary segregation
– Segregation of phosphorus and molybdenum
• Introduction, experiment, results and discussion, summary
G G 1 7 k J / m o l M o M o
PMo 0
Boundary concentration
Hardness
SEM
DBTT Fraction of Intergranular fracture
Machining
Metallographic analysis
Grain zize
Experimental procedure
2019/4/3
Experiment (3)
– Relationship between phosphorus boundary concentration and ductile-to-brittle transition temperature (DBTT)
• Introduction, experiment, results and discussion, summary

工学]材料科学基础英文版

• Atomic level: Arrangement of atoms in materials. • Nanostructure: the structure of material at a length-scale
of 100nm. • Microstructure: the structure of material at a length-scale
• Short-range atomic arrangements: the atoms of ions show a particular order only over relatively short distances.
7
2. Atomic Structure
• Atomic structure influences how atoms are bonded together.
13
• The complete set of quantum numbers for each of the 11 electrons in sodium.
3s1 electron 11 n=3, l=0, ml=0, ms=+1/2 or -1/2
{electron 10 n=2, l=1, ml=+1, ms= -1/2 electron 9 n=2, l=1, ml=+1, ms=+1/2
• Orbital (Azimuthal) quantum number l: describe the energy levels in each quantum shell. l=0,1,2,…,n-1. Magnetic quantum number ml: describes the number of energy levels for each orbital quantum number. –l, … +l

材料科学基础英文版

材料科学基础英文版Material Science Fundamentals。

Material science is an interdisciplinary field that explores the properties of materials and their applications in various industries. It combines elements of physics, chemistry, engineering, and biology to understand the behavior of materials at the atomic and molecular levels. This English version of the material science fundamentals aims to provide a comprehensive overview of the key concepts and principles in this field.1. Introduction to Material Science。

Material science is concerned with the study of materials and their properties. It encompasses the discovery, design, and development of new materials, as well as the investigation of existing materials for specific applications. The field is essential for the advancement of technology and innovation in various industries, including aerospace, automotive, electronics, and healthcare.2. Atomic Structure and Bonding。

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。

Sample exams from 2011 Exam I

这里给大家附上之前的Exam I之考试题，供大家学习参考！材料科学基础IV考试（一）试卷（A）┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆┆装订线2010 ~ 2011学年第二学期2011年3月1日8:00~9:35 AM学号：姓名：说明：1、本试卷共四部分题，总分是100。

2、本试卷共10页。

注：请同学们将答案写在装订线的右边（包括答题纸），谢谢！By signing below, I hereby promise that I have neither given nor received unauthorized aid on this exam. What is written on the exam paper has truely reflected my own effort to the best of my knowledge.Your name: _______________________ (Please print).Your signature: ____________________Date: ____________________________I. Please indicate if each of the following statement is true (T) or false (F) in the parenthesis provided after the statement. (10 points)( ) 1.The number of neutrons in an atom defines isotope number.( ) 2.In Bohr’s atomic model, an electron position is considered to be the probability of an electron’s being at various locations around the nucleus.( ) 3.Ionic, covalent, and metallic bonds are all non-directional bonds.( ) 4.An atom can covalently bond with at most 18-N’ other atoms, where N’ = number of valence electrons( ) 5.In order to form ionic bond, elements with similar electronegativity are required.( ) 6.Ionic bonding is directional, that is, the magnitude of the bond is equal in all directions around an ion.( ) 7.The energies are quantized, that is, continuous values of energy are allowed.( ) 8.In the Periodic Table, elements in Group 0 (the inert gas) have filled electron shells, and Group IA elements (the alkali metals) have one electron less than a filled electron shell. ( ) 9.From a plot of interatomic separation versus force for two atoms/ions, the equilibrium separation corresponds to the value at zero force.( ) 10.W ith metallic bonding, the valence electrons form a “sea of electrons” that is uniformly dispersed around the metal ion cores and acts as a form of glue for them.II. Fill in the blanks in the following statement. (10 points)1. A measure of how willing atoms are to accept electrons is termed as __________.322.The maximum number of electrons per shell is ______, given the principal quantum number n.3.Give the electron configuration of Cu+: ______________________________, given that theatomic number of Cu is 29.4.The electrons that occupy the outermost filled shell are called __________ electrons.5.The energy unit used to describe the energy lost / gained by an electron when it is takenthrough a potential difference of one volt is called ______________.6.Permanent dipole moments exist in some molecules by virtue of an asymmetrical arrangementof positively and negatively charged regions. Such molecules are termed _________ molecules.7.The two atomic models are Bohr and __________________ model.8.According to the Pauli exclusion principles, each electron state can accommodate no more than_________ electrons, which must have _________ spins.9.Relatively weak van der Waals bonds result from attractive forces between electric dipoles,which may be ____________ or permanent.III. Choose the best answer for the following questions or complete the sentences from A, B, C, or D given below. (20 points)1.Li, Na, K, etc, are examples of ___________ metals.(a) alkali (b) inert (c) halogen (d) rare earth2.An atom is composed of a nucleus and _______________ .(a) protons (b) neutrons (c) electrons (d) photons (e) phonons3.MgO is an example of _______________ bonding.(a) metallic (b) ionic (c) van der Walls (d) covalent4.The theory “Energy of electrons are quantized; that is, electrons are permitted to have onlyspecific value of energy.” were proposed by ____________.(a) Shrödinger (b) Bohr (c) L.de Broiglie (d) Pauli (e) Einstein5.Elements to the right of the Periodic Table have ________ electronegativity compared to thoseto the left of the Periodic Table.(a) larger (b) smaller (c) the same (d) comparable6.Subshells with one electron have ________ electronegativity than subshells with one missing(b) smaller (c) the same (d) comparable7.The predominant type of bonding for titanium (Ti) is __________.(a) metallic (b) ionic (c) van der Walls (d) covalent8.Electron volt is the energy lost / gained by an electron when it is taken through a potentialdifference of one volt, where 1 eV = ________ J.(a) 1.6 10-19(b) 1.6 1019(c) 1.9 10-16(d) 1.9 10169.When atoms are brought close to each other, according to the Pauli Exclusion Principle, whenthe electronic clouds surrounding the atoms starts to overlap, the energy of the system __________.(a) increases (b) decreases (c) does not change (d) reaches a plateau10.Which of the following bonding is not a type of Primary Bonding? ________.(a) Metallic (b) Ionic (c) Van der Walls (d) Covalent11.CH4 is an example of _______________ bonding.(a) metallic (b) ionic (c) van der Walls (d) covalent12.Electrons are _____________ in stead of ____________ in covalent bonding.(a) transferred…shared (b) shared…transferred(c) diploed…lost (d) lost…diploed13.1/12 of the atomic mass of the most common isotope of carbon atom that has 6 protons (Z=6)and six neutrons (N=6) is defined as ____________.(a) 1 gram (b) 1 amu (c) 1 mole (d) Avogadro’s Number14.What type(s) of electron subshell(s) does an L shell contain?(a) d (b) p(c) f(d) s(e) s and f(f) s and p(g) All of the above.15.Permanent dipole moments exist in some molecules by virtue of an asymmetrical arrangementof positively and negatively charged regions; such molecules are termed ________________.(a) dipoles (b) polar molecules (c) van der Walls bond(d) hydrogen bond (e) bond energy16.Among the following bonds, ____________ is the weakest.(a) fluctuating induced dipole bond (b)permanent dipole bonds(c) polar molecule-induced dipole bond (d) metallic bonds17.Materials with larger bond energy usually have a ________ melting temperature than materialswith lower bond energy.(b) smaller (c) the same (d) comparable18.The coefficient of thermal expansion increases with the _________of bond energy.(a) increase (b) decrease (c) change (d) fluctuating19.The two types of bonds in polymers are .(a) covalent, secondary (b) ionic, secondary (c) metallic, secondary(d) covalent, ionic (e) ionic, metallic20.The number of electrons in the M shell is _______________.(a) 2 (b) 32 (c) 8 (d) 18IV. Please solve the following problems with detailed steps. (60 points)1.Calculate the force of attraction between a K+ and an O2- ion the centers of which are separatedby a distance of 1.6 nm. (10 points)2.Silicon has three naturally-occurring isotopes: 92.23% of 28Si, with an atomic weight of 27.9769amu, 4.68% of 29Si, with an atomic weight of 28.9765 amu, 3.09% of 30Si, with an atomicweight of 29.9738 amu. On the basis of these data, confirm that the average atomic weight of Si is 28.0854 amu. (10 points)3. The net potential energy between two adjacent ions, N E , may be represented by the sum of theattractive energy-interatomic separation relationship and the repulsive energy-interatomic separation relationship, that is,nN r Br A ECalculate the bonding energy 0E in terms of the parameters A, B, and n. (15 points)4. Using Table 2.2 enclosed on the last page, determine the number of covalent bonds that arepossible for atoms of the following elements: germanium, phosphorus, selenium, and chlorine. (12 points)5.What type(s) of bonding would be expected for each of the following materials: brass (acopper-zinc alloy), rubber, barium sulfide (BaS), solid xenon (inert gas), bronze, nylon, and aluminum phosphide (AlP)? (13 points)Equation Sheet for Exam I, Fundamentals of Materials ScienceFall Semester, 20111. Relationship between atomic bonding energy and forceFdr E2. Attractive energy-interatomic separation relationshiprA E AThe constant A in the equation is equal to))((41210e Z e Z , where 0 is the permittivity of avacuum (121085.8 F/m), 1Z and 2Z are the valences of the two ion types, and e is the electronic charge (1910602.1 C).3. Repulsive energy-interatomic separation relationshipnB r B EThe value of n is approximately 8, and the value of constant B is fit experimentally. 4. Figures showing bonding energy and forces as a function of interatomic separation.5.Table showing electron configurations for some common elements.11。

高三英语材料科学单选题60题答案解析版

高三英语材料科学单选题60题答案解析版1.Metal is a kind of material that is usually hard and can conduct electricity. Which of the following is NOT a metal?A.IronB.CopperC.GlassD.Aluminum答案：C。

玻璃不是金属。

铁、铜、铝都是常见的金属。

本题考查材料科学中金属的概念。

2.In material science, plastic is a kind of material that is usually lightweight and can be easily molded. Which of the following is a characteristic of plastic?A.High melting pointB.Good electrical conductivityC.Easily deformedD.High hardness答案：C。

塑料通常容易变形。

塑料的熔点一般不高，导电性不好，硬度也不高。

本题考查塑料的特性。

3.Materials can be classified into different types. Which of the following is a classification of materials based on their physical properties?A.Metals and non-metalsB.Solids, liquids and gasesanic and inorganic materialsD.Conductors and insulators答案：B。

根据物理性质，材料可以分为固体、液体和气体。

A 是根据元素性质分类，C 是根据化学组成分类，D 是根据导电性分类。

本题考查材料的分类方法。

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 )。

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。

高三英语材料科学单选题60题

高三英语材料科学单选题60题1. The new material is highly _____, which makes it suitable for various applications.A. durableB. fragileC. flexibleD. rigid答案：A。

本题考查形容词的词义。

A 选项“durable”表示“耐用的”，新材料高度耐用，适合多种应用，符合语境。

B 选项“fragile”意为“易碎的”，与语境不符。

C 选项“flexible”指“灵活的”，在此处不合适。

D 选项“rigid”表示“僵硬的”，也不符合材料的特点。

2. The ______ of this alloy is its excellent heat resistance.A. featureB. propertyC. characteristicD. quality答案：B。

此题考查名词的辨析。

A 选项“feature”侧重指事物的“特征、特点”。

B 选项“property”常指物质的“属性、性能”，“heat resistance”是合金的一种性能，所以 B 选项更合适。

C 选项“characteristic”强调区别于其他事物的“特征、特性”。

D 选项“quality”指“质量、品质”，范围较宽泛。

3. Materials science focuses on the study of the ______ and behaviorof materials.A. structureB. formC. shapeD. pattern答案：A。

本题考查名词的含义。

A 选项“structure”意为“结构”，材料科学关注材料的结构和行为，这是最恰当的表述。

B 选项“form”侧重于“形式、形态”。

C 选项“shape”强调物体的“形状”。

D 选项“pattern”指“模式、图案”。

4. The ______ materials are essential for the development of advanced technologies.A. advancedB. primaryC. rawD. synthetic答案：C。

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

九年级化学材料科学英语阅读理解25题

九年级化学材料科学英语阅读理解25题1. In the passage, which of the following is a characteristic of metal materials?A. They are always softB. They are good conductors of electricityC. They cannot be meltedD. They are all magneticAnswer: B. Metals are generally known for their good conductivity of electricity. Option A is incorrect as many metals are hard. Option C is wrong because metals can be melted. And not all metals are magnetic, so option D is also incorrect.2. What kind of material is often used for making windows because it is transparent?A. WoodB. MetalC. GlassD. PlasticAnswer: C. Glass is a material that is often used for making windows due to its transparency. Wood is not transparent. Metals are not typically used for their transparency in window making. And while some plastics can be transparent, glass is the most common material for windows.3. Which material is usually lightweight and can be easily molded intodifferent shapes?A. StoneB. CeramicC. PlasticD. ConcreteAnswer: C. Plastic is usually lightweight and has the property of being easily molded into various shapes. Stone is heavy. Ceramics are brittle and not as easily molded. Concrete is heavy and not as pliable as plastic.4. The passage mentions that a certain material is very strong and resistant to heat. Which of the following could it be?A. CottonB. CeramicC. PaperD. RubberAnswer: B. Ceramics are known for their strength and heat resistance. Cotton is not heat - resistant in the same way. Paper is flammable. Rubber is not as strong and heat - resistant as ceramics.5. Which of the following materials is a natural polymer?A. NylonB. SilkC. PolyesterD. AcrylicAnswer: B. Silk is a natural polymer. Nylon, polyester and acrylic are all synthetic polymers.6. What property makes a material suitable for making electrical wires?A. High densityB. Low conductivityC. High conductivityD. High opacityAnswer: C. A material with high conductivity is suitable for making electrical wires as it allows the flow of electricity. High density is not relevant to wire making in this regard. Low conductivity would not allow the electricity to pass through. And opacity has no relation to wire making.7. Which of the following materials is brittle?A. MetalB. GlassC. RubberD. FabricAnswer: B. Glass is brittle, meaning it breaks easily when force is applied. Metals are generally malleable. Rubber is elastic. Fabric is flexible and not brittle.8. In the context of chemical materials, which one is often used as an insulator?A. CopperB. AluminumC. RubberD. IronAnswer: C. Rubber is often used as an insulator. Copper, aluminum and iron are metals which are good conductors, not insulators.9. Which material is made from clay and is fired at high temperatures?A. PlasticB. CeramicC. GlassD. WoodAnswer: B. Ceramics are made from clay and are fired at high temperatures. Plastic is made from synthetic polymers. Glass is made from silica etc. Wood is a natural material not made from clay.10. What is a common property of all polymers?A. They are all biodegradableB. They are made up of repeating unitsC. They are all conductorsD. They are all transparentAnswer: B. Polymers are made up of repeating units. Not all polymers are biodegradable. Most polymers are insulators not conductors. And not all polymers are transparent.11. Plastics are widely used in our daily life because they are _.A. easy to breakB. lightweight and durableC. difficult to produceD. not recyclable答案：B。