材料科学与工程专业英语第二版 翻译答案 老师整理的

United 1 材料科学与工程材料在我们的文化中比我们认识到的还要根深蒂固。

如交通、房子、衣物，通讯、娱乐和食物的生产，实际上，我们日常生活中的每一部分都或多或少地受到材料的影响。

历史上社会的发展、先进与那些能满足社会需要的材料的生产及操作能力密切相关。

实际上，早期的文明就以材料的发展程度来命名，如石器时代，铜器时代。

早期人们能得到的只有一些很有限的天然材料，如石头、木材、粘土等。

渐渐地，他们通过技术来生产优于自然材料的新材料，这些新材料包括陶器和金属。

进一步地，人们发现材料的性质可以通过加热或加入其他物质来改变。

在这点上，材料的应用完全是一个选择的过程。

也就是说，在一系列非常有限的材料中，根据材料的优点选择一种最适合某种应用的材料。

直到最近，科学家才终于了解材料的结构要素与其特性之间的关系。

这个大约是过去的60 年中获得的认识使得材料的性质研究成为时髦。

因此，成千上万的材料通过其特殊的性质得以发展来满足我们现代及复杂的社会需要。

很多使我们生活舒适的技术的发展与适宜材料的获得密切相关。

一种材料的先进程度通常是一种技术进步的先兆。

比如，没有便宜的钢制品或其他替代品就没有汽车。

在现代，复杂的电子器件取决于所谓的半导体零件.材料科学与工程有时把材料科学与工程细分成材料科学和材料工程学科是有用的。

严格地说，材料科学涉及材料到研究材料的结构和性质的关系。

相反，材料工程是根据材料的结构和性质的关系来设计或操纵材料的结构以求制造出一系列可预定的性质。

从功能方面来说，材料科学家的作用是发展或合成新的材料，而材料工程师是利用已有的材料创造新的产品或体系，和/或发展材料加工新技术。

多数材料专业的本科毕业生被同时训练成材料科学家和材料工程师。

“structure”一词是个模糊的术语值得解释。

简单地说，材料的结构通常与其内在成分的排列有关。

原子内的结构包括介于单个原子间的电子和原子核的相互作用。

在原子水平上，结构包括原子或分子与其他相关的原子或分子的组织。

A 高分子化学和高分子物理UNIT 1 What are Polymer?第一单元什么是高聚物？What are polymers? For one thing, they are complex and giant molecules and are different from low molecular weight compounds like, say, common salt. To contrast the difference, the molecular weight of common salt is only 58.5, while that of a polymer can be as high as several hundred thousand, even more than thousand thousands. These big molecules or ‘macro-molecules’ are made up of much smaller molecules, can be of one or more chemical compounds. To illustrate, imagine that a set of rings has the same size and is made of the same material. When these things are interlinked, the chain formed can be considered as representing a polymer from molecules of the same compound. Alternatively, individual rings could be of different sizes and materials, and interlinked to represent a polymer from molecules of different compounds.什么是高聚物？首先，他们是合成物和大分子，而且不同于低分子化合物，譬如说普通的盐。

Unit 2 Classification of MaterialsSolid materials have been conveniently grouped into three basic classifications: metals, ceramics, and polymers. This scheme is based primarily on chemical makeup and atomic structure, and most materials fall into one distinct grouping or another, although there are some intermediates. In addition, there are three other groups of important engineering materials —composites, semiconductors, and biomaterials.译文：译文：固体材料被便利的分为三个基本的类型：金属，陶瓷和聚合物。

固体材料被便利的分为三个基本的类型：金属，陶瓷和聚合物。

固体材料被便利的分为三个基本的类型：金属，陶瓷和聚合物。

这个分类是首先基于这个分类是首先基于化学组成和原子结构来分的，化学组成和原子结构来分的，大多数材料落在明显的一个类别里面，大多数材料落在明显的一个类别里面，大多数材料落在明显的一个类别里面，尽管有许多中间品。

尽管有许多中间品。

除此之外，此之外， 有三类其他重要的工程材料－复合材料，半导体材料和生物材料。

有三类其他重要的工程材料－复合材料，半导体材料和生物材料。

Composites consist of combinations of two or more different materials, whereas semiconductors are utilized because of their unusual electrical characteristics; biomaterials are implanted into the human body. A brief explanation of the material types and representative characteristics is offered next.译文：复合材料由两种或者两种以上不同的材料组成，然而半导体由于它们非同寻常的电学性质而得到使用；生物材料被移植进入人类的身体中。

Unit 1 Translation.1."材料科学〞涉与到研究材料的结构与性能的关系.相反,材料工程是根据材料的结构与性质的关系来涉与或操控材料的结构以求制造出一系列可预定的性质.2.实际上,所有固体材料的重要性质可以分为六类：机械、电学、热学、磁学、光学、腐蚀性.3.除了结构与性质,材料科学与工程还有其他两个重要的组成部分,即加工与性能.4.工程师或科学家越熟悉材料的各种性质、结构、性能之间的关系以与材料的加工技术,根据以上的原则,他或她就会越自信与熟练地对材料进行更明智的选择.5.只有在少数情况下,材料才具有最优或最理想的综合性质.因此,有时候有必要为某一性质而牺牲另一性能.6.Interdisciplinary dielectric constantSolid material<s> heat capacityMechanical property electromagnetic radiationMaterial processing elastic modulus7.It was not until relatively recent times that scientists came to understand the relationships between the structural elements of materials and their properties.8. Materials engineering is to solve the problem during the manufacturing and application of materials.9.10.Mechanical properties relate deformation to an applied load or force.Unit 21.金属是电和热很好的导体,在可见光下不透明；擦亮的金属表面有金属光泽.2.陶瓷是典型的导热导电的绝缘体,并且比金属和聚合物具有更高的耐热温度和耐恶劣环境性能.3.用于高科技领域的材料有时也被称为先进材料.4.压电陶瓷在电场作用下膨胀和收缩；反之,当它们膨胀和收缩时,他们也能产生一个电场.5.随着能够观察单个原子或者分子的扫描探针显微镜的出现,操控和移动原子和分子以形成新结构成为可能,因此,我们能通过一些简单的原子水平的构建就可以设计出新的材料.6.advanced materials ceramic materialshigh-performance materials clay mineralsalloy implantglass fibre carbon nanotube7.Metallic materials have large numbers of nonlocalized electrons and many properties of metals are directlyattributable to these electrons.8.Many of polymeric materials are organic compounds with very large molecular structures.9.Semiconductors hace electrical properties that are intermediate between the electrical conductors<viz. metalsand metal alloys> and insulators<viz. ceramics and polymers>.10.Biomaterials must not produce toxic substances and must be compatible with body tissues.Unit 31．金属的行为〔性质〕不同于陶瓷的行为〔性质〕,陶瓷的行为〔性质〕不同于聚合物的行为〔性质〕. 2．原子结构主要影响化学性质、物理性质、热学性质、电学性能、磁性能、光学性能.微结构和宏观结构虽也能影响这些性能但是他们主要影响力学性能和化学反应速率.3．金属的强度表明原子是通过强的键结合在一起的.4．元素的原子序数表明该元素的原子核内带正电的质子数.而原子的原子量则表明该原子核中质子数与中子数.5．Microstructure macrostructureChemical reaction atomic weightBalanced electrical charge positively charged proton6. 100 atoms form thousands of different substances ranging from the air we breathe to the metal used to supporttall buildings,7．The facts suggests that metallic atoms are held together bu strong bonds.8. Microstructure which includes features that cannot be seen with the naked eye,but using a microscope.Macrostructure includes features that can be seen with the naked eye.9. The atomic weight is the sum of proton number and neutron number in the nucleus.Unit 41．当密度小于水的密度时,物体将漂浮在水面上,当密度大于水的密度时,物体会沉降.相似的,当比重小于1时,物体将漂浮在水面上,当比重大于1时,物体会沉降.2．由于相互排斥而往相反的方向移动,导致磁通量密度比真空中小,这种材料为反磁性材料.3．使磁通量密度提高1倍以上小于或等于10倍的材料叫顺磁性材料,使磁通量密度提高10倍以上的材料叫铁磁性材料.4．某些铁磁性材料,特别是一些粉末状或夹层铁,钢或镍合金的相对导磁率可高达1000000.反磁性材料的相对导磁率小于1,但是到目前还没有哪种材料的相对导磁率远小于1.5．当顺磁性或铁磁性的芯插入线圈时,其磁感应系数等于相对磁导率乘以空芯时的磁感应系数. 6．specific gravity boiling point magnetic inductioncoefficient of thermal conductivity glass transition temperaturenon-ferrous metals linear coefficient of thermal expansionmass per unit of volume7. Properties that describe how a substance changes into a completely different substance are called chemicalproperties.8. Phase is a physical property of matter and matter can exist in four phases: solid, liquid, gas and plasma.9. At some temperature below the melting point, polymers start to lose their crystalline structure but the moleculesremain linked in chains, which tesults in a soft and pliable material.10. In engineering applications, permeability is often expressed in relative, rather than in absolute.Unit 51. 金属的力学性能决定了材料的使用X围与期望的服役寿命.2. 因此,一般多测几次以得到力学性能,报导的数值一般是平均值或者计算的统计最小值.3．材料的承载方式极大地影响了材料的力学性能,也决定了材料失效形式,以与在失效前是否有预警. 4．然而,受力弯曲时会产生一个应力分布,应力大小与到轴线的垂直距离有关.5．材料受到低于临界压力即屈服强度的力时,材料才会发生弹性形变.6．Test specimen static loading force normal axisEngineering strain critical stress yield strength stress areaStress- strain curve7. Temperatures below room temperature generally cause an increase in strength properties of metallic alloys;while ductility, fracture toughness, and elongation usually decrease.8. From the respective of what is happening within a material, stress is the internal distribution of forces withina body that balance and react to the loads applied to it.9. Engineering strain is defined as the amount of deformation in the direction of the applied force divided bythe initial length of the material.10. A material with high strength and high ductility will hace more toughness than a material with low strengthand high ductility.Unit 61. 随着影响我们星球上人类生存条件的社会问题的即将出现,材料科学与工程界有责任和机遇通过解决未来世界的需求—在能量、交通、住房、饮食、回收和健康方面的需求来改变世界.2. 不发达国家的人口增长率远高于1.4%的世界平均人口增长率.3. 全球能源使用的预算在2025以前将以每年1.7%速度增长,这比世界人口增长率快多了.4. 此外,发达地区的人均能量使用量是不发达地区人均能量使用量的九倍以上.5. gross domestic product materials science and engineering market economySocietal issues economic index sanitationGross national product popilation growth rate6. Some things that have been constant over time are human innovation and creativity, the engineer’s ability toaddress societal needs, and the entrepreneurial spirit of engineering.7. We have witnessed the re-shaping of our lives through revolutions that hace taken place in medicine,telecommunications, and transportantion industries.8. Eighteen percent of the world’s population lacks access to safe drinking water and nearly 40% has no access tosanitation.9. Materials and society are interlinked, and it is only rational that we should see a close relation between the MSEresearch agenda and societal issues that affect the human condition on the globe.Unit 71.从化学角度来说,金属是一类容易失电子以形成正离子的元素,它与其他金属原子形成金属键.2.金属键的无方向性被认为是金属具有延展性的主要原因.3.存在着共价键的晶体只有在原子之间的键断裂的情况下变形,从而导致晶体破裂.4.合金特别是为满足更高应用要求的合金比如喷气发动机,可能含有十种以上的元素.5.delocalized electron electronic structurealkaline earth metal chemical cellnuclear charge electric conductivity6.Metals are sometimes described as a lattice of positive ions surrounded by a cloud of delocalized electrons.7.Metals in general have superior electric and thermal conductivity, high luster and density, and the ability to bedeformed under stress without cleaving.8.An alloy is a mixture of two or more elements in solid solution in which the major component is a metal.bining different ration of metals as alloys modifies the properties of pure metals to produce desirablecharacteristics.Unit 81.超耐热合金的发展非常依赖于化学与加工的创新,并主要受到航空和能源工业的推动.2.抗蠕变性能主要取决于晶体结构内位错速度的放缓.3.超耐热合金在加工方面的发展使超耐热合金的操作温度大幅度提高.4.单晶高温合金是运用改进的定向凝固技术而形成单晶的,因此在材料中并无晶界.5.faced-centred cubic crystal structure turbine entry temperaturemetallic materials phase stabilitynuclear reactor synthesis of nanoparticle6.Superalloys typically hace an sustenitic faced-centered cubic crystal structure.7.Superalloys are used where there is a need for high temperature strength and corrosion/oxidation resistance.8.Superalloys are widely used in aircraft ,submarines, nuclear reactors and military electric motors.9.At high temperatures the gaseous aluminum chloride<or fluoride> is transfereed to the surface of part anddiffuseds inside.Unit 91.腐蚀过程从本质上说是一个电化学的过程,有着与电池相同的本质特征.2. 从矿物中提炼金属所需能源的问题与后续的腐蚀和能量释放直接相关.3. 当电子与中和的正离子〔如电解液中的氢离子〕发生反应时,阴极处电子得以平衡.4．Protective film circuitFree electron electron transferMetal cation anode reaction5.Some metals, such as gold and silver, can be found in the earth in their natural, metallic state, and they havelittle tendency to corrode.6.Oxidation is the process of stripping electrons from an atom and reduction occurs when an electron is addedto an atom.7.If the surface becomes wet, corrosion may take place through ionic exchange in the surface water layerbetween the anode and cathode.8.Corrosion is commonly classification based on the appearance of the corroded material.Unit 101. 我们要观察〔研究〕这些性能,看它们与我们所期望的陶瓷的组成有多匹配.2. 在高于玻璃化转变温度的高温下,玻璃不再具有脆性行为,而表现为粘稠液体.3. 它们显示出优异的力学性能、抗腐蚀/氧化性能,或电学、光学抑或是磁学性能.4. 一般认为,先进是最近100年才发展起来的,而传统的基于粘土的陶瓷早已在25000多年前就被使用了.5. the glass transition temperature ionic-covalent bondStress distribution coefficient of thermal expansionGlass optical fibre materials science and engineeringSolid-oxide fuel cells electron microscopy6. Diamond, which is classified as a ceramic, has the highest thermal conductivity of any known material.7. Ceramic are stronger in compression than in tension, whereas metals have comparable tensile and compressivestrength.8. Ceramics generally have low toughness, although combining them in composites can dramatically improvethis property.9. The functions of ceramic products are dependent on their chemical composition and microstructure, whichdetermines their properties.Unit 111. 材料科学与工程领域经常是根据四大方面—合成与加工,结构与组成,性质与性能之间的相互联系来定义的.2．我们不仅要考虑具有完美晶格和理想结构〔的情况〕,我们也要考虑材料中不可避免的结构缺陷的存在,甚至是无定形的.3. 通过热压可使孔径减小从而得到高密度产品.4. 在运输时,厂方要提供关于产品危害方面的信息.5. crystalline ceramics grain boundaryAlkaline earth oxide oxide additiveTriple point saturation magnetizationTelevision tube the color scale6.To understand the behavior and properties of any material, it is essential to understand its structure.7.The grain size is determined by the size of the initial powder particles and the way in which they wereconsolidated.8.Transparent or translucent ceramics require that we limit the scattering of light by pores and second-phaseparticles.9.Alumina ceramics are used as electrical insulators because of theie high electrical resistivity and lowdielectric constant.Unit 121. 材料的选择是任何组分设计至关重要的环节,尤其在植入体和其它医疗器械方面是特别重要的.2. 我们能进行承载应用的三种主要材料是金属、聚合物和陶瓷.3. 高密度、高纯度的氧化铝被大量的用于植入物,特别是在需要承载压力的髋关节修复和牙移植中.4. 在陶瓷或陶瓷复合材料中,氧化锆的磨损率远远高于氧化锆铝的磨损率.5. controlled reaction stress shieldingTotal hip prosthese strain-to-failure ratioMechanical stress flexural strengthMartensitic transformation6.Biomaterial is a non-viable material used in a medical device intended to interact with biological systems.7.These repairs become necessary when the existing part becomes diseased, damaged, or just simply wears out.8.Because of its low density, cancellous bone has a lower E and higher strain-to-failure ratio than cortical bone.9.Eliminating stress shielding, by reducing E, is one of the primary motivations for the development ofbioceramic composites.10.There are questions concerning the long-term effect of radiation emission from zirconia ceramics.Unit 131. 聚合物的俗名叫塑料,这个词指的是一大类具有许多性质和用途的天然材料和合成材料.2. 聚合物合成是一个把叫做单体的小分子通过共价键的结合形成链的过程.3. 支化聚合物分子是由一条带有一个或多个侧基或支链的主链组成.一些特殊的支化聚合物有星型聚合物、梳状聚合物和刷状聚合物.4. 某些生物聚合物是由一系列不同的胆识结构却相关的单体组成的,例如聚核苷酸是由核苷组成的.5. persistence length cross-linkPolar monomer nucleic acidPolymerization polyelectrolyte6.Most commercially important polymers today are entirely synthetic and produced in high volume, onappropriately scaled organic synthetic techniques.7.Some biological polymers are composed of a variety of different but structurally related monomers, suchas polynucleotides composed of nucleotide sbunits.8. A polymer molecule with a high degree of crosslinking is referred to as a polymer network.9.In polymers, however, the molecular mass may be expressed in terms of degree of polymerization,essentially the number of monomer Units which comprise the polymer.Unit 141. 大量合成的聚合物具有碳-碳骨架,这是因为碳原子具有与其它原子形成更强更稳定的键的优异性能.2. 它们在一定X围内软化,这与完好晶体相具有非常明确的熔点不同.3. 分子量取决于其合成时的条件,因此分子量可能分布很宽或分布很窄.4. Goodyear 很偶然的发现了在橡胶中加入硫磺并加热这个混合物能使橡胶变硬,对氧化和化学进攻能力的敏感性降低.5. thermosetting plastic cross-sectional areaPolymerization reaction double bondChemical composition carboxylic acidMelting point degradation by oxidation6.Polymer with different chemical composition has different physical and chemical property.7. A thermosetting plastic is shaped through irreversible chemical processes and therefore cannot be reshapedreadily.8.Natural rubber is not a useful polymer because it is too soft and too chemically reactive.9.Various substances may be added to polymers to provide protection against the effects of sunlight or againstdegradation by oxidation.Unit 151. 逐渐增强的环境意识促使包装薄膜与其加工既要方便使用又要具有环境友好的特点.2. 显而易见,实现这些性能对控制和改进机械性能和阻隔性能是非常重要的.3. 在羧酸、醇、醛、酮的含氧生物降解过程中,由水和热引发的过氧化反应可以使之降解成低摩尔质量的物质,这就是碳氢聚合物力学性能降低的主要原因.4. 用持久耐用的聚合物做短期使用的包装材料并不合理,另外也是因为包装材料被食物污染后再进行物理回收是不切合实际的.5. natural gas packaging materialsAroma compound bioplastic materialChemical structure the life cycle of biomass6. Bacteria , fungi, enzymes start the bioassimilation giving rise to biomass and CO2 that finally form the humus.7. The bioplastic aim is to imitate the life cycle of biomass, which includes conservation of fossil resources, waterand CO2 production.8. During the oxo-degradation of carboxylic acid, molecules of alcohols, aldehydes and ketones degradable withlow molar mass are produced by peroxidation initiated by heat or light .9. While most of the commercialized biopolymer materials are biodegradable, these are not fully compostable inreal composting conditions, which vary with temperature and relative humidity.Unit 161. 比如,多相金属在微观尺度上是复合材料.但一般意义上的复合材料是指通过键的作用使两种或多种不同的材料结合在一起的材料..2. 最常见的是,复合材料有一个连续的叫基体的本体相,还有一个分散的非连续的叫增强相的相.3. 先进材料采用了树脂与纤维的复合材料,一般为碳/石墨,凯芙拉或玻璃纤维与环氧树脂的复合材料.纤维具有高的硬度,而聚合物树脂基体能保持复合材料的结构.4. 如果复合材料设计和制备合理的话,复合材料就既具有增强相的强度又具有基体的韧性从而得到了性能的理想组合,这是任何一种组分单独存在时所具备的性能.5. composite material reinforcement materialFiberglass matrix materialStrengthening mechanism conventional material6. A composite is commonly defined as a combination of two or more distinct materials, each of which retainsits own distinctive properties, to create a new material with properties that cannot be achieved by any of the components acting alone.7.Carbon-epoxy composties are two thirds the weight of aluminum, and two and a half times as stiff.Composites are resistant to fatigue damage and harsh enviroments, and are repairable.8.According to the conception of composite , reinforced plastics, metal-matrix composites, ceramic-matrixcomposites and concrete are composites.9.In fiber-reinforced composites, the fiber is the primary load-bearing component. Fiberglass and carbon fibercomposites are examples of fiber-reinforced composites.Unit 171. 震荡、撞击或者重复的周期性应力能导致两层的界面处发生薄层分离,这种情况叫剥离.2.3. 事实上,工业材料既要质轻又要牢固的要求是推动复合材料发展的主要动力.4. 提到飞机,值得铭记的是复合材料不像金属〔如铝〕那样在压力的作用下会完全解体.5. orthotropic thermosetThermoplastic Young’s ModulusMechanical property constants extreme enviroment6.In contrast, isotropic material < for example, aluminuim or steel>, in standard wrought forms, typicallyhave the same stiffness regardless of the directional orientation of the applied forces and /or moments.7.The greatest advantage of composite materials is strength and stiffness combined with lightness.8.This makes them ideal for use in products thar are exposed to extreme enviroments such as boats,chemical-handling equipments and spacecrafts.posites will never totally replace tranditional materials like steel, but in many cased they are just whatwe need.Unit 181. 具有相分离的聚合物共混材料经常出现纳米尺度的相.2 在过去几十年里研究的基于溶胶-凝胶化学的有机-无机纳米复合材料已基本淡出纳米复合材料的研究.3. 理解粒子的性质随着尺寸降低到纳米级别而发生改变,这对于优化所得到的纳米复合材料很重要.4. 廉价石墨的生产尚未实现,石墨的广泛使用呕待石墨合成技术的突破.5. electro-optical property bactericidal propertyBlock copolymer interfacial phenomenaExfoliated graphene morphology control6.The field of nanotechnology is one of the most popular areas for current research and development inbasically all technical discillines.7.Nanoscale is considered where the dimensions of the particle, platelet or fiber modification are in the rangeof 1～100nm.8.These improvements are key to future aircraft and wind energy turbine applications.9.Nanostructured surfaces have been noted to yield superhydrophobic character and exceptional adhesion.。

United 1 材料科学与工程材料在我们的文化中比我们认识到的还要根深蒂固。

如交通、房子、衣物,通讯、娱乐和食物的生产,实际上,我们日常生活中的每一部分都或多或少地受到材料的影响。

历史上社会的发展、先进与那些能满足社会需要的材料的生产及操作能力密切相关。

实际上,早期的文明就以材料的发展程度来命名,如石器时代,铜器时代。

早期人们能得到的只有一些很有限的天然材料,如石头、木材、粘土等。

渐渐地,他们通过技术来生产优于自然材料的新材料,这些新材料包括陶器和金属。

进一步地,人们发现材料的性质可以通过加热或加入其他物质来改变。

在这点上,材料的应用完全是一个选择的过程。

也就是说,在一系列非常有限的材料中,根据材料的优点选择一种最适合某种应用的材料。

直到最近,科学家才终于了解材料的结构要素与其特性之间的关系。

这个大约是过去的60 年中获得的认识使得材料的性质研究成为时髦。

因此,成千上万的材料通过其特殊的性质得以发展来满足我们现代及复杂的社会需要。

很多使我们生活舒适的技术的发展与适宜材料的获得密切相关。

一种材料的先进程度通常是一种技术进步的先兆。

比如,没有便宜的钢制品或其他替代品就没有汽车。

在现代,复杂的电子器件取决于所谓的半导体零件.材料科学与工程有时把材料科学与工程细分成材料科学和材料工程学科是有用的。

严格地说,材料科学涉及材料到研究材料的结构和性质的关系。

相反,材料工程是根据材料的结构和性质的关系来设计或操纵材料的结构以求制造出一系列可预定的性质。

从功能方面来说,材料科学家的作用是发展或合成新的材料,而材料工程师是利用已有的材料创造新的产品或体系,和/或发展材料加工新技术。

多数材料专业的本科毕业生被同时训练成材料科学家和材料工程师。

“structure”一词是个模糊的术语值得解释。

简单地说,材料的结构通常与其内在成分的排列有关。

原子内的结构包括介于单个原子间的电子和原子核的相互作用。

在原子水平上,结构包括原子或分子与其他相关的原子或分子的组织。

材料科学与工程_专业英语_Uni...Unit 3 Structure-Property Relationships of MaterialsToday’s materials can be classified as metals and alloys, as polymers or plastics, as ceramics, or as composites; composites, most of which are man-made, actually are combinations of different materials.译文：当今的材料可以分为金属和合金，聚合物或者塑料，陶瓷或复合材料；复合材料，它们大多数是人造的，实际上是不同材料组合而成。

A pplica tion of these m ateria ls de pe nd on their pr ope rties; theref ore, w e ne ed to know w hat pr operties are re quired by the a pplica tion and to be a ble to re late those s pecifica tion to the m aterial.译文：这些材料的应用取决于它们的性质；因此，根据应用的场合，我们需要知道什么样的性质是必需的，我们需要能够把这些详细说明同材料联系起来。

For exam ple, a la dder m ust w ithsta nd a des ign loa d, the w eight of a pe rs on us ing the la dde r. H ow ever, the m ateria l property that ca n be m easured is s tre ngth, w hich is af f ecte d by the loa d a nd desig n dim ension. S tre ngth values m us t theref ore be applie d to dete rm ine d the la dde r dim ensions to e ns ure saf e us e. Therefore, in ge ne ral, the s truc tures of m etallic m aterials have ef fects on the ir prope rties.译文：比如，一个梯子必须能经受住设计的载荷，也就是使用这个梯子的人的重量。

ferrous alloys铁合金More than 90% by weight of the metallic materials used by human beings are ferrous alloy. This represents an immense family of engineering materials with a wide range of microstructures and related properties. The majority of engineering designs that require structural load support or power transmission involve ferrous alloys. As a practical matter, those alloys fall into two broad categories based on the carbon in the alloy composition. Steel generally contains between wc=0.05% and wc=4.5%.超过90%的重量的金属材料使用的人类是铁合金。

这是一个巨大的工程材料的家庭与广泛的微观结构和相关的属性。

大部分的工程设计,需要结构性的负载支持或电力传输涉及铁合金。

作为一个实际问题,这些合金分为两大类基于碳在合金成分。

钢一般包含在wc = 0.05%和wc = 4.5%。

Within the steel category,we shall other than carbon is used.A compositon of 5% total noncarbon high alloy steels. Those alloy additions are chosen carefully becouse they invariably bring with them sharply increased material costs. They are justified only by essential improvements in improvements such as higher strength or improved corrosion resistance在钢的类别,我们将使用碳。

Unit1：2.xx材料科学石器时代肉眼青铜器时代光学性质集成电路机械(力学)强度热导率1.材料科学指的是研究存于材料的结构和性能的相互关系。

相反，材料工程指的是，在基于材料结构和性能的相互关系的基础上，开发和设计预先设定好具备若干性能的材料。

2.实际上，固体材料的所有重要性质可以概括分为六类：机械、电学、热学、磁学、光学和腐蚀降解性。

3.除了结构和性质，材料科学和工程还有其他两个重要的组成部分：即加工和性能。

4.工程师与科学家越熟悉材料的结构-性质之间的各种相互关系以及材料的加工技术，根据这些原则，他或她对材料的明智选择将越来越熟练和精确。

5.只有在极少数情况下材料在具有最优或理想的综合性质。

因此，有必要对材料的性质进行平衡。

3.xxInterdispline dielectric constantSolid materials heat capacityMechanical properties electro-magnetic radiationMaterials processing elasticity modulus1.直到最近，科学家才终于了解材料的结构要素与其特性之间的关系。

It was not until relatively recent times that scientists came to understand the relationship between the structural elements of materials and their properties .2.材料工程学主要解决材料的制造问题和材料的应用问题。

Material engineering mainly solve the problems of materials processing and materials application.3.材料的加工过程不但决定了材料的结构，同时决定了材料的特征和性能。

材料科学与工程专业英语匡少平课后翻译答案精编W O R D版IBM system office room 【A0816H-A0912AAAHH-GX8Q8-GNTHHJ8】Alloy合金applied force作用力amorphous materials不定形材料artificial materials人工材料biomaterials生物材料biological synthesis生物合成biocompatibility生物相容性brittle failure脆性破坏carbon nanotub e碳纳米管carboxylic acid羟酸critical stress临近应力dielectric constant介电常数clay minera l粘土矿物cross-sectional area横截面积critical shear stress临界剪切应力critical length临界长度curing agent固化剂dynamic or cyclic loading动态循环负载linear coefficient of themal expansio n性膨胀系数electromagnetic radiation电磁辐射electrodeposition电极沉积nonlocalizedelectrons游离电子electron beam lithography电子束光刻elasticity 弹性系数electrostation adsorption静电吸附elastic modulus弹性模量elastic deformation弹性形变elastomer弹性体engineering strain工程应变crystallization 结晶fiber-optic光纤维Ethylene oxide环氧乙烷fabrication process制造过程glass fiber玻璃纤维glass transition temperature 玻璃化转变温度heat capacity热熔Hearing aids助听器integrated circuit集成电路Interdisplinary交叉学科intimate contact密切接触inert substance惰性材料implant移植individual application个体应用deformation局部形变mechanical strength机械强度mechanical attrition机械磨损Mechanical properties力学性Materials processing材料加工质mechanical behavior力学行为magnetic permeability磁导率magnetic hybrid technique混合技术induction磁感应mass per unit of volume单位体积质量monomer identity单体种类molecular mass分子量microsphere encapsulation technique微球胶囊技术macroscopical宏观的naked eye 肉眼nonlocalized nanoengineered materials纳米材料nanostructured materials纳米结构材料nonferrous metal有色金属线nucleic acid核酸nanoscale纳米尺度Nanotechnology纳米技术nanobiotechnology纳米生物技术nanocontact printing纳米接触印刷optical property光学性质optoelectronic device光电设备oxidation degradation 氧化降解piezoelectric ceramics压电陶瓷Relative density相对密度stiffnesses刚度sensor传感材料semiconductors半导体specific gravity比重shear 剪切Surface tention表面张力self-organization自组装static loading静载荷stress area应力面积stress-strain curves应力应变曲线sphere radius球半径submicron technique亚微米技术substrate衬底supramolecalar超分子sol-gel method溶胶凝胶法thermal/electrical conductivity 热/点导率thermoplastic materials热塑性材料Thermosetting plastic热固性塑料thermal motion热运动toughness test韧性试验tension张力torsion扭曲Tensile Properties拉伸性能Two-dimentional nanostructure二维纳米结构Tissue engineering组织工程transplantation of organs器官移植the service life使用寿命the longitudinal direction纵向the initial length of the materials初始长度the acceleration gravity重力加速度the normal vertical axis垂直轴the surface to volume ratio 比表面密度the burgers vector伯格丝矢量the mechanics and dynamics of tissues 组织力学和动力学phase transformation temperature相转变温度plastic deformation塑性形变Pottery陶瓷persistence length余晖长度polymer synthesis聚合物合成Polar monomer记性单体polyelectrolyte高分子电解质pinning point钉扎点plasma etching 等离子腐蚀pharmacological acceptability药理接受性pyrolysis高温分解ultrasonic treatment超射波处理yield strength屈服强度vulcanization硫化1-1:直到最近，科学家才终于了解材料的结构要素与其特性之间的关系。

Materials science材料科学Stone age石器时代Naked eye肉眼Bronze age铜器时代Optical property光学性能Integrated circuit集成电路Mechanical strength机械强度Thermal conductivity导热“Materials science” involves investigating the relationships that exist between the structures and properties of materials. In contrast ,”materials engineering “is ,on the basis of there structure property correlations ,designing or engineering the structure of a material that produce a predetermined set of properties。

，材料工程是根据材料的结构和性质的关系来设计或操纵材料的结构以求制造出一系列可预定的性质。

从功能方面来说，材料科学家的作用是发展或合成新的材料V irtually all important properties of solid materials may be grouped into six different categories: mechanical, electrical, thermal, magnetic, optical ,and deteriorative。

固体材料的所有重要的性质可以分成六个不同的种类，机械性能、电性能、热性能、磁性能、光性能和内耗。

In addition to structure and properties , two other important components are involved in the sciences and engineering of materials , namely“processing”and“performance”.除了组织性能之外，另外两个重要的性质也包括在材料科学和工程之中，即“加工”和“特性”The more familiar an engineer or scientist is with the various characteristics and structure-property relationship，as well as processing techniques of material，the more proficient and confident he or she will be to make judicious materials choices based on these criteria。

工程材料科学与设计原书第2版课后习题答案4—8章Solutions to Chapter 41. FIND: What material has a property that is hugely affected by a small impurity level?SOLUTION: Electrical conductivity spans a wide range. Incorporation of a few parts per million impurities can change electrical conductivity orders of magnitude. Small cracks in brittle materials decrease their tensile strength by orders of magnitude. Small additions of impurity can change the color of gems. COMMENTS: These are but a few examples. 2. COMPUTE: The temperature at which the vacancy concentration is one half thatof 25o C.GIVEN: C 2 = C C 25v C 35vo oEQUATION:⎪⎪⎭⎫⎝⎛RT Q - = C fv v expwhere C v = vacancy concentrationQ fv = activation energy for vacancy information R = gas constant 8.314 J/mole-KT = absolute temperatureIn the present problem C)25(C = C C);35(C = C o v 2v o v 1vand T 1 = 35 + 273 = 308KT 2 = 25 + 273 = 298K⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛RT Q + RT Q - = C CRT Q - RT Q -= C CRT Q = CRT Q = C 2fv 1fv 2v 1v2fv 1fv 2v1v 2fv 2v1fv 1vexp exp exp exp expalso C v(35o C) = 2C v(25o C)Thus, Solving for Q fv we get Q fv = 52893.5 J/mole.Using this value of Q fv , the C v (25o C) can be calculatedThe problem requires us to calculate the temperature at which the vacancy concentration is ½ C v (25o C).½ C v (25o C) = 2.675 x 10-10Thusfor solving T, we get: T = 288.63K or 15.63o C.3. COMPUTE:C)80( C 3 = (T) C ov vGIVEN: C) 80( C 41 = C) 25( C o v ov EQUATION:⎪⎭⎫⎝⎛298.R Q - C) 25( C Sv o v expDividing (1) by (2) we get:⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎥⎦⎤⎢⎣⎡91784Q 308 + Q 298- R 1 = 2R(298)Q + R(308)Q - = C)25( C C)25( C 2fv fv fv fv ov o v exp exp10 x 5.35 = C)25(C 298 x 8.3152893.5- = C)25(C 10-o v ov ⎪⎭⎫ ⎝⎛exp ⎪⎭⎫⎝⎛T x 8.3152893.5- = 10 x 2.67510-ex p⎪⎭⎫ ⎝⎛353.R Q - = C) 80( C ov ex pSolving for Q, we get:Q = 22033.56 J/mole= exp(-7.511)= 5.46 x 10-4The problem requires computing a temperature at which C v = 3C v (80o C).3C v (80o C) = 3 x 5.46 x 10-4= 1.63 x 10-3⎪⎭⎫⎝⎛T x 8.3122033.56- = 10 x 1.633-ex psolving for T, we get:T = 413.05K or 140.05o C4.5. FIND: Are Al and Zn completely soluble in solid solution?If Al-Zn system obeys all the Hume-Rothery rules. Then it is expected to show complete solubility.(i) The atomic radii of Al and Zn are 0.143nm and 0.133 nm respectively. Thedifference in their radii is 7.5% which is less than 15%.(ii) The electronegativities of Al and Zn are 1.61 an 1.65 respectively which arealso very similar.(iii) The most common valence of Al is +3 and +2 for Zn.(iv) Al has an FCC structure where Zn has a HCP structure.It appears that Al-Zn system obeys 3 out of 4 Hume-Rothery rules. In this case they are not expected to be completely soluble.⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛3531 - 2981 R Q - = 41= C) 80( C C) 25( C ov ov exp ⎪⎭⎫ ⎝⎛353 x 8.3122033.56- = C) 80(C ov ex p6. SHOW: The extent of solid solution formation in the following systems usingHume-Rothery Rules.(a) Al in NiSize: r(Ni) = 0.125nm; r(Al) = 0.143nm difference = 14.4%Electronegativity: Al = 1.61; Ni = 1.91Most Common Valence: Al3+; Ni2+Crystal Structure: Al: FCC; Ni:FCCThe crystal structure of Al and Ni are the same and the most common valencies are also comparable. However, the size difference is close to 15% and the difference is electronegativities is rather significant.Based on this, it appears that Ni and Al would not form a solid solution over theentire compositional range.(b) Ti in NiSize: r(Ti) = 0.147 nm, r(Ni) = 0.125nm difference = 17.6%Electronegativity: Ti: 1.54; Ni: 1.91Valence: Ti4+; Ni2+Crystal Structure: Ti:HCP; Ni FCCTi in Ni would not exhibit extensive solid solubility(c) Zn in FeSize r(Zn) = 0.133nm; r(Fe) - 0.124nm difference = 7.25%Electronegativity: Zn = 1.65; Fe = 1.83Most Common Valence: Zn2+; Fe2+Crystal Structure: An: HCP; Fe: BCCSince electronegativities and crystal structures are very different, Zn - Fe will notexhibit extensive solid solubility.(d) Si in AlSize r(Si) = 0.117 nm; r(Al) = 0.143nm; difference = 22.2%Electronegativity: (Si) = 1.90; Al = 1.61Valence: Si4+; A;3+Crystal Structure: Si: Diamond Cubic; Al: FCCSince the size difference is greater than 15%, and the crystal structures are different, Si-Al would not exhibit extensive solid solubility.(e) Li in AlSize r(li): 0.152, r(Al): 0.143; difference - 6.29%Electronegativity: Li: 0.98; Al: 1.61Most Common Valence: Li1+; Al3+Crystal Structure: Li:BCC; Al: FCCSince electronegativity and crystal structures are very different, Li-Al will not exhibit extensive solid solubility.(f) Cu in AuSize r(Cu) = 0.125nm; r(au) = 0.144nm; difference = 12.5% Electronegativity: Cu = 1.90; Au = 1.93Most Common Valence: Cu+; Au+Crystal Structure: Cu:FCC; Au:FCCCu-Au will exhibit extensive solid solubility.(g) Mn in FeSize r(Mn) = 0.112, r(Fe) = 0.124 difference = 10.71%Electronegativity: Mn 1.55; Fe 1.83Most Common Valence: Mn2+; Fe2+Crystal Structure: Mn:BCC; Fe BCCThe difference in electronegativity is high but Mn-Fe does obey the other 3Hume-Rothery rules. Therefore, it will form solid solutions but not over the entire compositional range.(h) Cr in FeSize r(Cr) = 0.125nm, Fe = 0.144nm difference = 12.5%Electronegativity: Cr = 1.66; Fe = 1.83Most Common Valence: Cr3+; Fe2+Crystal Structure: Cr:BCC; Fe:BCCCr in Fe will exhibit extensive solid solubility but not over the entire compositional range since it obeys only 3 of 4 Hume-Rothery rules.(i) Ni in FeSize r(Ni) = 0.125nm, r(Fe) = 0.124nm difference = 0.8%Electronegativity: Ni: 1.91; Fe 1.83Most Common Valence: Ni3+; Fe3+Crystal Structure: Ni:FCC; Fe: BCCNi and Fe obeys 3 of the 4 Hume-Rothery rules therefore, extensive solid solutionwill be exhibited but not over the entire compositional range.7. (a) When one attempts to add a small amount of Ni to Cu, Ni is the solute and Cuis the solvent.(b) Based on the relative sizes of Ni and Cu, radius of Ni = 0.128nm, radius of Cu =0.125nm, these two are expected to form substitutional solid solutions.(c) Ni and Cu will be completely soluble in each other because they obey all fourHume-Rothery rules.8. FIND: Predict how Cu dissolves in Al.DATA: Cu Alatomic radius (A) 1.28 1.43electronegativity 1.90 1.61valence 1+,2+ 3+crystal structure FCC FCCSOLUTION: All of Hume-Rothery's rules must be followed for a substitutionalsolution. In this case, the valences do not match. Cu will not go into substitutional positions in Al to a large extent.COMMENTS: This principle is often used to precipitation harden Al using Cu.9. What type of solid solution is expected to form when C is added to Fe?The radius of carbon atom is 0.077nm and that of an Fe atom is 0.124nm. The size difference between these two is ~61% which is much grater than ~15%. Thus,these two are not expected to form substitutional solid solution.If we compare the size ratio of C to Fe atoms with the size of tetrahedral andoctahedral interstitial sites in BCC iron, we find that C does not easily fit into either type of interstitial position. C, however, forms an interstitial solid solution with Fe but the solubility is limited.10. FIND: Calculate the activation for vacancy formation in Fe.GIVEN: The vacancy concentration at 727 C = 1000K is 0.00022.SOLUTION: We use equation 4.2-2 to solve this problem:C v = exp (-Q fv/RT)Solving for Q fv:Q fv = -RT ln C v = -(8.31 J/mole-K)(1000K) ln 0.00022 = 7.0 x 104 J/mole11. SHOW: A Schottky and Frenkel defect in MgF2 structuresA 2-D representation of the MgF2 structure containing a Schottky defect and aFrenkel defect is shown below.12. Explain why the following statement is incorrect: In ionic solids the number ofcation vacancies is equal to the number of anion vacancies.In ionic crystals, even in the presence of vacancies, the charge neutrality must bemaintained. Therefore, single vacancies do not occur in ionic crystals sinceremoval of a single ion would lead to charge imbalance. Instead the vacanciesoccur in a manner such that the anion: cation vacancy ratio render the solidelectrically neutral. This, however, does not mean that the anion vacancies areequal to cation vacancies. For example, a Schottky defect in MgCl2 or MgF2involves two Cl- or F- cation vacancies for every Mg2+ anion vacancy to maintainelectrical neutrality.The number of cation vacancies equals the number of anion vacancies only for thelimiting case where the chemical formula of the compound is MX.13. Calculate the number of defects created when 2 moles of NiO are added to 98 molesof SiO2. Also, determine the type of defect created.GIVEN: Neglect interstial vacanciesWe have 2 moles of NiO and 98 moles of SiO2. Since NiO is a 1:1 compound there are 2 moles of Ni2+ ions and 2 moles of O2- ions present. SiO2 on the other hand is a 1:2 compound; therefore, there are 98 moles of Si4+ and 196 moles of O2-. Thetotal number of each type of ion isN Ni = 2 molesN Si = 98 molesN O2 = 196 molesThe total number of moles of ions in the system isN T = N Ni + N Si + N O = 2 + 98 + 196 = 196 molesEach substitution of an Ni2+ for Si4+ results in a loss of 2 positive charges. If nointerstitials are created, this loss of positive charge is balanced by the creation ofanion vacancies. Charge neutrality requires one oxygen vacancy created for every Ni 2+ ion. Therefore, the number of oxygen vacancies isN Ov = N Ni = 2 molesThere are 2 moles of oxygen ion vacancies created with the addition of 2 moles of NiO to 98 moles of SiO 2. 14. Calculate the number of defects created when 1 mole of MgO is added to 99 moles ofAl 2O 3.MgO is a 1:1 compound, therefore there is 1 mole of Mg 2+ ions and 1 mole of O 2- ions in the system.From Al 2O 3, there are 198 moles of Al 3+ ions and 297 moles of O 2- ions in the system.Each substitution of an Mg 2+ ion for Al 3+ ion results in a loss of one positive charge. This loss of positive charge is balanced by oxygen vacancy. Charge neutrality requires one oxygen vacancy to be created for every two Mg 2+ ion 3. Therefore thenumber of oxygen ion vacancies created is0.5 moles of oxygen ion vacancies are created by the addition of 1 mole of MgO to 99 moles of Al 2O 3. 15.COMPUTE: Relative concentration of cation vacancies, anion vacancies and cation interstitials.GIVEN: Q Cv = 20kJ/moleQ Av = 40kJ/mole Q CI = 30kJ/moleASSUMPTION: assume room temperatureT = 298KConcentration of cation vacancies, C Cv is given by0.5moles = 21= 2N= N Mg O vSimilarly for anion vacancies and for cation interstitials16. (a) Describe a Schottky defect in U 2(b) Would you expect to find more cation or anion Frenkel defects in this compound? Why?UO 2 has a fluorite structure with U 4+ ions occupying FCC lattice sites and O 2- occupying tetrahedral interstitial sites.(a) A Schottky defect in UO 2 will involve one U 4+ cation vacancy and 2 O 2- anion vacancies.(b) In general cation Frenkel defects are more common than anion Frenkel defects because cations are usually smaller. In this case, the radii of U 4+ is 0.106nm and that of O 2- is 0.132nm. The U 4+ cation is smaller than the O 2- anion. However, the size difference is not very high. Still, cation Frenkel defects are expected to be more.17. Ionic compound Li 2O(a) Describe a Schottky defect (b) Describe a Frenkel defectLi 2O has an antifluorite structure. O 2- ions occupy FCC lattice sites and Li + occupies tetrahedral interstitial sites.10 x 3.108 = (-8.0763) = C 298 x 8.3120,000- = RT Q - = C 4-Cv cv Cv exp exp exp ⎪⎭⎫⎝⎛⎪⎭⎫⎝⎛10 x 9.6 = (-16.152) = 298 x 8.3140,000- = C 8-AV ex p ex p ⎪⎭⎫⎝⎛ 10 x 5.48 = (-12.114) = 298 x 8.3130,000- = C 6-CIex p ex p ⎪⎭⎫⎝⎛(a) A Schottky defect in Li2O involves 2 Li2+ cation vacancies and one O2- anion vacancy(b) The ionic radii of Li+ and O2- are 0.078nm and 0.132nm respectively. Thismaterial is most likely to exhibit cation Frenkel defect since the size of the cation is much smaller than the anion.18. DETERMINE:(a) Interstitial Na+ ions(b) Interstitial O2- ions(c) Vacant Na+ sites(d) Vacant O2- sites in Na2OGIVEN: r(Na+) = 0.098nmr(O2-) = 0.132nmNa2O structure is similar to antifluorite structure. Na+ ions occupy tetrahedralinterstitial sites and O2- ions occupy FCC lattice sites.Since the ratio of Na:0 is 2:1 for this materials, a Schottky defect results in 2 cation vacancies for every one anion vacancy.no. of vacant Na+ sites = 2 x no. of vacant O2- sitesA cation Frenkel defect is more likely to occur in this material(a) Interstitial Na+ ions = 1(b) Interstitial O2- ions = 0(c) Vacant Na+ sites = 2(d) Vacant O2- sites = 119. SOLVENT: AuSOLUTE: N, Ag or CsDETERMINE: (a) which element is most likely to form an interstitial solidsolution.(b) which element is most likely to form a substitutional solid solution.r(Au) = 0.144nmr(N) = 0.071nmr(Ag) = 0.144nmr(Cs) = 0.265nm(a) Based on atomic radii N is most likely to form are interstitial solid solutionwith Au as solvent.(b) Ag is most likely to form a substitutional solid solution because the size difference between Au & N and Au & Cs is more than 15%.In addition, Au and Ag have similar valence, and crystal structure. Theelectronegativities are not quite similar, but since Ag-Au system obeys 3 out of 4 of the Hume-Rothery rules, Ag is the most likely element with which Au forms a substitutional solid solution.Section 4.4 Diffusion20. Under what condition can Fick’s first law be used to solve diffusion problems.The Fick’s first law can be used to solve diffusion problems provided the concentration gradient does not change with time.21. GIVEN: 1 wt% B is added to Fe.FIND: (a) if B would be present as an interstitial impurity or substitutionalimpurity, (b) fraction of sites occupied by B atoms, (c) if Fe containing B were to be gas carburized, would the process be faster or slower than for Fe which has no B? Explain.r(B) = 0.097nm r(Fe) = 0.124nm(a) Based on the atomic radii B would be present as an interstitial impurity(b) amount of B present = 1 wt%As a basis of calculation assume 100gms of material.Determine the no. of moles of Fe and B present.Total no. of moles of Fe and B = 1.773 + 0.092 = 1.865 moles.Fraction of sites occupied by B atoms = 1.8650.092= mole fraction of B = 0.049Thus, B roughly occupies 5% of the sites.mole 0.092 = 10.8111= B of molesmoles 1.773 = 55.8599=Fe of wt mol.Feof gms= Fe of moles(c) If Fe containing B were to be gas carburized the process would be slower than for Fe which has no B simply because the presence of B atoms already in interstitial sites leave fewer sites for interstitial C to diffuse through. 22.Determine which type of diffusion would be easier (a) C in HCP Ti (b) N in BCC Ti (c) Ti in BCC Tir(C) << r(Ti) so we can predict that diffusion occurs via an interstitial mechanism r(N)<<r(Ti). In this case the diffusion also occurs via interstitial mechanism.Ti in BCC Ti is a case of self-diffusion and self-diffusion occurs via a vacancy mechanism. In general the activation energy for self diffusion is higher than interstitial mechanism because vacancy mechanism involves two steps. One is to create a vacancy and second is to promote a vacancy/atom exchange. Thus Ti in BCC Ti will be the slowest.The activation energy for diffusion via interstitial mechanism is just the energy necessary to move an atom into a neighboring interstitial site. An open crystal structure, as opposed to a dense structure, should have a lower activation energy. Between BCC Ti and HCP Ti, BCC Ti has a more open structure (lower APF) than HCP Ti.Thus, N in BCC Ti diffusion would be the easiest by virtue of its lowest activation energy.23. GIVEN:C 1 = 0.19 at % at surfaceC 2 = 0.18 at % at 1.2mm below the surfaceD = 4 x 10-14 m 2/sec a o = 4.049 A oCOMPUTE: Flux of copper atoms from surface to interior.We must first calculate the concentration gradient in terms of [copperatoms/cm 3/cm]. It can be calculated as follows:The concentration gradient is thencm / atoms 10 x 4.60 = 10 x 6.02 x 63.54] / 2.70) x [(0.0018= C cm / atoms 10 x 4.86 = 10 x 6.02 x 63.54] / 2.70) x [(.0019= CN x Cu] wt at / FCCAl) of density x [(a/oCu = C 319232319231AV24.FIND: Predict whether diffusion is faster in vitreous or crystalline silica.GIVEN: Diffusion is the movement of atoms through the material one step at a time. The ease of movement is in part determined by the amount of space that surrounds each atom. In more open or less dense structures, atoms have an increased chance of being able to squeeze past a neighbor into a new position. SOLUTION: Diffusion can be thought of as an Arrhenius process. The activation energy is that required to move an atom from one position to another, as shown in Fig. 2.3-2. In a crystal the activation energy will be greater than in a glass, since the density is higher and there is less free, or unoccupied, volume. Thus, we expect diffusion to be slower in crystal than in glasses at the same temperature.COMMENTS: When a noncrystalline material is raised to a temperature above the glass transition temperature, diffusion increases enormously. In metals this brings about rapid crystallization. In some ceramic and polymer systems, crystallization may be slow or absent.25.FIND: Do textile dyes more readily penetrate crystalline or noncrystalline regions? GIVEN: Most textile fibers are semicrystalline, containing both crystalline andnoncrystalline regions. The density of the noncrystalline regions is less than that of the crystalline regions. Often dyeing is conducted at a temperature at which the noncrystalline regions are above their glass transition temperature.SOLUTION: Dye penetration through the glass will be greater than that through the crystal; however, the rate of dyeing is not sufficiently high to be commercially feasible. The temperature must be raised so that the noncrystalline polymer is in the rubber state. Diffusion becomes rapid (radially inward) into the small fibers.secsec cm atoms10 x 8.5 = Jcm atoms 10 x 2.125-m cm 10 x m 10 x 4- =dx dc D - = Jcm / atoms 10 x 2.125- = 0.1210 x 2.55- =0.1210 x 4.86) - (4.60 = dx dc 29419224214-4191819⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛COMMENTS: One of the key lessons that dye houses learn is that a sufficient amount of noncrystalline poorly oriented polymer must be present in the fiber. The temperature of the dye bath needs to be above the glass transition temperature. Sometimes water and carriers are used to swell the noncrystalline regions to get yet a greater diffusion rate. The dyes may attach to the polymer using ionic bonds or covalent bonds. Unattached dye may wash out later. 26.CALCULATE: The factor by which the diffusion coefficient of Al in Al 2O 3 change when temperature is increased from 1800o C to 2000o C GIVEN: T 1 = 1800o C = 2073KT 2 = 2000o C = 2273KEQUATION: ⎪⎭⎫⎝⎛RT Q - D = D o ex pdividing (1) by (2), we getfrom table 4.4-1 of the text Q = 477kJ/mole and R = 8.31 J/mole-K⎪⎭⎫⎝⎛RT Q - D = D 1o 1exp ⎪⎭⎫⎝⎛RT Q - D = D 2o 2exp ⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛ T 1 - T 1 R Q - = RT Q + RT Q - = D D 212121exp expThus, the diffusion coefficient of Al in Al 2O 3 changes by a factor of 11.43 when thetemperature is increased from 1800o C to 2000o C.27. FIND: Temperature at which a specimen of Fe must becarburized for two hours to achieve the same diffusion result asat 900o C for 15 hrs.GIVEN: T 1 = 900o C = 1173K; Q = 84000 J/molet 1 = 15 hrs; D o = 2.00 x 10-6 m 2/sec. t 2 = 2 hrs; R = 8.31 J/mole-KThe value of flux J is in units of cm 2 per sec.Flux per cm 2 J f = Jx time 3600 x 15 x dxdcD - = J 11f(1)We need the same result in 2 hours.D 11.43 = D 0.0871= D 0.087 = D D0.087= [-2.436]=22731 - 20731 8.31477000- = D D 1122121exp exp ⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛dxdcD - = J./ m 10 x 3.62 =1173 x 8.3184000- 10 x 2.00 = RT Q - D = D 1210-6-o1sec exp exp ⎪⎭⎫ ⎝⎛⎪⎭⎫⎝⎛ 3600 x 2 x dx dc D - = J 22f,J = J 2f 1f dividing (1) by (2).28. GIVEN:D = 4 x 10-4 m 2/s @ 20o CC 1 = 2.2 x 10-3 k mol/m 3wall thickness = 3mm, diameter = 50cm height = 10cmCOMPUTE: Initial rate of mass loss through cylinder.Initially the concentration of He outside the cylinder, C 2, is zero. First, we need to convert the concentration of He from kmol/m 3 into (atoms/cm 3)/cm.C 1 = 2.2 x 10-3 kmol/m 3 = 2.2mol/m 3 = 2.2 x 10-6 mol/cm 3 = 2.2 x 10-6 mole/cm 3In terms of (atoms/cm 3)/cm0.00135 = T 10108.3-T 8.3184000- 10 x 2.00 = 10 x 2.7RT Q - D = D / m 10 x 72. = D x 7.5 = D 7.5 = D D7.5 x D D = 1226-9-2o229-2121221⎪⎭⎫⎝⎛⎪⎭⎫⎝⎛⎪⎭⎫⎝⎛exp exp exp sec C.1258 =1531K = T 6.60310108.3 = T6.603- = )1n(0.00135 = T 10108.3-o 222The concentration gradient isThe flux of atoms per second per cm 2 is obtained by using Fick’s first law ofdiffusionThe rate of mass loss is 1.766 x 1019atoms/cm 2 sec. The total surface area of the cylinder is 2πr(r+h) where r = radius and h = height.Total surface area = 2π x 25 (25 + 10) = 5497.79 cm 2The rate of mass loss per secondNote:(i ) The steady state mass loss is calculated because the initial rate of mass loss (i.e., rate of mass loss at time t = 0) is 0. (ii ) It is assumed that the curvature of the cylinder is large enough to calculate J using the expression for plate geometry. 29. Diffusion across a polymer membrane depends not only on size of the diffusing species but also the polarity of the diffusing species. A polar membrane may pass nonpolar species but serve as a barrier to polar species.Saran wrap contains highly polar atoms making it a polar membrane which serves as a barrier to water which is a polar compound. thus, there is no diffusion of water through the package unlike polyethylene, which is a nonpolar membrane and allows diffusion of water molecules which form ice. 30.COMPUTE: Temperature required to yield a carbon content of 0.5% at a depthcmatoms 10 x 1.324 = mole atoms 10 x 6.02 x cm mole102.2x = C 3182336-1 cm atoms/ 10 x 4.415- = 0.310 x 1.324 - 0 = dx dcxC - C = dx dc 4181812 .cm atoms/ 10 x 1.766 )10 x (-4.415 )10 x 10 x (4- = dxdcD - = J 2191844-sec secsec secmoles0.16 = .atoms 10 x 9.709 =cm 5497.79 x - cm atoms10 x 1.766 =222219of 0.4mn below the surface of the rod in 48 hours. GIVEN: Carbon concentration the interior = 0.2w/oCarbon concentration in the furnace = 1.0w/oBase material: HCP TiEQUATION: In this problem c(x, t) = 0.5wt%c o = 0.2 wt%c s = 1.0 wt%From figure 4.4-11, when⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛Dt 2x erf - 1 = ]c - c []c - t) (x, [cRT Q- D = D o s o o ex p0.625=0.375 - 1 = Dt .04 erf Dt .04 erf - 1 = 0.375Dt 2.04cm erf - 1 = 0.2 - 1.00.2 - 0.5⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡From Metals Handbook, Desk Edition, Pg. 28.66 for C diffusion in Ti, D o = 3.02 x 10-3 cm 2/sec, Q = 20,000 cal/mole = 83682 J/mole. 31.The diffusion process through vacancy-interchange mechanism depends on creation of vacancies and vacancy/atom interchange.At comparable homologous temperatures, for Ge and Cu the diffusion coefficientfor that material which has a higher vacancy concentration would be higher.A covalent bond as opposed to a metallic bond is stronger and directional. It isalso difficult to create vacancies in a covalently bonded material due to its strongbonding. Therefore, the activation energy for vacancy creation in a covalentlybonded material such as Ge is larger than Cu which has a weak metallic bond.The directional nature of a covalent bond places geometrical restrictions on thevacancy atom interchange which again results in an increase in the activationenergy.sec /cm 10 x 2.33 = DDt = 0.630.040.630 = Dt0.040.630 = Z0.625 = (Z) erf 28-2⎪⎭⎫ ⎝⎛ C 582 = T855K = T11.18-10065.2 = T11.18- = T10065.2-10 x 7.72 = T 10065.2- T x 8.31483682- 10 x 3.02 = 10 x 2.33 = D e 83682J/mol = mole 20,000cal/ = Q /cm 10 x 3.02 = D o 6-3-8-2-3o ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛exp exp secTherefore, at comparable temperatures the diffusion coefficient for Ge will belarger.32. FIND: Describe the energy and entropy in Fig. 4.4-5a, b, and c.SOLUTION: The order in part a is high. The materials is perfect. There is only one way to arrange the atoms in such a system. The entropy is low. In part b there is less order, more disorder, and the entropy has increased. Part c is nearly random. It has low order and high entropy. Energy contains a contribution from entropy: E = H -TS, where E is energy, T is absolute temperature, and S is entropy. Assuming all other contributions to energy change negligibly (T and H), the energy of part c is the low, part a is high and part b is intermediate.COMMENTS: What is shown in going from a to c is the entropy of mixing.33. GIVEN: After 10 hrs at 550o C an oxide layer of thickness 8 μm is formed.COMPUTE: Thickness after 100 hrs. Using the definition of effective penetration distance and equation 4.4-11 of text,with γ = 2 we have Dt 2 x eff ≈.In this case34. GIVEN: D w = 1.0 x 10-12 m 2/s (water)D dc = 1.0 x 10 (dye carrier)D d = 1.0 x 10-14 m 2/s (dye)COMPUTE:(a) Times required for the water, dye and carrier to penetrate to the center of the fiber. m25.3 = x 100hr 10hr = x m8t t = t D t D = x x100hr = t 10hr = t? = x m 8 = x 2eff 2eff2122111eff 1eff 212eff 1eff μμμ(b) Same as (a) but fiber diameter doubles(c) If thermal diffusivity of PET is secm 10 x 828-how long will it take for the heat to penetrate to the center of a 50μm diameter fiber.(a) using equation 4.4-11 of text with γ = 2.for water,for dye carriersimilarly for dye t = 6.25secs.(b) If the diameter fiber is doubled x eff = 50 x 10-6 mfor water, Dt 2 = x eff minutes2.60 or secs 156 = tt x 10 x 1.0 = 10 x 625.00t x 10 x 1.0 = 2)10 x (25.0m 10 x 25 = x/s m 10 x 1.0 = D 12-12-12-6-26-eff 2-12w ⎪⎪⎭⎫⎝⎛ minutes26.04 = secs 15.63 = tm 10 x 25 = x/s m 10 x 1.0 = D 6-eff 2-13dcsimilarly for dye carriert = 6250 secsand for dyet = 6.25 x 104 secs(c) Substituting D with D th , we can use the same equation to calculate the timerequired for heat to penetrate the center of fiber diameter = 50μm.Note: The units of thermal diffusivity is m 2/sec and notK - m Watt as printed in text35. FIND: How long will it take to case carburize a steel chain to a depth of 1/16 inch?GIVEN: It requires 4 hours to carburize a plate of similar composition to a depthof 1/16 inch.ASSUMPTIONS: All carburization conditions are the same in both treatments.SOLUTION: Equation 4.4-11 is used to solve the problem:secs.625 = 10 x 1.0)10 x 50 ( = tt x D = 2)10 x (5010 x 50 = x/s m 19 x 1.0 = D 12-26-26-6-eff 2-12wsecs0.00195 = t10x 81 x 210 x 25 = _tt x 10 x 8 = 210 x 25t x D 2 = x 8-6-28-6-2th eff ⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛。

仅供个人参考ForpesnauseonysudyandresachnofrcmmercauseA 高分子化学和高分子物理蚆UNIT 1 What are Polymer? 节什么是高聚物？第一单元袁What are polymers? For one thing, they are complex and giant molecules and are 蝿contrast To common salt. weight compounds like, say, different from low molecularthe difference, the molecular weight of common salt is only 58.5, while that ofa polymer can be as high as several hundred thousand, even more than thousand smaller of much ' are made up These thousands. big molecules or‘macro-moleculesmolecules, can be of one or more chemical compounds. To illustrate, imagine thatthings When these the same material. the same size and is made of of a set rings has from polymer as representing a the chain formed can be considered are interlinked, of be rings could same of the compound. Alternatively, individual moleculesfrom a represent polymer materials, and interlinked to different sizes and molecules of different compounds.什么是高聚物？首先，他们是合成物和大分子，而且不同于低分子化合物，譬如说普通蒇，而高聚物的分子量高于的盐。

1.“Materials science"involves investigating the relationships that exist between the structures and properties of materials. In contrast, "Materials engineering" involves, on the basis of these structur e-property correlations, designing or engineering the structure of a material to produce a predeter mined set of properties.“材料科学”涉及研究材料的结构和性能之间的关系。

相反，“材料工程”是指在这些结构和性能相关性的基础上，基于预期的性能来设计或生产有预定性能的材料。

2.Virtually all important Properties of solid materials may be grouped into six different categories: mechanical, electrical, thermal, magnetic, optical, and deteriorative实际上，固体材料的所有重要性质都可以分为六类：机械、电气、热、磁、光学和腐蚀性。

3.In addition to structure and properties, two other important components are involved in the scien ce and engineering of materials- namely“processing”and“performance”.除了结构和性能之外，材料科学和工程还涉及另外两个重要的组成部分，即“加工”和“性能”。

2.英译汉材料科学石器时代肉眼青铜器时代光学性质集成电路机械(力学)强度热导率1.材料科学指的是研究存于材料的结构和性能的相互关系。

相反，材料工程指的是，在基于材料结构和性能的相互关系的基础上，开发和设计预先设定好具备若干性能的材料。

2. 实际上，固体材料的所有重要性质可以概括分为六类：机械、电学、热学、磁学、光学和腐蚀降解性。

3. 除了结构和性质，材料科学和工程还有其他两个重要的组成部分：即加工和性能。

4. 工程师与科学家越熟悉材料的结构-性质之间的各种相互关系以及材料的加工技术，根据这些原则，他或她对材料的明智选择将越来越熟练和精确。

5. 只有在极少数情况下材料在具有最优或理想的综合性质。

因此，有必要对材料的性质进行平衡。

3. 汉译英Interdispline dielectric constantSolid materials heat capacityMechanical properties electro-magnetic radiationMaterials processing elasticity modulus1.直到最近，科学家才终于了解材料的结构要素与其特性之间的关系。

It was not until relatively recent times that scientists came to understand the relationship between the structural elements of materials and their properties .2.材料工程学主要解决材料的制造问题和材料的应用问题。

Material engineering mainly solve the problems of materials processing and materials application.3.材料的加工过程不但决定了材料的结构，同时决定了材料的特征和性能。

英语学习《材料科学与工程专业英语》《材料科学与工程专业英语》Unit1 Materials Science and Metallurgical EngineeringMaterials are properly more deep-seated in our culture than most of us realize. Trans -portation, housing, clothing, communication, recreation and food production--virtually every segment of our everyday lives is influenced to one degree or another by materials. Historically, the development and advancement of societies have been intimately tied to the members' abilities to produce and manipulate materials to fill their needs. In fact, early civilizations have been designated by the level of their materials development (i.e.Stone Age, Bronze Age).The earliest humans has access to only a very limited number of materials, those that occur naturally stone, wood, clay, skins, and so on. With time they discovered techniques for producing materials that had properties superior to those of the natural ones: these new materials included pottery and various metals. Furthermore, it was discovered that the properties of a material could be altered by heat treatments and by the addition of other substances. At this point, materials utilization was totally a selection process, that is, deciding from a given, rather limited set of materials the one that was best suited for an application by virtue of its characteristic. It was not until relatively recent times that scientists came to understand the relationships between the structural elements of materials and their properties. This knowledge, acquired in the past 60 years or so, has empowered them to fashion, to a large degree, the characteristics of materials. Thus, tens of thousands of differentmaterials have evolved with rather specialized characteristics that meet the needs of our modern and complex society.The development of many technologies that make our existence so comfortable has been intimately associated with the accessibility of suitable materials. Advancement in the under--standing of a material type is often the forerunner to the stepwise progression of a technology. For example, automobiles would not have been possible without the availability of inexpensive steel of some other comparable substitutes. In our contemporary era, sophisticated electronic devices rely on components that are made from what are called semiconducting materials.Materials Science EngineeringMaterials science is an interdisciplinary study that combines chemistry, physics, metallurgy, engineering and very recently life sciences. One aspect of materials science involves studying and designing materials to make them useful and reliable in the service of humankind. It strives for basic understanding of how structures and processes on the atomic scale result in the properties and functions familiar at the engineering level. Materials scientists are interested in physical and chemical phenomena acting across large magnitudes of space and time scales. In this regard it differs from physics of chemistry where the emphasis is more on explaining the properties of pure substances. In materials science there is also an emphasis on developing and using knowledge to understand how the properties of materials can be controllably designed by varying the compositions, structures, and the way in which the bulk and surfaces phase materials are processed.In contrast, materials engineering is, on the basis of those structure properties correlations, designing or engineering thestructure of a material to produce a predetermined set of properties. In other words, materials engineering mainly deals with the use of materials in design and how materials are manufactured."Structure" is a nebulous term that deserves some explanation. In brief, the structure of a material usually relates to the arrangement of its internal components. Subatomic structure involves electrons within the individual atoms and interactions with their nuclei. On an atomic level, structure encompasses the organization of atoms or molecules relative to one another. The next large structural realm, which contains large groups of atoms that are normally agglomerated together, is termed "microscopic" meaning that which is subject to direct observation using some type of microscope. Finally, structural elements that may be viewed with the naked eye are termed "macroscopic".The notion of "property" deserves elaboration. While in service use, all materials are exposed to external stimuli that evoke some type of response. For example, a specimen subject to forces will experience deformation; or a polished metal surface will reflect light. Property is a material trait in terms of the kind and magnitude of response to a specific imposed stimulus. Generally, definitions of properties are made independent of material shape and size.Virtually all important properties of solid materials may be grouped into six different categories; mechanical, electrical, thermal, magnetic, optical, and deteriorative. For each there is s characteristic type of stimulus capable of provoking different responses. Mechanical properties relate deformation to an applied load or force: examples include elastic modulus andstrength. For electrical properties, such as electrical conductivity and dielectric constant, the stimulus is an electric filed. The thermal behavior of solids can be represented in terms of heat capacity and thermal conductivity. Magnetic properties demonstrate the response of a material to the application of a magnetic field. For optical properties, the stimulus is electromagnetic or light radiation: index of refraction and reflectivity are representative optical properties. Finally, deteriorative characteristics indicate the chemical reactivity of materials.In addition to structure and properties, two other important components are involved in the science and engineering of materials, namely "processing" and "performance". With regard to the relationships of these four components, the structure of a material will depend on how it is processed. Furthermore, a material's performance will be a function of its properties. Thus, the interrelationship between processing, structure, properties, and performance is linear as follows:Processing→Structure→Properties→PerformanceWhy Study Materials Science and Engineering?Why do we study materials? Many an applied scientists or engineers, whether mechanical, civil, chemical, or electrical, will be exposed to a design problem involving materials at one time or another. Examples might include a transmission gear, the superstructure for a building, an oil refinery component, or an integrated circuit chip. Of course, materials scientists and engineers are specialists who are totally involved in the investigation and design of materials.Many times, a materials problem is to select the right material from many thousands available ones. There are severalcriteria on which the final decision is normally based. First of all, the in-service conditions must be characterized. On only rare occasion does a material possess the maximum or ideal combination of properties. Thus, it may be necessary to trade off one characteristic for another. The classic example involves strength and ductility; normally, a material having a high strength will have only a limited ductility. In such cases a reasonable compromise between two or more properties may be necessary.A second selection consideration is any deterioration of material properties that may occur during service operation. For example, significant reductions in mechanical strength may result from exposure to elevated temperatures or corrosive environments.Finally, probably the overriding consideration is economics. What will the finished product cost? A material may be found that has the ideal set of properties, but is prohibitively expensive. Here again, some compromise is inevitable. The cost of a finished piece also includes any expense incurred during fabrication.The more familiar an engineer or scientist is with the various characteristics and structure-property relationships, as well as processing techniques of materials, the more proficient and confident he or she will be to make judicious materials choices based on these criteria.(Selected from Materials Science and Engineering: AnIntroduction, by William D Callister,2002)New Words and Expressionspottery n. 陶瓷by virtue of 依靠（……力量），凭借，由于，因为empower vt.授权，准许，使能够empower sb.to do sth. 授权某人做某事forerunner n. 先驱（者），传令官，预兆stepwise a. 逐步地，分阶段地interdisciplinary a. 交叉学科的metallurgy n. 冶金学nebulous a. 星云的，云雾状的，模糊的，朦胧的agglomerate n. 大团，大块；a.成块的，凝聚的elaboration n. 详尽的细节，解释，阐述electrical conductivity 电导性，电导率dielectric constant 介电常数thermal conductivity 热导性，热导率heat capacity 热容refraction n. 衍射reflectivity n. 反射ductility n. 延展性corrosive a. 腐蚀的，蚀坏的，腐蚀性的；n. 腐蚀物，腐蚀剂overriding a. 最重要的；高于一切的prohibitive a. 禁止的，抵制的judicious a. 明智的criterion n. 标准，准则，尺度Notes1. It was not until relatively recent times that scientists came to understand the relationships between the structural elements of materials and their properties.这是一个强调句，强调时间。