材料科学与工程专业英语第二版1.2.7.10.13.16课后习题_翻译答案_

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材料科学与工程专业英语1(18单元课后翻译答案)-

材料科学与工程专业英语1（18单元课后翻译答案）-目睹了我们的生活通过发生在医药、电信和交通运输行业的革命得到了重塑。

8 .世界上80%的人口缺乏安全饮用水，近40%的人口没有卫生设施。

9 .材料和社会是相互联系的，我们应该看到微型企业研究议程和影响人类状况的社会问题之间的密切关系，这是合理的。

从化学角度来看，金属是一种容易失去电子形成正离子的元素，正离子与其他金原子形成金属键。

2.金属键的无方向性被认为是金属延展性的主要原因。

3.只有当原子间的键断裂时，带有共价键的晶体才会变形，导致晶体断裂。

4.合金，尤其是那些满足更高应用要求的合金，如喷气发动机，可能含有十多种元素。

5.离域电子电子结构碱土金属化学电池核电荷电导率。

金属有时被描述为被离域电子云包围的正离子晶格。

7 .金属通常具有优异的导电性和导热性、高密度和在应力下变形而不裂开的能力。

8 .合金是两种或多种元素在固溶体中的混合物，其中主要成分是金属。

9 .将不同比例的金属结合在一起作为超级合金的发展严重依赖于化学和加工创新，主要由航空和能源行业推动。

2.抗蠕变性主要取决于晶体结构中位错速度的减缓。

3.高温合金加工技术的发展大大提高了高温合金的工作温度。

4.单晶高温合金是采用改进的定向凝固技术形成的，因此材料中没有晶界。

5.面心立方晶体结构涡轮入口温度金属材料相稳定性核反应堆纳米粒子的合成。

超级合金通常具有悬浮面心立方晶体结构。

7 .在需要高温强度和腐蚀/氧化的地方使用超级合金电阻。

8 .超级合金广泛用于航空潜艇、核反应堆和军用电动机。

9 .在高温下，气态明矾腐蚀过程本质上是一个电化学过程，具有与电池相同的基本特性。

2.从矿物中提取金属所需的能量问题与随后的腐蚀和能量释放直接相关。

3.当电子与中和的正离子(如电解质中的氢离子)反应时，阴极的电子必须平衡。

4.保护膜电路自由电子转移金属阳离子阳极反应5。

一些金属如金和银可以在地球上以天然金属状态存在，它们几乎不容易腐蚀。

材料科学与工程专业英语第二版1.2.7.10.13.16课后习题_翻译答案_可编辑】

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。

材料科学与工程专业英语第二版课文翻译(1,2,3,10)

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。

(完整版)高分子材料工程专业英语第二版课文翻译(基本全了

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。

材料科学与工程专业英语1-18单元课后翻译答案

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

材料科学与工程专业英语课文翻译(1,2,3,10).

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。

材料科学专业英语第二章翻译

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。

材料科学与工程专业英语 (1).

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。

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第二版翻译

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。

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课后习题答案unit1-7

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。

《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;。

工程材料科学与设计原书第2版课后习题答案4—8章

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 ⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛。

材料科学与工程专业英语1（18单元课后翻译答案）-

材料科学与工程专业英语1（18单元课后翻译答案）-材料科学与工程专业英语1（18单元课后翻译答案）-目睹了我们的生活通过发生在医药、电信和交通运输行业的革命得到了重塑。

8 .世界上80%的人口缺乏安全饮用水，近40%的人口没有卫生设施。

9 .材料和社会是相互联系的，我们应该看到微型企业研究议程和影响人类状况的社会问题之间的密切关系，这是合理的。

从化学角度来看，金属是一种容易失去电子形成正离子的元素，正离子与其他金原子形成金属键。

2.金属键的无方向性被认为是金属延展性的主要原因。

3.只有当原子间的键断裂时，带有共价键的晶体才会变形，导致晶体断裂。

4.合金，尤其是那些满足更高应用要求的合金，如喷气发动机，可能含有十多种元素。

5.离域电子电子结构碱土金属化学电池核电荷电导率。

金属有时被描述为被离域电子云包围的正离子晶格。

7 .金属通常具有优异的导电性和导热性、高密度和在应力下变形而不裂开的能力。

8 .合金是两种或多种元素在固溶体中的混合物，其中主要成分是金属。

9 .将不同比例的金属结合在一起作为超级合金的发展严重依赖于化学和加工创新，主要由航空和能源行业推动。

2.抗蠕变性主要取决于晶体结构中位错速度的减缓。

3.高温合金加工技术的发展大大提高了高温合金的工作温度。

4.单晶高温合金是采用改进的定向凝固技术形成的，因此材料中没有晶界。

5.面心立方晶体结构涡轮入口温度金属材料相稳定性核反应堆纳米粒子的合成。

超级合金通常具有悬浮面心立方晶体结构。

7 .在需要高温强度和腐蚀/氧化的地方使用超级合金电阻。

8 .超级合金广泛用于航空潜艇、核反应堆和军用电动机。

9 .在高温下，气态明矾腐蚀过程本质上是一个电化学过程，具有与电池相同的基本特性。

2.从矿物中提取金属所需的能量问题与随后的腐蚀和能量释放直接相关。

3.当电子与中和的正离子(如电解质中的氢离子)反应时，阴极的电子必须平衡。

4.保护膜电路自由电子转移金属阳离子阳极反应5。

材料科学与工程专业英语课后答案

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”。

材料科学与工程_专业英语_Unit_2_Classification_of_Materials译文

Solid 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：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，尽管有许多中间品。

除此之外，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。

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。

材料科学与工程专业英语第二版翻译答案老师整理的知识分享

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。

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》

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。

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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。

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。

Materials processing process determines not only their structure but also their characteristic and performance.4.材料的力学性能与其所受外力或负荷而导致的形变有关。

Material mechanical properties is relative with(relates with) its deformation coming from(resulting from, due to ) outside(applied) force or load.2.英译汉复合材料游离电子先进材料刚度、刚性半导体生物材料智能材料纳米工程材料1.金属导电、导热能力特别强，对可见光不透明；一个抛光的金属表面具有光泽。

2.陶瓷是典型的绝热、绝缘体，在对高温和苛刻环境的抵抗力方面，优于金属和高聚物。

3.应用与高技术领域的材料有时候被称为先进材料。

4.响应外加电场(或电压)，压电陶瓷会膨胀和收缩；相反的，当尺寸改变时，压电陶瓷也会产生一个电场。

5.随着扫面探针显微镜的问世，这种显微镜允许观察单个原子或者分子，使得操作和移动原子和分子形成新的结构、基于简单原子水平上设计新材料成为可能。

3.汉译英Advangced materials ceramic materialsHigh-performance materials clay mineralsAlloys implantGlass fiber carbon nanotube1、金属元素有许多有利电子，金属材料的许多性质可直接归功于这些电子。

Metallic materials have large numbers of nonlocalized electrons，many properties of metals are directly attributable to these electrons.2、许多聚合物材料是有机化合物，并具有大的分子结构。

Many of polymers are organic compounds,and they have very large molecular structures.3、半导体材料的电性特征介于导体材料（如金属、金属合金）与绝缘体（陶瓷材料和聚合体材料）之间。

Semiconductors have electrical properties that are intermediate between the electrical conductors ( viz. metals and metal alloys ) and insulators ( viz. ceramics and polymers ).4、生物材料不能产生毒性，并且不许与人体组织互相兼容。

Biomaterials must not produce toxic substances and must be compatible with body tissues.Unit7：2英译汉导带碱金属原子半径轨道能量重叠离子晶格固溶体1.化学上，金属被定义为这样一种元素：容易失去电子行程正离子、容易和其他金属原子形成金属键。

2.金属键的非方向性被认为是金属延展性的主要原因。

3.共价键晶体只有打破原子间价键才能发生形变，因此导致了晶体破碎。

4.为一些高性能场合应用，如喷气式发动机，特殊设计的合金可以包含10种以上元素。

3 汉译英delocalized electrons electrical structurealkali-earth metals electrochemical cellnuclear charge electrical conductivity1、金属有时被描述为由游离电子团包围的正离子晶格。

Metals are sometimes described as a lattice of positive ions surrounded by a cloud of delocalized electrons.2、通常地，金属具有良好的导电性和导热性，具有金属光泽，密度较大，并且具有在压力下变形而不会断裂的能力。

Metals in general have superior electric and thermal conductivity, high luster and density and the ability to be deformed under stress without cleaving.3、合金是指两种或两种以上的元素形成的固溶体混合物，其中主要组分为金属。

An alloy is a mixture of two or more elements in solid solution in which the major component is a metal.4、不同比率的金属结合成为合金可以改变纯金属的性质，从而产生所需要的性能。

Combining different ratios of metals as alloys modifies the properties of pure metals to produe desirable characteristics.Unit10：2.英译汉原材料的提纯长链烷烃玻璃烧杯粘性液体火花塞绝缘子玻璃陶瓷计算机辅助过程控制表面分析方法1.我们将审视几个性质，会看到这些性质与我们对陶瓷构成的预判是何等匹配啊。

2.高温下(高于玻璃转变温度)，玻璃不在表现出脆性行为，而是表现得像粘状液体一样。

3.它们表现出优异的力学性能、抗氧化腐蚀性能、或者电学、光学磁学性能。

4.尽管传统的粘土基陶瓷已经使用了25000多年，先进陶瓷仅仅是近100年内发展起来的。

3.汉译英Glass transiton temperature covalent ionic bondingStress distribution thermal expansion coefficientGlass fiber materials science and engineeringSolid-oxide fuell cells electron microscope1.Diamond, which is classified as a ceramic, has the highest thermal conductivity ofany known materials.2.Ceeramics are stronger in compression than in tension, whereas metals havecomparable tensile and compressive strengths.3.Ceramics generally have low toughness, although combining them incomposites can dramatically improve this property.4.The functions of ceramic products are dependent on their chemical compositonand microstructure, which determines their properties.Unit13：2.英译汉聚合物合成天然聚合物材料单体确认生物合成链长持久(余辉)长度1. 尽管聚合物这个属于在通用意义上一般指塑料，这个术语实际上也包含一大类天然和合成材料，这些材料有不同的性质和用途。

2. 聚合物合成是这样一个过程：将许多被称为单体的小分子连接成以共价键结合的链状物的过程。

3. 接枝聚合物分子由带有一个或者多个侧链的主链构成，特殊类型接枝聚合物包含星型形状、梳子形状、刷子形状。

4. 一些生物型聚合物由许多不同但结构上有联系的单体组成，如多聚核苷酸由核苷亚单体组成。

3.汉译英Persistent length cross bonding Polar monomer nucleic acid Polyreaction polyelectrolyte1. 现在商业上重要的聚合物是通过有机合成技术制备的。

Most commercially important polymers today are entirely synthetic and produced in high volume, on appropriately scaled organic synthetic techniques. 2. 某些生物聚合物是由一系列不同的但是结构却是相关的单体，例如：聚核苷酸是由核苷组成的。

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