《材料科学与工程基础》英文影印版习题及思考题及答案

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《材料科学与工程基础》习题和思考题及答案

《材料科学与工程基础》习题和思考题及答案

《材料科学与工程基础》习题和思考题及答案第二章2-1.按照能级写出N、O、Si、Fe、Cu、Br原子的电子排布(用方框图表示)。

2-2.的镁原子有13个中子,11.17%的镁原子有14个中子,试计算镁原子的原子量。

2-3.试计算N壳层内的最大电子数。

若K、L、M、N壳层中所有能级都被电子填满时,该原子的原子序数是多少?2-4.计算O壳层内的最大电子数。

并定出K、L、M、N、O壳层中所有能级都被电子填满时该原子的原子序数。

2-5.将离子键、共价键和金属键按有方向性和无方向性分类,简单说明理由。

2-6.按照杂化轨道理论,说明下列的键合形式:(1)CO的分子键合(2)甲烷CH的分子键合 24(3)乙烯CH的分子键合(4)水HO的分子键合 242(5)苯环的分子键合(6)羰基中C、O间的原子键合 2-7.影响离子化合物和共价化合物配位数的因素有那些?2-8.试解释表2-3-1中,原子键型与物性的关系?332-9.0?时,水和冰的密度分别是1.0005 g/cm和0.95g/cm,如何解释这一现象?+2-10.当CN=6时,K离子的半径为0.133nm(a)当CN=4时,半径是多少?(b)CN=8时,半径是多少?32-11.(a)利用附录的资料算出一个金原子的质量?(b)每mm 的金有多少个原子?(c)根据金21的密度,某颗含有10个原子的金粒,体积是多少?(d)假设金原子是球形(r=0.1441nm),Au21并忽略金原子之间的空隙,则10个原子占多少体积?(e)这些金原子体积占总体积的多少百分比?2+2-2-12.一个CaO的立方体晶胞含有4个Ca离子和4个O离子,每边的边长是0.478nm,则CaO的密度是多少?2-13.硬球模式广泛的适用于金属原子和离子,但是为何不适用于分子? 2-14.计算(a)面心立方金属的原子致密度;( b)面心立方化合物NaCl的离子致密度(离子半径r+=0.097,r-=0.181);(C)由计算结果,可以引出什么结论? NaCl 4702-15.铁的单位晶胞为立方体,晶格常数 a=0.287nm,请由铁的密度算出每个单位晶胞所含的原子个数。

材料科学与工程基础知识点(打印版)英汉双语版

材料科学与工程基础知识点(打印版)英汉双语版

Fundamentals of Materials Science and Engineering材料科学与工程基础知识点复习第一章绪论一、学习目的:材料科学家或工程技术人员经常遇到的问题是设计问题,而设计问题主要涉及机械、民用、化学和电。

而这些领域都要涉及到选择材料问题。

如何选择材料是非常重要的,选材包含两方面一个是满足性能要求,另一方面是成本低,即所谓“合理选材”。

材料的性能与其成分和内部的组织结构密切相关,材料的组织结构与加工过程有关。

本课程的目的就在于掌握加工过程和材料的组织结构以及性能之间的关系。

为今后进行材料设计和合理选材打下理论基础。

二、本章主要内容1、简介材料的发展史2、材料科学与工程的含义和内容3、材料的分类4、先进材料5、现代材料的需求三、重要术语和概念metal: 金属ceramic: 陶瓷polymer: 聚合物Composites: 复合材料Semiconductors: 半导体Biomaterials: 生物材料Processing: 加工过程Structure: 组织结构Properties: 性质Performance: 使用性能Mechanical properties: 力学性能Electrical properties: 电性能Thermal behavior: 热性能Magnetic properties: 磁性能Optical properties: 光性能Deteriorative characteristics: 老化特性第二章原子结构与化学键一、学习目的我们在自然界中观察到各种现象,归根结底是物质的不同表现形式,也就是说物质构成了世界。

自然界中所有物体均由化学元素及其化合物所组成,同样,各种固体材料也都是由一种或多种元素的原子结合而成的。

学习物质的原子结构和化学键合,是认识和研究各类材料在结构与性能方面所表现出来的个性和共性的基础,也是正确认识和理解材料的性能的重要依据。

材料科学与工程基础知识点(打印版)英汉双语版

材料科学与工程基础知识点(打印版)英汉双语版

Fundamentals of Materials Science and Engineering材料科学与工程基础知识点复习第一章绪论一、学习目的:材料科学家或工程技术人员经常遇到的问题是设计问题,而设计问题主要涉及机械、民用、化学和电。

而这些领域都要涉及到选择材料问题。

如何选择材料是非常重要的,选材包含两方面一个是满足性能要求,另一方面是成本低,即所谓“合理选材”。

材料的性能与其成分和内部的组织结构密切相关,材料的组织结构与加工过程有关。

本课程的目的就在于掌握加工过程和材料的组织结构以及性能之间的关系。

为今后进行材料设计和合理选材打下理论基础。

二、本章主要内容1、简介材料的发展史2、材料科学与工程的含义和内容3、材料的分类4、先进材料5、现代材料的需求三、重要术语和概念metal: 金属ceramic: 陶瓷polymer: 聚合物Composites: 复合材料Semiconductors: 半导体Biomaterials: 生物材料Processing: 加工过程Structure: 组织结构Properties: 性质Performance: 使用性能Mechanical properties: 力学性能Electrical properties: 电性能Thermal behavior: 热性能Magnetic properties: 磁性能Optical properties: 光性能Deteriorative characteristics: 老化特性第二章原子结构与化学键一、学习目的我们在自然界中观察到各种现象,归根结底是物质的不同表现形式,也就是说物质构成了世界。

自然界中所有物体均由化学元素及其化合物所组成,同样,各种固体材料也都是由一种或多种元素的原子结合而成的。

学习物质的原子结构和化学键合,是认识和研究各类材料在结构与性能方面所表现出来的个性和共性的基础,也是正确认识和理解材料的性能的重要依据。

《材料科学与工程基础》习题和思考题及答案

《材料科学与工程基础》习题和思考题及答案

《材料科学与工程基础》习题和思考题及答案第二章2-1.按照能级写出N、O、Si、Fe、Cu、Br原子的电子排布(用方框图表示)。

2-2.的镁原子有13个中子,11.17%的镁原子有14个中子,试计算镁原子的原子量。

2-3.试计算N壳层内的最大电子数。

若K、L、M、N壳层中所有能级都被电子填满时,该原子的原子序数是多少?2-4.计算O壳层内的最大电子数。

并定出K、L、M、N、O壳层中所有能级都被电子填满时该原子的原子序数。

2-5.将离子键、共价键和金属键按有方向性和无方向性分类,简单说明理由。

2-6.按照杂化轨道理论,说明下列的键合形式:(1)CO2的分子键合(2)甲烷CH4的分子键合(3)乙烯C2H4的分子键合(4)水H2O的分子键合(5)苯环的分子键合(6)羰基中C、O间的原子键合2-7.影响离子化合物和共价化合物配位数的因素有那些?2-8.试解释表2-3-1中,原子键型与物性的关系?2-9.0℃时,水和冰的密度分别是1.0005 g/cm3和0.95g/cm3,如何解释这一现象?2-10.当CN=6时,K+离子的半径为0.133nm(a)当CN=4时,半径是多少?(b)CN=8时,半径是多少?2-11.(a)利用附录的资料算出一个金原子的质量?(b)每mm3的金有多少个原子?(c)根据金的密度,某颗含有1021个原子的金粒,体积是多少?(d)假设金原子是球形(r Au=0.1441nm),并忽略金原子之间的空隙,则1021个原子占多少体积?(e)这些金原子体积占总体积的多少百分比?2-12.一个CaO的立方体晶胞含有4个Ca2+离子和4个O2-离子,每边的边长是0.478nm,则CaO的密度是多少?2-13.硬球模式广泛的适用于金属原子和离子,但是为何不适用于分子?2-14.计算(a)面心立方金属的原子致密度;(b)面心立方化合物NaCl的离子致密度(离子半径r Na+=0.097,r Cl-=0.181);(C)由计算结果,可以引出什么结论?2-15.铁的单位晶胞为立方体,晶格常数a=0.287nm,请由铁的密度算出每个单位晶胞所含的原子个数。

《材料科学与工程基础》习题和思考题及答案

《材料科学与工程基础》习题和思考题及答案

《材料科学与工程基础》习题和思考题及答案《材料科学与工程基础》习题和思考题及答案第二章 2-1.按照能级写出N、O、Si、Fe、Cu、Br原子的电子排布(用方框图表示)。

2-2.的镁原子有13个中子,11.17%的镁原子有14个中子,试计算镁原子的原子量。

2-3.试计算N壳层内的最大电子数。

若K、L、M、N壳层中所有能级都被电子填满时,该原子的原子序数是多少? 2-4.计算O壳层内的最大电子数。

并定出K、L、M、N、O壳层中所有能级都被电子填满时该原子的原子序数。

2-5.将离子键、共价键和金属键按有方向性和无方向性分类,简单说明理由。

2-6.按照杂化轨道理论,说明下列的键合形式:(1)CO2的分子键合(2)甲烷CH4的分子键合(3)乙烯C2H4的分子键合(4)水H2O的分子键合(5)苯环的分子键合(6)羰基中C、O间的原子键合 2-7.影响离子化合物和共价化合物配位数的因素有那些? 2-8.试解释表2-3-1中,原子键型与物性的关系? 2-9.0℃时,水和冰的密度分别是1.0005 g/cm3和0.95g/cm3,如何解释这一现象? 2-10.当CN=6时,K+离子的半径为0.133nm(a)当CN=4时,半径是多少?(b)CN=8时,半径是多少?2-11.(a)利用附录的资料算出一个金原子的质量?(b)每mm3的金有多少个原子?(c)根据金的密度,某颗含有1021个原子的金粒,体积是多少?(d)假设金原子是球形(rAu=0.1441nm),并忽略金原子之间的空隙,则1021个原子占多少体积?(e)这些金原子体积占总体积的多少百分比? 2-12.一个CaO的立方体晶胞含有4个Ca2+离子和4个O2-离子,每边的边长是0.478nm,则CaO的密度是多少? 2-13.硬球模式广泛的适用于金属原子和离子,但是为何不适用于分子?2-14.计算(a)面心立方金属的原子致密度;(b)面心立方化合物NaCl的离子致密度(离子半径rNa+=0.097,rCl-=0.181);(C)由计算结果,可以引出什么结论? 2-15.铁的单位晶胞为立方体,晶格常数 a=0.287nm,请由铁的密度算出每个单位晶胞所含的原子个数。

《材料科学与工程基础》习题和思考题及答桉.doc

《材料科学与工程基础》习题和思考题及答桉.doc

《材料科学与工程基础》习题和思考题及答案第二章2.1.按照能级写出N、0、Si、Fe、Cu、Br原子的电子排布(用方框图表示)。

2・2.的镁原子有13个中子,11.17%的镁原子有14个中子,试计算镁原子的原子量。

2.3.试计算N壳层内的最大电子数。

若K、L、M、N壳层中所有能级都被电子填满时,该原子的原子序数是多少?2-4.计算O壳层内的最大电子数。

并定出K、L、N、O壳层中所有能级都被电子填满时该原子的原子序数。

2-5.将离子键、共价键和金属键按有方向性和无方向性分类,简单说明理由。

2-6.按照杂化轨道理论,说明下列的键合形式:(1)CO?的分子键合(2)甲烷CH’的分子键合(3)乙烯C2H4的分子键合(4)水HQ的分子键合(5)苯环的分子键合(6)城基中C、。

间的原子键合2.7.影响离子化合物和共价化合物配位数的因素有那些?2-8.试解释表2.3-1中,原子键型与物性的关系?2-9.0°C时,水和冰的密度分别是1.0005 g/cm3和0.95g/cm3,如何解释这一现象?2.10.当CN=6时,K+离子的半径为0.133nm(a)当CN=4时,半径是多少? (b)CN=8时,半径是多少?2-11 .(a)利用附录的资料算出一•个金原子的质量?(b)每mm3的金有多少个原子?(c)根据金的密度,某颗含有10由个原子的金粒,体积是多少?(d)假设金原子是球形(E=0.1441nm), 并忽略金原子之间的空隙,则10刀个原子占多少体积?(e)这些金原子体积占总体积的多少百分比?12.—个CaO的立方体晶胞含有4个Ca*离子和4个O?-离子,每边的边长是0.478nm, 则CaO的密度是多少?-硬球模式广泛的适用于金属原子和离子,但是为何不适用于分子?14.计算(a)面心立方金属的原子致密度;(b)面心立方化合物NaCl的离子致密度(离子半径r Na+=0.097,心=0.181);(C)由计算结果,可以引出什么结论?2-15.铁的单位品胞为立方体,品格常数a=0.287nm,请由铁的密度算出每个单位品胞所含的原子个数。

材料科学与工程基础知识点(打印版)英汉双语版

材料科学与工程基础知识点(打印版)英汉双语版

Fundamentals of Materials Science and Engineering材料科学与工程基础知识点复习第一章绪论一、学习目的:材料科学家或工程技术人员经常遇到的问题是设计问题,而设计问题主要涉及机械、民用、化学和电。

而这些领域都要涉及到选择材料问题。

如何选择材料是非常重要的,选材包含两方面一个是满足性能要求,另一方面是成本低,即所谓“合理选材”。

材料的性能与其成分和内部的组织结构密切相关,材料的组织结构与加工过程有关。

本课程的目的就在于掌握加工过程和材料的组织结构以及性能之间的关系。

为今后进行材料设计和合理选材打下理论基础。

二、本章主要内容1、简介材料的发展史2、材料科学与工程的含义和内容3、材料的分类4、先进材料5、现代材料的需求三、重要术语和概念metal: 金属ceramic: 陶瓷polymer: 聚合物Composites: 复合材料Semiconductors: 半导体Biomaterials: 生物材料Processing: 加工过程Structure: 组织结构Properties: 性质Performance: 使用性能Mechanical properties: 力学性能Electrical properties: 电性能Thermal behavior: 热性能Magnetic properties: 磁性能Optical properties: 光性能Deteriorative characteristics: 老化特性第二章原子结构与化学键一、学习目的我们在自然界中观察到各种现象,归根结底是物质的不同表现形式,也就是说物质构成了世界。

自然界中所有物体均由化学元素及其化合物所组成,同样,各种固体材料也都是由一种或多种元素的原子结合而成的。

学习物质的原子结构和化学键合,是认识和研究各类材料在结构与性能方面所表现出来的个性和共性的基础,也是正确认识和理解材料的性能的重要依据。

材料科学与工程专业英语课后习题答案unit1-7

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)_南京航空航天大学中国大学mooc课后章节答案期末考试题库2023年

材料科学与工程基础(英文)_南京航空航天大学中国大学mooc课后章节答案期末考试题库2023年

材料科学与工程基础(英文)_南京航空航天大学中国大学mooc课后章节答案期末考试题库2023年1.The driving force for steady-state diffusion is the __________.答案:concentration gradient2.Diffusion coefficient is with the increasing diffusion temperature.答案:exponentially increased;3.Due to , alloys are usually than pure metals of the solvent.答案:solid solution strengthening, stronger;4.The finer the grains, the larger the , and .答案:strength, hardness, toughness;5.With plastic deformation,the increase of dislocationdensity will result in .答案:higher strength;6.In general, Brinell Hardness test is to measure thematerial’s hardness.答案:relatively softer7.Yield strength is corresponding to the occurrenceof deformation.答案:noticeable plastic8.Strain Hardening is also named as .答案:work hardening9.Vacancy diffusion is usually interstitial one.答案:slower than10.Edge and screw dislocations differ in what way?答案:angle between Burgers vector and line direction.11. A ____ may form when impurity atoms are added to a solid, in which case theoriginal crystal structure is retained and no new phases are formed.答案:solid solution12.One explanation for why graphite powder acts so well as a “solid lubricant”is .答案:carbon atoms in graphite are covalently bonded within planar layers but have weaker secondary bonds between layers13.Substitutional atom (impurity) is an example of ______.答案:point defect14.Interstitial solid solution belongs to .答案:finite solid solution;15.The atomic packing factor for FCC is .答案:0.7416.The coordination number of BCC crystal structure is .答案:817.The crystal structure of Cu is ?答案:FCC18.How many atoms does the face centered cubic unit cell contain?答案:Four19.If the electron configuration of Fe is 1s2 2s2 2p6 3s2 3p6 3d6 4s2, then theelectron configurations for the Fe3+ is 1s2 2s2 2p6 3s2 _____.答案:3p6 3d520.Bonds in most metals are referred to as ______.答案:Non-directional21.Covalent bonding occurs as a result of _________ sharing.答案:electron22.Which of the following is NOT an example of primary bonding?答案:Van der Waals23.Atomic weight (A) of an element corresponds to the weighted average of theatomic masses of the atom’s naturally occurring ___________.答案:isotopes24.The point on a phase diagram where the maximum number of allowablephases are in equilibrium is .答案:eutectic point25.Sterling silver (92.5%Ag/7.5%Cu) is an example of ___________.答案:Solid solution26.Engineering stress-strain curve and true stress-strain curve are equal up to .答案:Yeild point27.Among thefollowingtypical transformations of austenite in steels,____________transformation is diffusionless.答案:martensitic28.The heat-treatable aluminum alloy can be strengthened by .答案:Both of above29.In the as-quenched state, martensite is very hard and so brittle that a heattreatment known as must be accomplished sequently.答案:tempering30.During heat treatment of steel, austenite transforms into martensite by .答案:quenching31.Which of the following plane has the highest planar density for fcc.答案:(111)32.Which of the following describes recrystallization?答案:Diffusion dependent with no change in phase composition33.Heating the cold-worked metal progresses in three stages: .答案:recovery, recrystallization, grain growth;34.Strength is increased by making dislocation motion .答案:difficult35.The boundary above which only liquid phase exist is called _________.答案:liquidus36.We have an annealed carbon steel which has hardness of 150HBS. Supposewe know the hardness of Pearlite is 200HBS and the hardness of Ferrite is 80HBS, determine the carbon amount of this steel.答案:0.45%37.The maximum solubility of C in γ-austenite - solid solution is .答案:2.1438.In a plain steel that contains 0.2 percentage carbon, we should expect: .答案:a 25% pearlite and 75% pro-eutectoid ferrite39. A copper-nickel alloy is high-temperature heat treated; the diffusion of Cuinto Ni and Ni into Cu regions is referred to as _____________________.答案:Inter-diffusion40.The phase diagram of Sn-Pb alloy is called .答案:Eutectic phase diagram。

《材料科学与工程基础》英文影印版习题及思考题及答案

《材料科学与工程基础》英文影印版习题及思考题及答案

《材料科学与工程基础》英文习题及思考题及答案第二章习题和思考题Questions and Problems2.6 Allowed values for the quantum numbers ofelectrons are as follows:The relationships between n and the shell designationsare noted in Table 2.1.Relative tothe subshells,l =0 corresponds to an s subshelll =1 corresponds to a p subshelll =2 corresponds to a d subshelll =3 corresponds to an f subshellFor the K shell, the four quantum numbersfor each of the two electrons in the 1s state, inthe order of nlmlms , are 100(1/2 ) and 100(-1/2 ).Write the four quantum numbers for allof the electrons inthe L and M shells, and notewhich correspond to the s, p, and d subshells.2.7 Give the electron configurations for the followingions: Fe2+, Fe3+, Cu+, Ba2+,Br-, andS2-.2.17 (a) Briefly cite the main differences betweenionic, covalent, and metallicbonding.(b) State the Pauli exclusion principle.2.18 Offer an explanation as to why covalently bonded materials are generally lessdense than ionically or metallically bonded ones.2.19 Compute the percents ionic character of the interatomic bonds for the followingcompounds: TiO2 , ZnTe, CsCl, InSb, and MgCl2 .2.21 Using Table 2.2, determine the number of covalent bonds that are possible foratoms of the following elements: germanium, phosphorus, selenium, and chlorine.2.24 On the basis of the hydrogen bond, explain the anomalous behavior of waterwhen it freezes. That is, why is there volume expansion upon solidification?3.1 What is the difference between atomic structure and crystal structure?3.2 What is the difference between a crystal structure and a crystal system?3.4Show for the body-centered cubic crystal structure that the unit cell edge lengtha and the atomic radius R are related through a =4R/√3.3.6 Show that the atomic packing factor for BCC is 0.68. .3.27* Show that the minimum cation-to-anion radius ratio for a coordinationnumber of 6 is 0.414. Hint: Use the NaCl crystal structure (Figure 3.5), and assume that anions and cations are just touching along cube edges and across face diagonals.3.48 Draw an orthorhombic unit cell, and within that cell a [121] direction and a(210) plane.3.50 Here are unit cells for two hypothetical metals:(a)What are the indices for the directions indicated by the two vectors in sketch(a)?(b) What are the indices for the two planes drawn in sketch (b)?3.51* Within a cubic unit cell, sketch the following directions:.3.53 Determine the indices for the directions shown in the following cubic unit cell:3.57 Determine the Miller indices for the planesshown in the following unit cell:3.58Determine the Miller indices for the planes shown in the following unit cell: 3.61* Sketch within a cubic unit cell the following planes:3.62 Sketch the atomic packing of (a) the (100)plane for the FCC crystal structure, and (b) the (111) plane for the BCC crystal structure (similar to Figures 3.24b and 3.25b).3.77 Explain why the properties of polycrystalline materials are most oftenisotropic.5.1 Calculate the fraction of atom sites that are vacant for lead at its meltingtemperature of 327_C. Assume an energy for vacancy formation of 0.55eV/atom.5.7 If cupric oxide (CuO) is exposed to reducing atmospheres at elevatedtemperatures, some of the Cu2_ ions will become Cu_.(a) Under these conditions, name one crystalline defect that you would expect toform in order to maintain charge neutrality.(b) How many Cu_ ions are required for the creation of each defect?5.8 Below, atomic radius, crystal structure, electronegativity, and the most commonvalence are tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated.Which of these elements would you expect to form the following with copper:(a) A substitutional solid solution having complete solubility?(b) A substitutional solid solution of incomplete solubility?(c) An interstitial solid solution?5.9 For both FCC and BCC crystal structures, there are two different types ofinterstitial sites. In each case, one site is larger than the other, which site isnormally occupied by impurity atoms. For FCC, this larger one is located at the center of each edge of the unit cell; it is termed an octahedral interstitial site. On the other hand, with BCC the larger site type is found at 0, __, __ positions—that is, lying on _100_ faces, and situated midway between two unit cell edges on this face and one-quarter of the distance between the other two unit cell edges; it is termed a tetrahedral interstitial site. For both FCC and BCC crystalstructures, compute the radius r of an impurity atom that will just fit into one of these sites in terms of the atomic radius R of the host atom.5.10 (a) Suppose that Li2O is added as an impurity to CaO. If the Li_ substitutes forCa2_, what kind of vacancies would you expect to form? How many of thesevacancies are created for every Li_ added?(b) Suppose that CaCl2 is added as an impurity to CaO. If the Cl_ substitutes forO2_, what kind of vacancies would you expect to form? How many of thevacancies are created for every Cl_ added?5.28 Copper and platinum both have the FCC crystal structure and Cu forms asubstitutional solid solution for concentrations up to approximately 6 wt% Cu at room temperature. Compute the unit cell edge length for a 95 wt% Pt-5 wt% Cu alloy.5.29 Cite the relative Burgers vector–dislocation line orientations for edge, screw, andmixed dislocations.6.1 Briefly explain the difference between selfdiffusion and interdiffusion.6.3 (a) Compare interstitial and vacancy atomic mechanisms for diffusion.(b) Cite two reasons why interstitial diffusion is normally more rapid thanvacancy diffusion.6.4 Briefly explain the concept of steady state as it applies to diffusion.6.5 (a) Briefly explain the concept of a driving force.(b) What is the driving force for steadystate diffusion?6.6Compute the number of kilograms of hydrogen that pass per hour through a5-mm thick sheet of palladium having an area of 0.20 m2at 500℃. Assume a diffusion coefficient of 1.0×10- 8 m2/s, that the concentrations at the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.6.7 A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at 1200℃and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 6×10-11m2/s, and the diffusion flux is found to be 1.2×10- 7kg/m2-s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m3. How far into the sheet from this high-pressure side will the concentration be 2.0 kg/m3?Assume a linear concentration profile.6.24. Carbon is allowed to diffuse through a steel plate 15 mm thick. Theconcentrations of carbon at the two faces are 0.65 and 0.30 kg C/m3 Fe, whichare maintained constant. If the preexponential and activation energy are 6.2 _10_7 m2/s and 80,000 J/mol, respectively, compute the temperature at which the diffusion flux is 1.43 _ 10_9 kg/m2-s.6.25 The steady-state diffusion flux through a metal plate is 5.4_10_10 kg/m2-s at atemperature of 727_C (1000 K) and when the concentration gradient is _350kg/m4. Calculate the diffusion flux at 1027_C (1300 K) for the sameconcentration gradient and assuming an activation energy for diffusion of125,000 J/mol.10.2 What thermodynamic condition must be met for a state of equilibrium to exist? 10.4 What is the difference between the states of phase equilibrium and metastability?10.5 Cite the phases that are present and the phase compositions for the followingalloys:(a) 90 wt% Zn–10 wt% Cu at 400℃(b) 75 wt% Sn–25wt%Pb at 175℃(c) 55 wt% Ag–45 wt% Cu at 900℃(d) 30 wt% Pb–70 wt% Mg at 425℃(e) 2.12 kg Zn and 1.88 kg Cu at 500℃(f ) 37 lbm Pb and 6.5 lbm Mg at 400℃(g) 8.2 mol Ni and 4.3 mol Cu at 1250℃.(h) 4.5 mol Sn and 0.45 mol Pb at 200℃10.6 For an alloy of composition 74 wt% Zn–26 wt% Cu, cite the phases presentand their compositions at the following temperatures: 850℃, 750℃, 680℃, 600℃, and 500℃.10.7 Determine the relative amounts (in terms of mass fractions) of the phases forthe alloys and temperatures given inProblem 10.5.10.9 Determine the relative amounts (interms of volume fractions) of the phases forthe alloys and temperatures given inProblem 10.5a, b, and c. Below are given theapproximate densities of the various metalsat the alloy temperatures:10.18 Is it possible to have a copper–silveralloy that, at equilibrium, consists of a _ phase of composition 92 wt% Ag–8wt% Cu, and also a liquid phase of composition 76 wt% Ag–24 wt% Cu? If so, what will be the approximate temperature of the alloy? If this is not possible,explain why.10.20 A copper–nickel alloy of composition 70 wt% Ni–30 wt% Cu is slowly heatedfrom a temperature of 1300_C .(a) At what temperature does the first liquid phase form?(b) What is the composition of this liquid phase?(c) At what temperature does complete melting of the alloy occur?(d) What is the composition of the last solid remaining prior to completemelting?10.28 .Is it possible to have a copper–silver alloy of composition 50 wt% Ag–50 wt%Cu, which, at equilibrium, consists of _ and _ phases having mass fractions W_ _0.60 and W_ _ 0.40? If so, what will be the approximate temperature of the alloy?If such an alloy is not possible, explain why.10.30 At 700_C , what is the maximum solubility (a) of Cu in Ag? (b) Of Ag in Cu?第三章习题和思考题3.3If the atomic radius of aluminum is 0.143nm, calculate the volume of its unitcell in cubic meters.3.8 Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomicweight of 55.85 g/mol. Compute and compare its density with the experimental value found inside the front cover.3.9 Calculate the radius of an iridium atom given that Ir has an FCC crystal structure,a density of 22.4 g/cm3, and an atomic weight of 192.2 g/mol.3.13 Using atomic weight, crystal structure, and atomic radius data tabulated insidethe front cover, compute the theoretical densities of lead, chromium, copper, and cobalt, and then compare these values with the measured densities listed in this same table. The c/a ratio for cobalt is 1.623.3.15 Below are listed the atomic weight, density, and atomic radius for threehypothetical alloys. For each determine whether its crystal structure is FCC,BCC, or simple cubic and then justify your determination. A simple cubic unitcell is shown in Figure 3.40.3.21 This is a unit cell for a hypotheticalmetal:(a) To which crystal system doesthis unit cell belong?(b) What would this crystal structure be called?(c) Calculate the density of the material, given that its atomic weight is 141g/mol.3.25 For a ceramic compound, what are the two characteristics of the component ionsthat determine the crystal structure?3.29 On the basis of ionic charge and ionic radii, predict the crystal structures for thefollowing materials: (a) CsI, (b) NiO, (c) KI, and (d) NiS. Justify your selections.3.35 Magnesium oxide has the rock salt crystal structure and a density of 3.58 g/cm3.(a) Determine the unit cell edge length. (b) How does this result compare withthe edge length as determined from the radii in Table 3.4, assuming that theMg2_ and O2_ ions just touch each other along the edges?3.36 Compute the theoretical density of diamond given that the CUC distance andbond angle are 0.154 nm and 109.5°, respectively. How does this value compare with the measured density?3.38 Cadmium sulfide (CdS) has a cubic unit cell, and from x-ray diffraction data it isknown that the cell edge length is 0.582 nm. If the measured density is 4.82 g/cm3 , how many Cd 2+ and S 2—ions are there per unit cell?3.41 A hypothetical AX type of ceramic material is known to have a density of 2.65g/cm 3 and a unit cell of cubic symmetry with a cell edge length of 0.43 nm. The atomic weights of the A and X elements are 86.6 and 40.3 g/mol, respectively.On the basis of this information, which of the following crystal structures is (are) possible for this material: rock salt, cesium chloride, or zinc blende? Justify your choice(s).3.42 The unit cell for Mg Fe2O3 (MgO-Fe2O3) has cubic symmetry with a unit celledge length of 0.836 nm. If the density of this material is 4.52 g/cm 3 , compute its atomic packing factor. For this computation, you will need to use ionic radii listed in Table 3.4.3.44 Compute the atomic packing factor for the diamond cubic crystal structure(Figure 3.16). Assume that bonding atoms touch one another, that the angle between adjacent bonds is 109.5°, and that each atom internal to the unit cell is positioned a/4 of the distance away from the two nearest cell faces (a is the unit cell edge length).3.45 Compute the atomic packing factor for cesium chloride using the ionic radii inTable 3.4 and assuming that the ions touch along the cube diagonals.3.46 In terms of bonding, explain why silicate materials have relatively low densities.3.47 Determine the angle between covalent bonds in an SiO44—tetrahedron.3.63 For each of the following crystal structures, represent the indicated plane in themanner of Figures 3.24 and 3.25, showing both anions and cations: (a) (100)plane for the rock salt crystal structure, (b) (110) plane for the cesium chloride crystal structure, (c) (111) plane for the zinc blende crystal structure, and (d) (110) plane for the perovskite crystal structure.3.66 The zinc blende crystal structure is one that may be generated from close-packedplanes of anions.(a) Will the stacking sequence for this structure be FCC or HCP? Why?(b) Will cations fill tetrahedral or octahedral positions? Why?(c) What fraction of the positions will be occupied?3.81* The metal iridium has an FCC crystal structure. If the angle of diffraction forthe (220) set of planes occurs at 69.22°(first-order reflection) when monochromatic x-radiation having a wavelength of 0.1542 nm is used, compute(a) the interplanar spacing for this set of planes, and (b) the atomic radius for aniridium atom.4.10 What is the difference between configuration and conformation in relation topolymer chains? vinyl chloride).4.22 (a) Determine the ratio of butadiene to styrene mers in a copolymer having aweight-average molecular weight of 350,000 g/mol and weight-average degree of polymerization of 4425.(b) Which type(s) of copolymer(s) will this copolymer be, considering thefollowing possibilities: random, alternating, graft, and block? Why?4.23 Crosslinked copolymers consisting of 60 wt% ethylene and 40 wt% propylenemay have elastic properties similar to those for natural rubber. For a copolymer of this composition, determine the fraction of both mer types.4.25 (a) Compare the crystalline state in metals and polymers.(b) Compare thenoncrystalline state as it applies to polymers and ceramic glasses.4.26 Explain briefly why the tendency of a polymer to crystallize decreases withincreasing molecular weight.4.27* For each of the following pairs of polymers, do the following: (1) state whetheror not it is possible to determine if one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.(a) Linear and syndiotactic polyvinyl chloride; linear and isotactic polystyrene.(b) Network phenol-formaldehyde; linear and heavily crosslinked ci s-isoprene.(c) Linear polyethylene; lightly branched isotactic polypropylene.(d) Alternating poly(styrene-ethylene) copolymer; randompoly(vinylchloride-tetrafluoroethylene) copolymer.4.28 Compute the density of totally crystalline polyethylene. The orthorhombic unitcell for polyethylene is shown in Figure 4.10; also, the equivalent of two ethylene mer units is contained within each unit cell.5.11 What point defects are possible for MgO as an impurity in Al2O3? How manyMg 2+ ions must be added to form each of these defects?5.13 What is the composition, in weight percent, of an alloy that consists of 6 at% Pband 94 at% Sn?5.14 Calculate the composition, in weight per-cent, of an alloy that contains 218.0 kgtitanium, 14.6 kg of aluminum, and 9.7 kg of vanadium.5.23 Gold forms a substitutional solid solution with silver. Compute the number ofgold atoms per cubic centimeter for a silver-gold alloy that contains 10 wt% Au and 90 wt% Ag. The densities of pure gold and silver are 19.32 and 10.49 g/cm3 , respectively.8.53 In terms of molecular structure, explain why phenol-formaldehyde (Bakelite)will not be an elastomer.10.50 Compute the mass fractions of αferrite and cementite in pearlite. assumingthat pressure is held constant.10.52 (a) What is the distinction between hypoeutectoid and hypereutectoid steels?(b) In a hypoeutectoid steel, both eutectoid and proeutectoid ferrite exist. Explainthe difference between them. What will be the carbon concentration in each?10.56 Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 727_C(a) What is the proeutectoid phase?(b) How many kilograms each of total ferrite and cementite form?(c) How many kilograms each of pearlite and the proeutectoid phase form?(d) Schematically sketch and label the resulting microstructure.10.60 The mass fractions of total ferrite and total cementite in an iron–carbon alloyare 0.88 and 0.12, respectively. Is this a hypoeutectoid or hypereutectoid alloy?Why?10.64 Is it possible to have an iron–carbon alloy for which the mass fractions of totalferrite and proeutectoid cementite are 0.846 and 0.049, respectively? Why orwhy not?第四章习题和思考题7.3 A specimen of aluminum having a rectangular cross section 10 mm _ 12.7 mmis pulled in tension with 35,500 N force, producing only elastic deformation. 7.5 A steel bar 100 mm long and having a square cross section 20 mm on an edge ispulled in tension with a load of 89,000 N , and experiences an elongation of 0.10 mm . Assuming that the deformation is entirely elastic, calculate the elasticmodulus of the steel.7.7 For a bronze alloy, the stress at which plastic deformation begins is 275 MPa ,and the modulus of elasticity is 115 Gpa .(a) What is the maximum load that may be applied to a specimen with across-sectional area of 325mm, without plastic deformation?(b) If the original specimen length is 115 mm , what is the maximum length towhich it may be stretched without causing plastic deformation?7.8 A cylindrical rod of copper (E _ 110 GPa, Stress (MPa) ) having a yield strengthof 240Mpa is to be subjected to a load of 6660 N. If the length of the rod is 380 mm, what must be the diameter to allow an elongation of 0.50 mm?7.9 Consider a cylindrical specimen of a steel alloy (Figure 7.33) 10mm in diameterand 75 mm long that is pulled in tension. Determine its elongation when a load of 23,500 N is applied.7.16 A cylindrical specimen of some alloy 8 mm in diameter is stressed elasticallyin tension. A force of 15,700 N produces a reduction in specimen diameter of 5 _ 10_3 mm. Compute Poisson’s ratio for this material if its modulus of elasticity is 140 GPa .7.17 A cylindrical specimen of a hypothetical metal alloy is stressed in compression.If its original and final diameters are 20.000 and 20.025 mm, respectively, and its final length is 74.96 mm, compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 105 Gpa and 39.7 GPa,respectively.7.19 A brass alloy is known to have a yield strength of 275 MPa, a tensile strength of380 MPa, and an elastic modulus of 103 GPa . A cylindrical specimen of thisalloy 12.7 mm in diameter and 250 mm long is stressed in tension and found to elongate 7.6 mm . On the basis of the information given, is it possible tocompute the magnitude of the load that is necessary to produce this change inlength? If so, calculate the load. If not, explain why.7.20A cylindrical metal specimen 15.0mmin diameter and 150mm long is to besubjected to a tensile stress of 50 Mpa; at this stress level the resulting deformation will be totally elastic.(a)If the elongation must be less than 0.072mm,which of the metals in Tabla7.1are suitable candidates? Why ?(b)If, in addition, the maximum permissible diameter decrease is 2.3×10-3mm,which of the metals in Table 7.1may be used ? Why?7.22 Cite the primary differences between elastic, anelastic, and plastic deformationbehaviors.7.23 diameter of 10.0 mm is to be deformed using a tensile load of 27,500 N. It mustnot experience either plastic deformation or a diameter reduction of more than7.5×10-3 mm. Of the materials listed as follows, which are possible candidates?Justify your choice(s).7.24 A cylindrical rod 380 mm long, having a diameter of 10.0 mm, is to besubjected to a tensile load. If the rod is to experience neither plastic deformationnor an elongation of more than 0.9 mm when the applied load is 24,500 N,which of the four metals or alloys listed below are possible candidates?7.25 Figure 7.33 shows the tensile engineering stress–strain behavior for a steel alloy.(a) What is the modulus of elasticity?(b) What is the proportional limit?(c) What is the yield strength at a strain offset of 0.002?(d) What is the tensile strength?7.27 A load of 44,500 N is applied to a cylindrical specimen of steel (displaying thestress–strain behavior shown in Figure 7.33) that has a cross-sectional diameter of 10 mm .(a) Will the specimen experience elastic or plastic deformation? Why?(b) If the original specimen length is 500 mm), how much will it increase inlength when t his load is applied?7.29 A cylindrical specimen of aluminumhaving a diameter of 12.8 mm and a gaugelength of 50.800 mm is pulled in tension. Usethe load–elongation characteristics tabulatedbelow to complete problems a through f.(a)Plot the data as engineering stressversusengineering strain.(b) Compute the modulus of elasticity.(c) Determine the yield strength at astrainoffset of 0.002.(d) Determine the tensile strength of thisalloy.(e) What is the approximate ductility, in percent elongation?(f ) Compute the modulus of resilience.7.35 (a) Make a schematic plot showing the tensile true stress–strain behavior for atypical metal alloy.(b) Superimpose on this plot a schematic curve for the compressive truestress–strain behavior for the same alloy. Explain any difference between thiscurve and the one in part a.(c) Now superimpose a schematic curve for the compressive engineeringstress–strain behavior for this same alloy, and explain any difference between this curve and the one in part b.7.39 A tensile test is performed on a metal specimen, and it is found that a true plasticstrain of 0.20 is produced when a true stress of 575 MPa is applied; for the same metal, the value of K in Equation 7.19 is 860 MPa. Calculate the true strain that results from the application of a true stress of 600 Mpa.7.40 For some metal alloy, a true stress of 415 MPa produces a plastic true strain of0.475. How much will a specimen of this material elongate when a true stress of325 MPa is applied if the original length is 300 mm ? Assume a value of 0.25 for the strain-hardening exponent n.7.43 Find the toughness (or energy to cause fracture) for a metal that experiences bothelastic and plastic deformation. Assume Equation 7.5 for elastic deformation,that the modulus of elasticity is 172 GPa , and that elastic deformation terminates at a strain of 0.01. For plastic deformation, assume that the relationship between stress and strain is described by Equation 7.19, in which the values for K and n are 6900 Mpa and 0.30, respectively. Furthermore, plastic deformation occurs between strain values of 0.01 and 0.75, at which point fracture occurs.7.47 A steel specimen having a rectangular cross section of dimensions 19 mm×3.2mm (0.75in×0.125in.) has the stress–strain behavior shown in Figure 7.33. If this specimen is subjected to a tensile force of 33,400 N (7,500lbf ), then(a) Determine the elastic and plastic strain values.(b) If its original length is 460 mm (18 in.), what will be its final length after theload in part a is applied and then released?7.50 A three-point bending test was performed on an aluminum oxide specimenhaving a circular cross section of radius 3.5 mm; the specimen fractured at a load of 950 N when the distance between the support points was 50 mm . Another test is to be performed on a specimen of this same material, but one that has a square cross section of 12 mm length on each edge. At what load would you expect this specimen to fracture if the support point separation is 40 mm ?7.51 (a) A three-point transverse bending test is conducted on a cylindrical specimenof aluminum oxide having a reported flexural strength of 390 MPa . If the speci- men radius is 2.5 mm and the support point separation distance is 30 mm ,predict whether or not you would expect the specimen to fracture when a load of 620 N is applied. Justify your prediction.(b) Would you be 100% certain of the prediction in part a? Why or why not?7.57 When citing the ductility as percent elongation for semicrystalline polymers, it isnot necessary to specify the specimen gauge length, as is the case with metals.Why is this so?7.66 Using the data represented in Figure 7.31, specify equations relating tensilestrength and Brinell hardness for brass and nodular cast iron, similar toEquations 7.25a and 7.25b for steels.8.4 For each of edge, screw, and mixed dislocations, cite the relationship between thedirection of the applied shear stress and the direction of dislocation line motion.8.5 (a) Define a slip system.(b) Do all metals have the same slip system? Why or why not?8.7. One slip system for theBCCcrystal structure is _110__111_. In a manner similarto Figure 8.6b sketch a _110_-type plane for the BCC structure, representingatom positions with circles. Now, using arrows, indicate two different _111_ slip directions within this plane.8.15* List four major differences between deformation by twinning and deformationby slip relative to mechanism, conditions of occurrence, and final result.8.18 Describe in your own words the three strengthening mechanisms discussed inthis chapter (i.e., grain size reduction, solid solution strengthening, and strainhardening). Be sure to explain how dislocations are involved in each of thestrengthening techniques.8.19 (a) From the plot of yield strength versus (grain diameter)_1/2 for a 70 Cu–30 Zncartridge brass, Figure 8.15, determine values for the constants _0 and ky inEquation 8.5.(b) Now predict the yield strength of this alloy when the average grain diameteris 1.0 _ 10_3 mm.8.20. The lower yield point for an iron that has an average grain diameter of 5 _ 10_2mm is 135 MPa . At a grain diameter of 8 _ 10_3 mm, the yield point increases to 260MPa. At what grain diameter will the lower yield point be 205 Mpa ?8.24 (a) Show, for a tensile test, thatif there is no change in specimen volume during the deformation process (i.e., A0 l0 _Ad ld).(b) Using the result of part a, compute the percent cold work experienced bynaval brass (the stress–strain behavior of which is shown in Figure 7.12) when a stress of 400 MPa is applied.8.25 Two previously undeformed cylindrical specimens of an alloy are to be strainhardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are 16 mm and11 mm, respectively. The second specimen, with an initial radius of 12 mm, musthave the same deformed hardness as the first specimen; compute the secondspecimen’s radius after deformation.8.26 Two previously undeformed specimens of the same metal are to be plasticallydeformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular is to remain as such. Their original and deformeddimensions are as follows:Which of these specimens will be the hardest after plastic deformation, and why?8.27 A cylindrical specimen of cold-worked copper has a ductility (%EL) of 25%. Ifits coldworked radius is 10 mm (0.40 in.), what was its radius beforedeformation?8.28 (a) What is the approximate ductility (%EL) of a brass that has a yield strengthof 275 MPa ?(b) What is the approximate Brinell hardness of a 1040 steel having a yieldstrength of 690 MPa?8.41 In your own words, describe the mechanisms by which semicrystalline polymers(a) elasticallydeform and (b) plastically deform, and (c) by which elastomerselastically deform.8.42 Briefly explain how each of the following influences the tensile modulus of asemicrystallinepolymer and why:(a) molecular weight;(b) degree of crystallinity;(c) deformation by drawing;(d) annealing of an undeformed material;(e) annealing of a drawn material.8.43* Briefly explain how each of the following influences the tensile or yieldstrength of a semicrystalline polymer and why:(a) molecular weight;。

材料科学与工程基础英文版第五版课后习题

材料科学与工程基础英文版第五版课后习题

Homework 11.1 What are materials? List eight commonly encountered engineering materials. Answer1.1: Materials are substances of which something is composed or made. Steels, aluminum alloys, concrete, wood, glass, plastics, ceramics and electronic materials.1.2 What are the main classes of engineering materials?Answer1.2: Metallic, polymeric, ceramic, composite, and electronic materials are the five main classes.1.3 What are some of the important properties of each of the five main classes of engineering materials?Answer1.3:Metallic Materials• many are relatively strong and ductile at room temperature• some have good strength at high temperature• most have relatively high electrical and thermal conductivitiesPolymeric Materials• generally are poor electrical and thermal conductors• most have low to medium strengths• most have low densities• most are relatively easy to process into final shape• some are transparentCeramic Materials• generally have high ha rdness and are mechanically brittle• some have useful high temperature strength• most have poor electrical and thermal conductivitiesComposite Materials• have a wide range of strength from low to very high• some have very high strength-to-weight ratios (e.g. carbon-fiber epoxy materials)• some have medium strength and are able to be cast or formed into a variety of sha (e.g. fiberglass-polyester materials)• some have useable strengths at very low cos t (e.g. wood and concrete)Electronic Materials• able to detect, amplify and transmit electrical signals in a complex manner• are light weight, compact and energy efficient1.8 What are nanomaterials? What are some proposed advantages of using nanomaterials over their conventional counterparts?Answer1.8: Are defined as materials with a characteristic length scale smaller than 100 nm. The length scale could be particle diameter, grain size in a material, layer thicknessin a sensor, etc. These materials have properties different than that at bulk scale or at themolecular scale. These materials have often enhanced properties and characteristics because of their nano-features in comparison to their micro-featured counterparts. The structural, chemical, electronic, and thermal properties (among other characteristics) are often enhanced at the nano-scale.Homework 2Chapter 3, Problem 4What are the three most common metal crystal structures? List five metals that have each of these crystal structures. Chapter 3, Solution 4The three most common crystal structures found in metals are: body-centered cubic (BCC), face-centered cubic (FCC), and hexagonal close-packed (HCP). Examples of metals having these structures include the following. BCC:iron,α-vanadium, tungsten, niobium, and chromium.FCC: copper, aluminum, lead, nickel, and silver. HCP: magnesium, titanium,α-zinc, beryllium, and cadmium.Chapter 3, Problem 5For a BCC unit cell, (a) how many atoms are there inside the unit cell, (b) what is the coordination number for the atoms, (c) what is the relationship between the length of the side a of the BCC unit cell and the radius of its atoms, and (d) APF = 0.68 or 68%Chapter 3, Solution 5(a) A BCC crystal structure has two atoms in each unit cell. (b) A BCC crystal structure has a coordination number of eight . (c) In a BCC unit cell, one complete atom and two atom eighths toucheach other along the cube diagonal. This geometry translates into the relationship 4.R =Chapter 3, Problem 6For an FCC unit cell, (a) how many atoms are there inside the unit cell, (b) What is the coordination number for the atoms, (c) 24R a =, and (d) what is the atomic packing factor?Chapter 3, Solution 6(a) Each unit cell of the FCC crystal structure contains four atoms. (b) The FCC crystal structure has acoordination number of twelve . (d) By definition, the atomic packing factor is given as:volume of atoms in FCC unit cellAtomic packing factor volume of the FCC unit cell=These volumes, associated with the four-atom FCC unit cell, are33416433atoms V R R ππ⎡⎤==⎢⎥⎣⎦and 3unit cellV a =where a represents the lattice constant. Substitutinga =33unit cellV a ==The atomic packing factor then becomes,3316APF (FCC unit cell)3632R R ππ⎛⎫⎛⎫== ⎪ ⎪ ⎪⎝⎭⎝⎭=0.74 Chapter 3, Problem 7For an HCP unit cell (consider the primitive cell), (a) how many atoms are there inside the unit cell, (b) What is the coordination number for the atoms, (c) what is the atomic packing factor, (d) what is the ideal c/a ratio for HCP metals, and (e) repeat a through c considering the “larger” cell.Chapter 3, Solution 7The primitive cell has (a) two atoms/unit cell; (b) The coordination number associated with the HCP crystal structure is twelve . (c)the APF is 0.74 or 74%; (d) The ideal c/a ratio for HCP metals is 1.633; (e) all answers remain the same except for (a) where the new answer is 6.Homework 3 Chapter 3, Problem 25Lithium at 20︒C is BCC and has a lattice constant of 0.35092 nm. Calculate a value for the atomic radius of a lithium atom in nanometers.Chapter 3, Solution 25For the lithium BCC structure, which has a lattice constant of a = 0.35092 nm, the atomic radius is,R ===0.152 nmPalladium is FCC and has an atomic radius of 0.137 nm. Calculate a value for its lattice constant a in nanometers.Chapter 3, Solution 27Letting a represent the FCC unit cell edge length and R the palladium atomic radius,Chapter 3, Problem 31 Draw the following directions in a BCC unit cell and list the position coordinates of the atoms whose centers are intersected by the direction vector: (a ) [100] (b ) [110] (c ) [111]Chapter 3, Solution 31Chapter 3, Solution 324 or R a R ====0.387 nm(1, 0, 0)yxzyxz[111]x = +1y = -1 z = -1 x = +1 y = -1 z = 0(a) (b)[110]x = -½ y = 1(c)x = – ⅓ y = – ⅓(d)A cubic plane has the following axial intercepts: . What are the Miller indicesof this plane?Chapter 3, Solution 46Given the axial intercepts of (⅓, -⅔, ½), the reciprocal intercepts are:Multiplying by 2 to clear the fraction, the Miller indices areChapter 3, Problem 50Determine the Miller indices of the cubic crystal plane that intersects the following positioncoordinates:Chapter 3, Solution 50First locate the three position coordinates as shown. Next, connect points a and b and extend the line to point d . Complete the plane by connecting point d to c and point c to b . Using (1, 0, 1) as the plane origin, x = -1, y = 1 and z = –1. The intercept reciprocals are thusThe Miller indices are121332, , a b c ==-=11313,, 2.2x y z ==-=.(634)1122(, 0, ); (0,0,1); (1,1,1).1111,1, 1.x y z =-==-.(111)a(½, 0, ½ )。

材料科学与工程基础知识点(打印版)英汉双语版

材料科学与工程基础知识点(打印版)英汉双语版

Fundamentals of Materials Science and Engineering材料科学与工程基础知识点复习第一章绪论一、学习目的:材料科学家或工程技术人员经常遇到的问题是设计问题,而设计问题主要涉及机械、民用、化学和电。

而这些领域都要涉及到选择材料问题。

如何选择材料是非常重要的,选材包含两方面一个是满足性能要求,另一方面是成本低,即所谓“合理选材”。

材料的性能与其成分和内部的组织结构密切相关,材料的组织结构与加工过程有关。

本课程的目的就在于掌握加工过程和材料的组织结构以及性能之间的关系。

为今后进行材料设计和合理选材打下理论基础。

二、本章主要内容1、简介材料的发展史2、材料科学与工程的含义和内容3、材料的分类4、先进材料5、现代材料的需求三、重要术语和概念metal: 金属ceramic: 陶瓷polymer: 聚合物Composites: 复合材料Semiconductors: 半导体Biomaterials: 生物材料Processing: 加工过程Structure: 组织结构Properties: 性质Performance: 使用性能Mechanical properties: 力学性能Electrical properties: 电性能Thermal behavior: 热性能Magnetic properties: 磁性能Optical properties: 光性能Deteriorative characteristics: 老化特性第二章原子结构与化学键一、学习目的我们在自然界中观察到各种现象,归根结底是物质的不同表现形式,也就是说物质构成了世界。

自然界中所有物体均由化学元素及其化合物所组成,同样,各种固体材料也都是由一种或多种元素的原子结合而成的。

学习物质的原子结构和化学键合,是认识和研究各类材料在结构与性能方面所表现出来的个性和共性的基础,也是正确认识和理解材料的性能的重要依据。

材料科学与工程专业英语课后答案

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”。

材料科学基础英文版习题

材料科学基础英文版习题

What is the difference between Vickers and Knoop hardness tests? Chapter 6, Problem 13 Describe the slip mechanism that enables a metal to be plastically deformed without fracture. Chapter 6, Problem 14
ChapБайду номын сангаасer 6,Problem 15 (a) What are the principal slip planes and slip directions for FCC metals? (b) What are the principal slip planes and slip directions for BCC metals? (c) What are the principal slip planes and slip directions for HCP metals? Chapter 6, Problem 16 Describe the deformation twinning process that occurs in some metals when they are plastically deformed. Chapter 6, Problem 17 What is the difference between the slip and twinning mechanisms of plastic deformation of metals? Chapter 6,Problem 18 What other types of slip planes are important other than the basal planes for HCP metals with low c/a ratios? Chapter 6,Problem 19 What important role does twinning play in the plastic deformation of metals with regard to deformation of metals by slip? Chapter 6, Problem 20 Define the critical resolved shear stress for a pure metal single crystal? What happens to the metal from the macroscale point of view and behavior point of view once critical resolved shear stress is exceeded? Chapter 6, Problem 21 By what mechanism do grain boundaries strengthen metals? Chapter 6, Problem 22 (a) What is solid-solution strengthening? Describe the two main types, (b) What are two
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《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;。

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