材料科学基础2版余永宁 (8)

合集下载

材料科学基础(第2版)

《材料科学基础（第2版）》共分10章，教学时数不变；删去《材料科学基础（第1版）》中的热加工专业编写的金属材料与热处理部分，并对原书第六章、第七章及第十章重新编写。第六章改为“材料的凝固与气相沉积”。传统的冶金工程只讲材料的凝固，已不能适应材料科学的发展，所以在第六章中增加了气固转就，并介绍了沉积法在一些重要领域的应用。
教学资源
教学资源
《材料科学基础（第2版）》开通了有数字资源的下载，其中包含《材料科学基础（第2版）》（各章知识点）（教学大纲）、《材料科学基础（第2版）》（测验题及其参考答案）、《材料科学基础（第2版）》（多媒体电子课件）（PPT课件）。
作者简介
作者简介
石德珂Байду номын сангаас西安交通大学材料系副教授。
全书共10章，包含材料结构概论、晶体结构、高分子材料结构、晶体缺陷、相结构与相图等材料科学与工程专业的技术知识，还阐述材料的成分、组织结构与性能间关系的基本原理。
成书过程
成书过程
《材料科学基础（第1版）》是遵照1996年12月全国高校材料工程类专业教学指导委员会的决定编写的。当时考虑到专业改造与发展，将该书定为原“金属材料及热处理”和原“金属热加工”两类专业共用的技术基础课教材，教学时数为100至120，几年的教学实践表明，多数院校的热加工专业已将该课程时数减少了许多，以致无法共用。故这次按“材料科学与工程”专业的要求修订《材料科学基础（第2版）》。
2019年12月2日，《材料科学基础（第2版）》由石德珂主编，机械工业出版社出版。
内容简介
内容简介
《材料科学基础（第2版）》除导论外共十章，内容包括：材料结构概论、晶体结构、高分子材料结构、晶体缺陷、相结构与相图、材料的凝固与气相沉积、扩散与固态相变、变形与断裂、材料的电子结构与物理性能及材料概论。

武汉理工大学材料科学基础(第2版)课后习题和答案

武汉理工大学材料科学基础(第2版)课后习题和答案第一章绪论1、仔细观察一下白炽灯泡，会发现有多少种不同的材料？每种材料需要何种热学、电学性质？2、为什么金属具有良好的导电性和导热性？3、为什么陶瓷、聚合物通常是绝缘体？4、铝原子的质量是多少？若铝的密度为2.7g/cm3，计算1mm3中有多少原子？5、为了防止碰撞造成纽折，汽车的挡板可有装甲制造，但实际应用中为何不如此设计？说出至少三种理由。

6、描述不同材料常用的加工方法。

7、叙述金属材料的类型及其分类依据。

8、试将下列材料按金属、陶瓷、聚合物或复合材料进行分类：黄铜钢筋混凝土橡胶氯化钠铅-锡焊料沥青环氧树脂镁合金碳化硅混凝土石墨玻璃钢9、Al2O3陶瓷既牢固又坚硬且耐磨，为什么不用Al2O3制造铁锤？第二章晶体结构1、解释下列概念晶系、晶胞、晶胞参数、空间点阵、米勒指数（晶面指数）、离子晶体的晶格能、原子半径与离子半径、配位数、离子极化、同质多晶与类质同晶、正尖晶石与反正尖晶石、反萤石结构、铁电效应、压电效应.2、（1）一晶面在x、y、z轴上的截距分别为2a、3b、6c，求出该晶面的米勒指数；（2）一晶面在x、y、z 轴上的截距分别为a/3、b/2、c，求出该晶面的米勒指数。

3、在立方晶系的晶胞中画出下列米勒指数的晶面和晶向：（001）与[210]，（111）与[112]，（110）与[111]，（322）与[236]，（257）与[111]，（123）与[121]，（102），（112），（213），[110]，[111]，[120]，[321]4、写出面心立方格子的单位平行六面体上所有结点的坐标。

5、已知Mg2+半径为0.072nm，O2-半径为0.140nm，计算MgO晶体结构的堆积系数与密度。

6、计算体心立方、面心立方、密排六方晶胞中的原子数、配位数、堆积系数。

7、从理论计算公式计算NaC1与MgO的晶格能。

MgO的熔点为2800℃，NaC1为80l℃, 请说明这种差别的原因。

《材料科学基础2》课程简介和教学大纲

《材料科学基础2》课程简介课程编号：02024036课程名称：材料科学基础2 ［5E］ /Fundamentals of MaterialsScience 2学分：2. 5学时：40适用专业：无机非金属材料建议修读学期：第5学期先修课程：物理化学，材料科学基础1 ［无］考核方式与成绩评定标准：闭卷考试教材与主要参考书目：Ll］无机材料学基础，张其土，华东理工大学出版社［2］无机材料科学基础，陆佩文，武汉理工大学出版社［3］材料科学基础，张联盟，武汉理工大学出版社内容概述：本课程是无机非金属材料工程专业本科生的重要专业基础课，是一门理论性很强、涉及面广的课程，是本专业的专业课开设前所必须学的课程。

本课程是使学生掌握材料的组成、结构与性能之间的相互关系和变化规律，掌握材料的结构、物性和化学反应的规律及其相互的联系，为今后从事夏杂的技术工作和开发新型材料打下良好的基础。

The course of fUndamentals of materials science, which is highly theoretical, and almost involves all the sides of materials science, is an important fundamental one for the students majoring in inorganic materials science and engineering. Thus it is set to be taught before other specialized courses. It aims at allowing the students to master the relations between materials compositions, structures and properties, and to establish a good theoretical base for the research and development of new materials in the future.《材料科学基础2》［无］教学大纲课程编号：02024036课程名称：材料科学基础2 /Fundamentals of Materials Science 2学分：2. 5学时：40适用专业：无机非金属材料建议修读学期：第5学期先修课程：物理化学，材料科学基础1 ［无］一、课程性质、目的与任务【课程性质】本课程是无机非金属材料工程专业(建材方向、陶瓷与耐火材料方向)本科生的重要专业基础课，是一门理论性很强、涉及面广的课程，是本专业的专业课开设前所必须学的课程。

《材料科学基础2料》课程简介和教学大纲

《材料科学基础2 ［料］》课程简介课程编号：02034019课程名称：材料科学基础B2∕Fundamental of Material Science B2学分：3学时：48适用专业：材料科学与工程建议修读学期：5先修课程：物理化学，材料科学基础1考核方式与成绩评定标准：课程考核成绩采用平时成绩+期终考试成绩相结合的方式，平时成绩占课程考核成绩的20%,平时成绩考核采用考勤、作业和课堂提问相结合的方式；期终考试成绩占课程考核成绩的80%教材与主要参考书目：【教材】材料科学基础（第三版），胡廉祥、蔡南、戎咏华，上海交通大学出版社，2010 【参考书目】1.材料科学基础，余永宁，高等教育出版社，20062.材料科学基础，潘金生，清华大学出版社，20113. Fundamentals of Materials Science and Engineering: An Integrated Approach, William D. Callister, David G. Rethwisch, 2012 内容概述：材料科学基础2的课程教学内容主要为材料的形变和再结晶，单组元相图及纯金属的凝固，二元相图，三元相图以及材料的亚稳态和功能特性等。

通过本课程的学习，可使学生掌握固体材料变形的基本方式、金属及合金强化机理；掌握结晶的基本过程、热力学条件、形核及长大规律、凝固理论的应用；掌握相图的基本知识，二元相图的基本类型，分析与使用方法，熟练应用铁碳相图；掌握三元相图类型、分析方法、等温截面、变温截面等。

为后续专业课的学习打下牢固的基础。

The contents of Fundamental of Material Science 2are the deformation and recrystallization of materials, single phase diagram and pure metals solidification, two phases diagram, three phases diagram, the metastable phase and functional characteristics of materials, etc. By studying of this course, the students can master the basic deformation methods of materials, the strengthening mechanism of metals and alloys, the basic processing of crystallization, thermodynamic of crystallization, nucleation and growth of crystal, the application of solidification theory, the basicknowledge of phase diagram, the basic types of two phases diagram, the application of Fe-C phase diagram, the analysis of three phases diagram, isothermal section and variable section. This course lays a solid foundation for the following courses.《材料科学基础2［料］》教学大纲课程编号：02034019课程名称：材料科学基础B2∕Fundamental of Material Science B2学分：3学时：48适用专业：材料科学与工程建议修读学期：5先修课程：物理化学，材料科学基础1一、课程性质、目的与任务【课程性质】本课程是材料科学与工程专业的主要专业基础课之一。

材料科学基础第二版

材料科学基础第二版材料科学是一门研究材料的性质、结构、制备和应用的学科，它涉及到物质的基本特性和相互作用，对于现代工业和科技的发展起着至关重要的作用。

本书《材料科学基础第二版》旨在系统介绍材料科学的基本理论和知识，帮助读者全面了解材料科学的基本概念和原理，为相关专业的学生和科研人员提供一本全面而深入的参考书籍。

第一章从材料科学的基本概念和发展历程入手，介绍了材料科学的研究对象、基本特征以及其在工程技术中的应用。

通过对材料科学的起源和发展进行梳理，读者可以更好地理解材料科学的学科内涵和研究意义。

第二章主要介绍了材料的结构与性能。

材料的性能直接受其结构的影响，因此了解材料的结构对于预测和改善材料的性能至关重要。

本章详细介绍了晶体结构、非晶态结构以及材料的力学性能、热学性能等方面的知识，为读者提供了全面的材料结构与性能的基础知识。

第三章涉及了材料的制备与加工技术。

材料的制备和加工是材料科学的重要内容之一，它直接影响着材料的性能和应用。

本章主要介绍了材料的制备方法、加工工艺以及相关的材料表征技术，为读者提供了全面了解材料制备与加工技术的知识基础。

第四章讨论了材料的性能测试与评价。

材料的性能测试是材料科学研究的重要手段，通过对材料性能的测试和评价，可以全面了解材料的特性和应用潜力。

本章详细介绍了材料性能测试的方法、技术以及测试结果的分析与评价，为读者提供了全面了解材料性能测试与评价的知识基础。

第五章介绍了材料的应用与发展。

材料的应用是材料科学研究的最终目的，本章主要介绍了材料在工程技术、电子材料、光学材料、生物材料等方面的应用，并展望了材料科学的未来发展方向。

通过对材料科学基础的系统介绍，本书旨在帮助读者全面了解材料科学的基本理论和知识，为相关专业的学生和科研人员提供一本全面而深入的参考书籍。

希望本书能够成为读者学习和研究材料科学的重要工具，为材料科学的发展做出贡献。

材料科学基础2版余永宁 (6)

平衡态下的点缺陷浓度是很低的，金属晶体在接近熔点温度时，空位浓度约在103104范围，在低温时就更低了。
虽然空位浓度不高，但是，它在物质的输运过程起很大的作用。在另一些场合，例如在辐照条件下，受高能粒子碰击；离子注入表面、冷加工以及从高温快冷（淬火）等情况，会产生大量的非平衡的点缺陷，特别在高能粒子的持续碰击会做成材料的肿胀（空位凝聚称空洞）等损伤，甚至可以把原来的晶态变成非晶态。
元素
W Mo
-Fe
Ni Cu Au Al Pb Zn Cd
晶体结构
b.c.c b.c.c b.c.c f.c.c f.c.c f.c.c f.c.c f.c.c h.c.p h.c.p
hf , eV 3.7 3.2 1.5 1.6 1.22 0.96 0.68 0.58 0.53 0.41
sf/k
2 — — — 2 1 1 2 — —
6.1.2 离子晶体中的点缺陷
ቤተ መጻሕፍቲ ባይዱ
离子晶体至少含两种带反号电荷的离子，各种离子组成各自的亚点阵，离子晶体可看作是这些亚点阵穿插所构成。这样，离子晶体的点缺陷与金属晶体中的不同，它的差异来源于它的多原子性、缺陷带电和晶体保持电中性的要求。
引入Kröger-Vink符号描述
Z反映在Y位置相对与正常离
材料科学基础
北京科技大学材料科学与工程学院
编制
第6章
有序介质中的点缺陷和线缺陷
完美和无缺陷的晶体只是一个理论上的概念，晶体永远是不完整的。不完整性所占的分数（原子分数或体积分数）非常小，但对于材料的很多性能来说，起主要作用的却是这些非完整性，晶体的完整性只居于次要的地位。
按缺陷引起严重畸变的范围大小缺陷分
空位的形成焓Hf ；同时，也使得空位周围的原子振动模式有所改变，从而相应改变了熵值，这就是空位的形成熵Sf 。所以，一个空位的形成自由能Gf为

材料科学基础课后习题答案(部分)_第2版_西安交通大学_石德珂主编

9.半径为r1的位错环。10.自己看
11.（1）能，因为能量降低
（2）
b
a 3
111 ,
Frank不全位错
15.
在
111
面内：a 2
110
a 6

211
层错
a 6
121
在111
面内：a 2
110
a 6
121
层错
a 6
211
12.自己看 13.不用看 14自己看 16-20不用看
6.
x 0.25 0.7333 x y
12.（1）0.097+0.181=0.278
(2)R 0.2782 2 0.393nm (3) 0.2782 2 0.2782 0.481nm (4) (0.278 2)2 0.2782 0.622nm
13.
m V
(35.453 22.99) 4 6.021023 (0.278 2)3 1021
6.69 Fe3C% 100% 55.17% 44.83%
CV
Aexp( EV kT
)
每个原子的质量是： 107.9g / mol 6.021023个 / mol
1.791022
g
/
个
1cm3的原子
数为
： 9.58g / cm3 1.791022 g /
个
5.351022 个
/
cm3
CV
2.N N0 6.021023 9.58106 5.341028 / m3
M
107.9
ne exp
N
kT
kT ln ne
N
1.381023 1073ln
3.6 1023 5.34 1028

材料科学基础——塑性变形

滑移带
Slip band
单晶体的塑性变形
Plastic deformation of single crystals 3. 临界分切应力 (Critical resolving shear stress)
6.2
单晶体的塑性变形
Plastic deformation of single crystals 临界分切应力
单晶体的塑性变形
Plastic deformation of single crystals 2. 滑移带与滑移线(Slip band and Slip line)
6.2
Al 单晶
单晶体的塑性变形
Plastic deformation of single crystals 滑移线
6.2
Slip line
单晶体的塑性变形
Plastic deformation of single crystals
6.2
扭折(Kink)
单晶体的塑性变形
Plastic deformation of single crystals
6.2
6.2.2 滑移(Slip)
滑移：外力作用下晶体的一部分相对于另一部分沿一定晶面和晶向发生滑动位移，且不破坏晶体内部原子排列规律性的塑性变形机制。
成分和组织：金属越纯，塑性越好
晶粒度：晶粒细小，强度、塑韧性均好外因：温度：低温易脆断应力状态和裂纹：微裂纹大，拉应力状态，易脆断
应变速率：应变速率大，易发生脆性断裂
单晶体的塑性变形
Plastic deformation of single crystals
6.2
滑移 (Slip)
孪生(Twin)
即使完全消除装置的各种阻力其摆动振幅也会

材料科学基础第二版答案

材料科学基础第二版答案材料科学基础是材料科学与工程专业的入门课程，它为学生提供了材料科学的基本概念、原理和知识体系。

本文档将为您提供材料科学基础第二版的答案，希望能够对您的学习和教学有所帮助。

第一章，材料科学基础概论。

1. 什么是材料科学？材料科学是研究材料的结构、性能、制备和应用的学科，它涉及金属、陶瓷、高分子材料等各种材料的研究和开发。

2. 材料的分类有哪些？材料可以分为金属材料、无机非金属材料和有机高分子材料三大类，每一类又可以进一步细分。

3. 材料的性能指标有哪些？材料的性能指标包括力学性能、物理性能、化学性能、热学性能等多个方面。

第二章，晶体结构。

1. 什么是晶体？晶体是由原子或分子按一定的规则排列而成的固体，具有规则的几何形状和周期性的结构。

2. 晶体结构的分类有哪些？晶体结构可以分为离子晶体、共价晶体、金属晶体和分子晶体四种类型，每一种类型都有其特定的结构特点和性质。

3. 晶体缺陷对材料性能有何影响？晶体缺陷会对材料的机械性能、热学性能、电学性能等产生影响，了解晶体缺陷对材料设计和制备具有重要意义。

第三章，材料的物理性能。

1. 材料的密度如何影响其性能？材料的密度直接影响其质量和体积，对材料的力学性能、热学性能等有重要影响。

2. 材料的热膨胀系数是什么？材料的热膨胀系数是材料在温度变化时长度变化的比例，对材料的热胀冷缩性能有重要影响。

3. 材料的导热性能和电导率有何关系？材料的导热性能和电导率都与材料内部的电子、原子结构密切相关，了解二者之间的关系对材料的应用具有指导意义。

第四章，材料的力学性能。

1. 材料的弹性模量是什么？材料的弹性模量是材料在受力时表现出的弹性变形能力，是衡量材料刚度的重要参数。

2. 材料的屈服强度和抗拉强度有何区别？材料的屈服强度是材料在受力时开始产生塑性变形的应力值，而抗拉强度是材料在拉伸断裂时所承受的最大应力值。

3. 材料的硬度测试方法有哪些？材料的硬度测试方法包括布氏硬度、洛氏硬度、维氏硬度等多种方法，每种方法都有其适用的范围和特点。

工程材料科学与基础余永宁答案

Chapter 6 - Problem Solutions1. FIND: Compare the structure of a glass to that of the liquid.GIVEN: Both are noncrystalline.SOLUTION: The structure of a glass is essentially that of a frozen liquid. There is SRO but no LRO. Since the glass is at a lower temperature than the liquid of the same composition andmost materials contrast as they are cooled, the density of the glass is usually less than that of the liquid. The density of the glass is usually considerably greater than that of the crystal, however.Density is mass per unit volume, and the units we see most often are g/cm3. The most common units of reciprocal density are cm3/g. This is volume per unit mass, or specific volume. It isfrequently used by chemical physicists, who study noncrystalline materials.2.3. FIND: How does C p of an amorphous material change as the temperature is increased throughthe glass transition temperature?SOLUTION: Heat capacity is an intensive property, one that does depend on the bulkproperties of the material. Most thermodynamic intensive properties behave exactly the sameas (molar) volume in the vicinity of the glass transition temperature. Hence, heat capacitychanges slope through the T g.4.5. FIND: Estimate the volume thermal expansion coefficient of a glass.GIVEN: Its linear thermal expansion coefficient in the melt is 10 x 10-6 o C-1.ASSUMPTIONS: The material behaves typically, so that the thermal expansion coefficient of the glass is about 1/3 of that of the melt.SOLUTION: The linear thermal expansion coefficient is αth and the volumetric thermalexpansion coefficient is αv.αth(glass) ≈αth(melt)/3 and αv(glass) ≈ 3αth(glass)Hence,αv(glass) ≈ 3αth(melt) = 10 x 10-6 o C-16. FIND: Derive the relationship between αth and αv: αth/αv = 1/3SOLUTION: consider a cube of materials, length 1 on a side. With heat, the materialsexpands isotropically to length 1+δ1. We do the problem first using differentials.Now with deltas: V = (ι + δι)3 = 13 + 3ι2δι + ...⇒∆V ≈ 3ι2δι. Therefore,9. FIND: Is T g a temperature or range of temperatures?SOLUTION: Although we often cite a glass transition temperature, the glass transition occurs over a range of temperatures. T g is rate sensitive and structure sensitive; it depends on the rate of heating or cooling and on the local structure which is statistically variable in a glass or melt. 10. FIND: The temperature range for transitions that involve units smaller than a mer.SKETCH:SOLUTION: In a polymer, the repeat unit is a mer. If the repeat unit gains mobility, then the entire molecule and all of its parts are mobile. It requires less thermal input, kT, to excitesmaller units, such as rotation of the ring side group in the polystyrene shown in the figure.Hence, such transition are sub-T g.COMMENTS: Transitions in the crystalline regions can occur above T g and below T m.11. FIND: Design a rubber gasket for use in outer space.GIVEN: Outer space can fluctuate between cold and hot, and the atmosphere depends on the location in space. Solar radiation is very strong in space.ASSUMPTIONS: We'll design for space being a vacuum.SOLUTION: Fortunately the seal will probably never see solar radiation, so this is not a design problem. That is fortunate, since flexible materials (polymers) do not stand up to radiation.(Look at what happens to your skin when you expose it to bright sunshine.) Temperature can also be a problem. High temperatures can soften, melt or degrade polymers. Low temperature can change a rubber to a glass, making a material that is flexible at room temperature to brittle at high temperature. Rubber seals that have become glassy may break or leak. Metal seals will decrease in volume as the temperature is decreased. If your choice is a rubber gasket, then itneeds to remain flexible at use temperature. Liquid oxygen is very cold and embrittles manymaterials.12. FIND: Provide examples of materials that behave like silly putty.GIVEN: Silly putty is used at a temperature that it sometimes behaves in a fluid-like manner and sometimes in a solid-like manner.SOLUTION: There are many such examples, but they are sometimes hard to identify. Consider these materials:1. Plumbers putty2. Rubber mounts for car engines and vibrating machines3. Rubber bumpers4. Water. (Consider jumping off a 100 foot cliff into a water-filled quarry)5. Bullet-proof vest (Textile-like in ordinary use and bullet-proof when necessary)COMMENTS: For a number of applications we need materials with similar but differentproperties - a material that flows under low stress but does not flow under its own weight(Bingham plastic). Plumbers putty is an example, but it also shows the characteristics of sillyputty.13. FIND: Is it unique that motor oils do not thin with increasing temperature?SOLUTION: Recall equation 6.3-5b:η = ηo exp (Q/RT).This equation tells you that as temperature increases, viscosity decreases exponentially. Ifmotor oil does not behave this way, then it behaves in an unusual manner.COMMENTS: Motor oil, in fact, does behave in an unusual manner. It's viscosity isessentially constant with temperature! Let me try to explain how this is accomplished. Oilcontains polymer molecules in a solvent. The molecules do not like the solvent all that much, so they tend to ball up or coil somewhat tightly on themselves. As the temperature is increased, the interaction between solvent and polymer changes. The polymer begins to like the solvent, so it uncoils. The polymer molecules then become entangled in one another, raising the viscosity,which counteracts the normal decrease with increasing temperature.14. FIND: Whether it is more difficult to obtain a GIVEN: shear strain rate with a high viscosityfluid or a low viscosity fluid.ASSUMPTIONS: The oils behave in a similar fashion in a stress field.SKETCH:SOLUTION:Newton’s Law of Viscosity states that shear stress is proportional to velocitygradient. The constant of proportionality is viscosity: τ = η(dv/dx). Since the velocity gradient is invariant in this problem, the shear stress varies with viscosity. The higher viscosity oil willrequire a larger stress to maintain the plate velocity. Skotch® tape is a polymer film with a thin coating of an oil-like material. It is simply the thinness of the oil film and its viscosity thatprevents slippage between substrate and film.COMMENTS: When a large stress is required to shear a fluid, then much work is lost so that heat is generated in the fluid.15. FIND: Explain how viscosity changes as a material is crystallized or solidifies as a glass.SOLUTION: Let us first consider what occurs when a material like molasses or honey is cooled.As cooling proceeds, the material gets "thicker and thicker". Technically, we mean that theviscosity decreases with temperature. Eventually the material is so hard that we say it is frozen.What we mean really is that we are below the glass transition temperature. Thus, viscositydecreases many orders of magnitude as molasses, or any materials that does not crystallize, iscooled from the fluid-like state to the rock-hard state. Now consider a material that crystallizes, say, water, since we are all familiar with it. As water is cooled, it changes viscosity very little. At 0︒C both water and ice coexist. Below 0︒C only ice exists. Thus, the viscosity of H2O changes orders of magnitude at 0︒C. Unlike materials that do not crystallize, viscosity changes orders of magnitude over a very narrow temperature range - less than a degree.16. FIND: Show the atactic and isotactic configurations of PP. Which is more likely to besemicrystalline?GIVEN: The structure of PP is shown in Table 6.4-1. The structure of atactic and isotacticpolymer is shown in Fig. 6.4-5.SKETCH: i-PP has the methyl groups all on the same side. (The H side groups are not shown, but each carbon is bonded to 4 atoms. You should mentally visualize all the H atoms.)C CCCCCCCCCH3H3H3H3H3a-PP has the methyl groups appearing randomly on one side or the other.C CCCCC CCCC H3C H3C H33C H3SOLUTION: Since the methyl groups, which are large and bulky, are all on the same side in i-PP, the molecules can be efficiently packed together (when the molecules are stretched out). Hence, i-PP is semicrystalline. It mechanical properties are generally good up to about the meltingtemperature, 160︒C. a-PP is noncrystalline, since the molecules cannot be packed together into a unit cell. It's properties are limited by its glass transition temperature, which is about 0︒C.COMMENTS: A-PP is a useless gummy substance. All the PP in use today is i-PP. It isused in huge quantities.17. FIND: Show the stereo isomers of PAN.GIVEN:: PAN is poly(vinyl cyanide).SKETCH: (H are not shown)isotacticsyndiotacticatacticCOMMENTS: Commercially available PAN is atactic.18. FIND: Which polymer is more likely to be semicrystalline, PVdF (which has 2 F's) or PVF (avinyl polymer, which has 1 F)?SOLUTION: Note that the PVdF [poly(vinylidene fluoride)] is symmetric. Symmetricmolecules are easier to pack than nonsymmetric ones. PVdF is semicrystalline. PVF, like PVC, is noncrystalline.19. FIND: Estimate the glass transition temperature of PET.GIVEN: It's melting temperature is about 255︒C = 528K.SOLUTION: The melting temperature of nonsymmetric polymers is about 2/3 T m, when themelting temperature is in absolute degrees. Hence, T g≈ 2/3 x 528K = 353K = 79︒C.COMMENTS: The T g of PET is up to 40︒C higher than 80︒C, depending on the specificpolymer characteristics.20. FIND: Calculate the number of mers or degree of polymerization is a sample of PP with amolecular weight of 150,000 g/mole.DATA: PP is a vinyl polymer with a methyl side group.SOLUTION: There are 3 C and 6 H per repeat unit ⇒ MW = 3 x 12 + 6 x 1 = 42 g/mer.Hence, number of mers = 150,000 g/mole / 42 g/mole of mers = 3571 mers.COMMENTS: We always ignore chain end groups in these sorts of calculations because the effect is negligible.21. FIND: Calculate the MW of a cellulose mer. If a molecule of cotton has a MW of 9,000g/mole, how many mers are joined?GIVEN: The structure of cellulose is shown in Fig. 6.4-3a.DATA: According to Appendix A, the atomic weight of C is 12.01, H 1.01, and O 16.00g/mole.SOLUTION: Count the number of atoms of each type per mer: 10 O, 12 C, and 8 H. Thus, the MW of a mer is 10 x 16 + 12 x 12.01 + 8 x 1.01 = 312.20 g/mole. A molecule of cotton with a MW of 9,000 g/mole has 9,000 g/mole / 312.2 g/mole 29 mers joined.COMMENTS: This is a typical MW of cotton, 9,000 g/mole. It is formed by joining onlyabout 29 mers.22. FIND: How will the addition of pentaerythritol affect the crystallinity and glass transitiontemperature of PET?GIVEN: Pentaerythritol is tetra functional.SOLUTION: Any branching agent will disrupt the ability of the molecules to pack efficiently.Hence, crystallinity or the potential for crystallinity will be reduced. Crosslinking makesmolecular motion more difficult, so it raises the glass transition temperature.23. FIND: How does radiation that cleaves covalent bonds effect crystallinity?SOLUTION: Order must be perfect or near perfect in order to have a crystal. Cleavingbonds in crystals destroys the balance of order. Some bonds are broken, creating a difference in the bond arrangement than exists in the virgin crystal.COMMENTS: Organic molecules form crystals readily under appropriate conditions, but the structure of the crystal and the unit cell parameters do not resemble those of similar polymers.24. FIND: Does the fact that PP, crystallized under quiescent conditions, is characterized byMaltese cross patterned spherulites that fill the entire sample imply 100% crystallinity?SOLUTION: Spherulites are aggregates of crystalline and noncrystalline material. Thus, acompletely spherulitic sample is semicrystalline.25. FIND: Explain why amorphous PET that is hot stretched becomes opaque and slowly colddrawn amorphous PET may remain transparent.DATA: The glass transition temperature of PET is about 100o C.SOLUTION: Stretching a limited extent at room temperature does not induce crystallization in PET, whereas stretching above T g and orienting the more mobile molecules into a positionsimilar to those occupied by the molecules in a crystal likely induces crystallization.COMMENTS: In making a very strong PET fiber it is necessary to orient the molecules atfirst without inducing crystallization. Subsequent drawing steps are typically carried out atprogressively higher temperatures.26. FIND: How many O are in a mer of cellulose?GIVEN: The structure of cellulose is shown in Fig. 6.4-3a.SOLUTION: Count the O: There are 2 in the ether positions, connecting the rings; there are3 as OH groups on each of the 2 rings; and there is one as part of each of the 2 rings. Addthem up: 2 + 2 x 3 + 1 x 2 = 10 O per mer.COMMENTS: They are extremely important in providing cellulose its properties!27. FIND: Why are CaO and Na2O added to SiO2 in most applications for silicate glasses?SOLUTION: According to Table 6.5-1, neither CaO nor Na2O are glass forming systems.Rather, they are network modifiers, as stated in Table 6.5-2. They loosen the silicate network, lowering the glass transition temperature significantly. Thus, they are added to silica to reduce the cost of raw materials and, more importantly, the cost of processing.COMMENTS: The T g of silica in on the order of 1000︒C and that of soda-lime -silicate can be on the order of 500︒C.28. FIND: Is lead oxide a good glass former?SOLUTION:Zachariasen’s rules state that the metal should have a coordination number of 3 or 4. The valence of lead is such the PbO is the oxide predicted. Hence, PbO is not a goodglass former. It is an intermediate.29. FIND: Are epoxies and thermoset polyesters semicrystalline or noncrystalline?GIVEN: Both are highly crosslinked, transparent, hard and brittle.SOLUTION: Highly crosslinked polymers are always noncrystalline and below T g at roomtemperature. They are glasses. The crosslinks prevent crystallization.COMMENTS: On of the problems with some composite matrices is that they are brittle and not tough. Thermoplastic matrices are being developed for various demanding applications.High molecular weight thermoplastics, which are semicrystalline polymers, have high viscosity.It is difficult to get them to flow into the spaces between fibers.30. FIND: How can you detect when a glassy metal crystallizes?SOLUTION: Heat is released when crystallization occurs. If the metal were like a coin in your pocket, it might burn you. Use a calorimeter to quantify the effect. X-ray diffraction will alsoshow when crystallization occurs. The density or specific volume of the material changes withcrystallization.COMMENTS: Many other techniques can also be used to detect crystallization.31. FIND: Will mixtures, actually solutions, of PbO and SiO2 be good glass formers?GIVEN: SiO2 is a good glass former; PbO is not.SOLUTION: Since they form a solution, we expect the solution, which is even more complex than either component, to be even slower to crystallize. Thus, PbO-SiO2 mixtures should beexcellent glass formers so long as the PbO content is not high.32.33. FIND: Explain the role of B and Si in Metglas®.GIVEN: Metglas® is an iron based amorphous metal alloy.SOLUTION: The additives hinder crystallization, by increasing the viscosity of the melt,thereby reducing the diffusion coefficient, and by increasing the size of the unit cell, therebymaking it necessary for atoms to move farther to their crystallographic positions.34. FIND: How does modulus change as molecules are aligned along a fiber's axis?GIVEN: A single molecule is bonded by covalent forces. A collection of molecules is boned by both primary and weaker secondary bonds. In a fiber with all molecules aligned along thefiber axis, forces are transmitted along covalent bonds only.SKETCH:loFnl ot anxei sArei bgcul eoMml al i gnrai sMSOLUTION: The figure can best be read by beginning a large values on the abscissa. As the alignment increases, the modulus increases. With near zero misalignment, the molecules arepacked together in a parallel fashion and stresses are carried by covalent bonds. The fiberrequires a large stress to achieve significant deformation.COMMENTS: Pound for pound, organic fibers with this morphology have better mechanical properties than do metals.35. FIND: What polymer would you select for use as a flexible gasket on a liquid nitrogen tank?GIVEN: Liquid nitrogen boils at a very low temperature.SOLUTION: You need a material that remains flexible even down to liquid nitrogentemperatures and one that does not react with nitrogen. Some silicone rubbers (thermosetelastomers) are currently used for this application.36.37. FIND: Why is the modulus of Spectra® PE roughly 30 times greater than that of sandwich bagPE?GIVEN: Both are PE.SOLUTION: The difference is modulus is chiefly a result of the very high molecular orientation in Spectra® PE fiber and virtually no molecular orientation in sandwich bag PE.38. FIND: Predict interesting properties of poly(dimethyl siloxane).SKETCH:SOLUTION: The polymer contain no carbon in the backbone. It is inorganic. The methylside groups preclude crystallization, so the material is amorphous. The molecule is symmetricand its T g is well below room temperature. Hence, it is a rubber. Lightly crosslinked, it hasexcellent elastomeric properties. Because of its chemistry, it is chemically inert in mostenvironments, as well as stable to moderate temperatures.COMMENTS: A low grade of the materials is sold as RTV silicone rubber.39. FIND: Calculate the end-to-end separation of PS molecules.GIVEN: The DP is 5000DATA: 1 = 1.54ASKETCH:SOLUTION: We use equation 6.6-2: L = [m12(1 + cosθ')]½ to approximate the end-to-end separation. There are 2 bonds, each 1 = 1.54A per mer. The backbone is all carbon, so thefactor (1 +cosθ')/(1-cosθ') is 2. Substituting gives:L = [2 x 5000 x 1.542 x 2]1/2 = 218ACOMMENTS: This is a lower bound, largely because of the terms neglected in equation 6.6-2. A more sophisticated calculation shows the value is about 300A.40. FIND: Calculate the end-to-end separation of 150,000 g/mole a-PS.GIVEN: a-PS does not crystallize. PS is a vinyl polymer with a backbone of all C and a side group, as shown in Fig. 6.4-1.ASSUMPTIONS: The chains have no net molecular orientation. The chain end separation is governed by random flight statistics.DATA: The molecular weight of the mer is 8 x C + 5 x H = 8 x 12.01 g/mole + 5 x 1.01g/mole = 101.13 g/mole. θ in equation 6.6-2 is 109.5︒ and l is 1.54 A, as shown in Table A as twice the covalent radius of C.SOLUTION: The number of mers in a 150,000 g/mole samples of PS is:150,000 g/mole ÷ 101.13 g/mole of mers = 1483 mers.For each mer there are 2 C-C bonds, as shown in Fig. 6.4-1. Equation 6.6-2 can be used toestimate the end-to-end separation: (The molecule will look as is depicted in Fig. 6.6-8b.) end to-end separation = l mA A 1115429661109511095168+-=+-=cos cos .cos .cos .θθ.41.FIND: Show the change in modulus with temperature for a semicrystalline polymer. GIVEN: T g = 0o C and T m = 160o C.SKETCH:SOLUTION: The modulus will decrease at T g and fall to a very low value at T m .COMMENTS: This is how PP behaves. To lessen the impact of T g , polymer scientists and engineers attempt to increase crystallinity as much as possible.42. FIND: Show how the molecular weight between crosslinks affects the mechanical propertiesof an epoxy.SOLUTION: Increasing the molecular weight between crosslinks is equivalent to decreasing the crosslink density. As the crosslink density decreases, the modulus decreases and the elongation-to-break increases. Strength changes are not straightforward to predict. Usually, strength increases with crosslink density up to a point.COMMENTS: Too high or Too low a crosslink density leads to a mechanically inferior product.43. FIND: How might you make a fiber from crosslinked rubber? from a thermoplastic elastomer? GIVEN: Crosslinked rubber is one molecule. It does not melt and flow. The moleculescannot slide past one another irreversibly. Thermoplastic elastomers have temporary crosslinks. Upon heating, the molecules can slide past one another irreversibly.SOLUTION: It can indeed be difficult to make a fiber from crosslinked rubber. In fact, to make a fiber using latex (natural) rubber, the crosslinking is induced after fiber formation.Rubber bands, which are essentially thick fibers, are generally not round in cross-section. They are slit sheets, so the fibers are square or rectangular in cross-section. Thermoplastic fiber can be melt- or solution-formed directly. Lycra is a thermoplastic elastomer.44. FIND: Why do fabrics shrink when washed in hot water?GIVEN: Fiber are composed of aligned polymer molecules.SKETCH:SOLUTION: When heat is applied the aligned molecules seek to crystallize or coil onthemselves. When they move to coil, the fiber shrinks.COMMENTS: Most of the shrinkage occurs during the first heat. Hence, the fiber can beheat set to minimize subsequent shrinkage.45. FIND: Explain why the modulus of rubber is much lower than that characteristic of ceramic oroxide glass.SOLUTION: Rubbers are polymers that undergo conformational changes or straightening out of the coiled polymer chains in the fluid-like state with stress. Ceramics and oxide glassesrespond to stress by attempting to push or pull atoms out of their energy wells. Conformationchanges are much easier to achieve. Rubber elasticity is entropy driven. Hookean elasticity isenergy driven.46. FIND: Calculate the end-to-end separation of PP (a) coiled on itself and (b) completelystretched out.GIVEN: The molecular weight is 150,000 g/mole.DATA: 1 = 1.54ASKETCH:SOLUTION: We need to determine the number of bonds in the molecule for both parts and b.MW molecule/MW mer = number of mers or DP. MW mer = 3 x 12 + 6 x 1 = 42 g/mole.Therefore, DP = 150,000/42 = 3571 twice that many bonds(a) Using equation 6.6-2,L = [m12(1+cosθ')/(1-cosθ')½ = [2 x 3571 x 1.542 x 2]1/2 = 184ANote that in the hydrocarbon chain the entire cos factor is 2 and that there are 2 bonds per mer in all vinyl polymers.(b) Using equation 6.6-1,L ext = mlcos(θ/s) = 2 x 3571 x 1.54 x cos(109.5o/2) = 6348A。

无机材料科学基础教程(第二版)课后答案

第一章晶体几何基础1-1 解释概念：等同点：晶体结构中，在同一取向上几何环境和物质环境皆相同的点。

空间点阵：概括地表示晶体结构中等同点排列规律的几何图形。

结点：空间点阵中的点称为结点。

晶体：内部质点在三维空间呈周期性重复排列的固体。

对称：物体相同部分作有规律的重复。

对称型：晶体结构中所有点对称要素（对称面、对称中心、对称轴和旋转反伸轴）的集合为对称型，也称点群。

晶类：将对称型相同的晶体归为一类，称为晶类。

晶体定向：为了用数字表示晶体中点、线、面的相对位置，在晶体中引入一个坐标系统的过程。

空间群：是指一个晶体结构中所有对称要素的集合。

布拉菲格子：是指法国学者 A.布拉菲根据晶体结构的最高点群和平移群对称及空间格子的平行六面体原则，将所有晶体结构的空间点阵划分成14种类型的空间格子。

晶胞：能够反应晶体结构特征的最小单位。

晶胞参数：表示晶胞的形状和大小的6个参数（a、b、c、α 、β、γ ）.1-2 晶体结构的两个基本特征是什么？哪种几何图形可表示晶体的基本特征？解答：⑴晶体结构的基本特征：①晶体是内部质点在三维空间作周期性重复排列的固体。

②晶体的内部质点呈对称分布，即晶体具有对称性。

⑵14种布拉菲格子的平行六面体单位格子可以表示晶体的基本特征。

1-3 晶体中有哪些对称要素，用国际符号表示。

解答：对称面—m，对称中心—1，n次对称轴—n，n次旋转反伸轴—n螺旋轴—ns ，滑移面—a、b、c、d1-5 一个四方晶系的晶面，其上的截距分别为3a、4a、6c，求该晶面的晶面指数。

解答：在X、Y、Z轴上的截距系数：3、4、6。

截距系数的倒数比为：1/3:1/4:1/6=4:3:2晶面指数为：（432）补充：晶体的基本性质是什么？与其内部结构有什么关系？解答：①自限性：晶体的多面体形态是其格子构造在外形上的反映。

②均一性和异向性：均一性是由于内部质点周期性重复排列，晶体中的任何一部分在结构上是相同的。

异向性是由于同一晶体中的不同方向上，质点排列一般是不同的，因而表现出不同的性质。

清华大学材料科学基础目录及第一章题解

《金属学原理》习题解答北京科技大学余永宁目录第一章．晶体学 3 第二章．晶体结构19 第三章．相图22 第四章．金属和合金中的扩散45 第五章．凝固56 第六章．位错65 第七章．晶态固体的表面和界面79 第八章．晶体的塑性形变86 第九章．回复和再结晶92 第十章．固态转变98第1章1. 把图1-55的图案抽象出一个平面点阵。

解：按照等同点的原则，右图(图1-55)黑线勾画出的点阵就是由此图案抽象出的平面点阵。

2. 图1-56的晶体结构中包含两类原子，把这个晶体结构抽象出空间点阵，画出其中一个结构基元。

解：下右图(图1-56)的结构单元是由一个黑点和一个白点组成，按照等同点原则，抽象除的空间点阵如下左图所示，它的布拉喇菲点阵是面心立方。

3. 在图1-57的平面点阵中，指出哪些矢量对是初基矢量对。

请在它上面再画出三个不同的初基矢量对。

解：根据初基矢量的定义，由它们组成的平面初基单胞只含一个阵点，右图(图1-57)中的①和②是初基矢量对，③不是初基矢量对。

右图的黑粗线矢量对，即④、⑤和⑥是新加的初基矢量对。

4. 用图1-58a 中所标的a 1和a 2初基矢量来写出r 1，r 2，r 3和r 4的平移矢量的矢量式。

用图1-58b 中所标的初基矢量a 1，a 2和a 3来写出图中的r 矢量的矢量式。

解：右图(图1-58)a 中的a 1和a 2表示图中的各矢量：r 1=a 1+2a 2 r 2=-2a 2 r 3=-5a 1-2a 2 r 4=2a 1-a 2右图b 中的a 1、a 2和a 3表示图中的r 矢量： r =-a 1+a 2+a 35. 用矩阵乘法求出乘积{2[100]⋅4[001]}的等价操作，再求{4[001]⋅2[100]}的等价操作，这些结果说明什么? 解：因−−=100010001}2{]100[−=100001010}4{]001[{2[100]⋅4[001]}的等价操作为−−−= −⋅−−=⋅100001010100001010100010001}4{}2{]001[]100[这组合的操作和}2]011[{操作等效。

北京科技大学材料科学基础的一些建议

2016年北京科技大学材料科学基础的一些建议这几天才开始在意这个考研论坛，原来没怎么在意过。

我12年考的是《材料科学基础》，129分，听老师说130以上就一个，所以材科基算是前几名吧。

我本身也是北科材料的，给一些帖子作了一些回复后感觉可以小建议归纳一下放在一块，供有需要的同学复习参考（仅仅个人的小想法，有异议的话可以交流，但不要攻击）：1．关于本科生的笔记、课件、试卷、作业题个人认为，本科生的笔记是没啥用的，我自己也记过一些笔记，在考研题的答题过程中几乎派不上用场。

因为本科生用的是余永宁版的《材科基》，偏重计算，偏重一些深度比较大的知识点，因此不适合考研用，考研大部分都是一些基础性的东西。

有人问北科有木有啥专业课辅导班，很抱歉的说，北科自己是没有专业课辅导班的，办辅导班的都是一些考研机构，我看过他们的讲义，是关于每年都涉及到的一些考点，个人认为含金量不高。

2．关于看哪本辅导书一般来说有3本，上交胡赓祥的《材料科学基础》、宋维锡的《金属学》、余永宁的《材料科学基础》。

首先说一下余永宁的材科基，这是一本很让人头疼的书，当初学的时候得一句话一句话地分析是什么意思（绝对没有夸张），到现在为止，我估计已经来回看过5、6遍差不多了。

考研的时候我是在8月份把余永宁的材科基过了一遍（我是从7月25号左右正式开始考研复习的），然后就再也没有系统地看过第二遍，只是偶尔在做题的时候感觉金属学或者上交的那本书讲的不够好或者没有的时候再拿来看看参考一下。

这本书号称北京学院路四大天书之一，可知它的难度，所以如果是本校的话学材科专业的建议最好看一遍，其他材料专业的以及外校的最好就不要看了，看了你会得不到一丝的愉悦反而更加痛苦。

（我在这吐一下槽，老余的书逻辑性没有上交那版好，知识点都列在那，但是学的时候就是前后串不起来。

老余今年78左右了吧，现在还坚持给我们上课，它教了50多年的金属学，所以可知他的书里面有多少岁月的沉淀）然后就是上交那版的《材料科学基础》，这本书是我极力推荐大家重点看的。

材料科学工程基础_余永宁_第一章晶体学(习题)

13. 在单胞中画出(010)、(110)、 (121) 、(312)等晶面，画出 [111] 、 [123] 、 [110] 和[211]等晶向。
3
14. 用四轴坐标系画出六方晶系的 (112 0) 、(1012) 、(1011) 等晶面及 [112 0] 、[2113] 、[312 5] 等晶向。
6
查看标准极图的低指数面间的夹角关系定出三个极点的面指数。 32.图 1-63 是一个立方系单晶体，它的取向是上题标定的取向。晶体的 A 面和投影面平行，
B 面和 Y 轴成 70°(如图示)，pq 是 A 面和 B 面的交线，与 X 轴平行。一个晶面和 A 面及 B 面相交的迹痕与 pq 的夹角分别是 α=55°和 β=157.5°，求此面的面指数(hkl)。
15.写出图 1-62 中晶向的四轴坐标晶向指数。
16.列出三斜、单斜、正交及四方系中{210}面族包含面的数目及其指数；列出六方系中 {2130} 面族包含面的数目及其指数。。
17. (110 )、(112 )、( 312 )面是否同属一个晶带？如是，求出晶带轴的方向指数。
18. 下列的晶面：(234)、( 20 1 )、(1 1 1)、(241)、( 221 )、( 432 )、(101)、(010)和( 432 )中有哪些面属于同一个晶带？求出晶带轴。解：两个晶面一定有交线，交线一定是这两个晶面的晶带轴，这些情况在这里不讨论。不用计算的方法，可以用晶带定律试探求出晶带轴。试探得[102]是( 20 1 )、( 221 )、( 432 ) 和(010)面的晶带轴。
5. 用矩阵乘法求出乘积{2[100]⋅4[001]}的等价操作，再求{4[001]⋅2[100]}的等价操作，这些结果说明什么?