《工程材料科学与设计》(james p. schaffer)chapter-07.ppt

Fundamentals of Materials Science and Engineering Chapter two原子结构及原子间作用力1.了解所学的两种原子模型，并能区别其不同。

玻尔模型1913年，年轻的丹麦物理学家玻尔在总结当时最新的物理学发现（普朗克黑体辐射和量子概念、爱因斯坦光子论、卢瑟福原子带核模型等）的基础上建立了氢原子核外电子运动模型，提出了原子结构理论上的三点假设（1）任意轨道上绕核运动，而是在一些符合一定量子化条件的轨道上运动；（2）电子轨离核越远，原子所含的能量越高，电子尽可能处在离核最近的轨道上；（3）只有电子从较高能级跃迁到较低能级时，原子才会以光子形式释放能量。

玻而尔理论解释了原子发光现象但无法解释精细结构和多原子、分子或固体的光谱，存在局限性。

量子力学模型量子力学是建立在微观世界的量子性和微粒运动统计性基本特征上，在量子力学处理氢原子核外电子的理论模型中，最基本的方程叫做薛定谔方程，是由奥地利科学家薛定谔（E.Schrödinger 1887-1961）在1926年提出来的。

薛定谔方程是一个二阶偏微分方程，它的自变量是核外电子的坐标，它的因变量是电子波的振幅（ψ）。

给定电子在符合原子核外稳定存在的必要、合理的条件时，薛定谔方程得到的每一个解就是核外电子的一个定态，它具有一定的能量，具有一个电子波的振幅随坐标改变的的函数关系式ψ=f(x,y,z)，称为振幅方程或波动方程。

2.能够描述有关电子能量的量子力学法则。

能量最低原理，Pauli不相容原理，Hund规则。

4.(a)能够简单描述离子键，共价键，金属键，氢键和范德华键。

(b)能够列出以这些化学键结合的典型物质。

离子键：原子之间发生电子转移，形成正、负离子，并通过静电作用而形成的化学键。

离子键的本质是静电作用，无方向性、无饱和性。

离子键程度与元素的电负性有关。

共价键：不同原子依靠共享电子，或原子轨道的最大重叠而结合形成的化学键为共价键。

工程材料基础书
以下是一些关于工程材料基础的推荐书籍：
1. 《材料力学基础（原理与展开）》（Fundamentals of Materials Mechanics）- William F. Hosford
这本书介绍了工程材料的基本原理和力学行为，涵盖了弹性、塑性、蠕变、疲劳和断裂等方面的内容。

2. 《材料科学与工程》（Materials Science and Engineering: An Introduction）- William D. Callister Jr., David G. Rethwisch
这本书是材料科学与工程的经典教材之一，介绍了工程材料的结构、性质和应用，覆盖了金属、陶瓷、聚合物和复合材料等各类材料。

3. 《材料学导论》（Introduction to Materials Science and Engineering）- William D. Callister Jr., David G. Rethwisch
这本教材提供了关于工程材料科学与工程的基础知识，涵盖材料结构、晶体缺陷、相变、力学行为等方面内容。

4. 《材料科学工程导论》（Introduction to Materials Science for Engineers）- James F. Shackelford
这本书针对工程师介绍了材料科学的关键原理和应用，包括材料结构与性能、材料选择和设计等方面。

这些书籍都是在工程材料基础领域具有良好声誉的教材，适用于大学生、研究生以及从事相关学科研究或工程实践的人员。

值得注意的是，根据个人需求和背景，选择适合自己的教材和参考资料可能会更加有针对性。

材料学有关书籍材料学是一门研究材料的性质、结构、制备、性能、应用以及在各个领域中的发展与应用的学科。

对材料学的学习和研究不仅是工科学生和研究人员必备的背景知识，也对从事材料相关行业和研究的人员来说是非常重要的。

下面我将介绍一些关于材料学的经典和权威的书籍，这些书籍对于对材料学感兴趣的读者来说是非常有帮助的。

1.《材料科学基础》- William D. Callister这本书是对材料科学和工程的经典教材，深入浅出地介绍了材料的总体性质、组成和结构、热力学以及材料的力学行为等方面的内容。

书中配有大量的图片和例子，帮助读者更好地理解和应用所学知识。

2.《材料科学与工程概论》- William F. Smith, Javad Hashemi这本书是指导学生了解和理解材料科学和工程领域的基本概念、原理和应用的教材。

书中详细介绍了各种材料的性质、制备、处理、性能和应用等方面，并提供了一些实际案例和实验数据，帮助学生进行学习和研究。

3.《材料科学导论》- James F. Shackelford这本书对材料科学的起源、发展和应用进行了全面的介绍。

书中不仅介绍了材料的基本性质和分类，还涵盖了材料的结构、热力学、力学、电磁性能等方面的内容，并提供了一些应用案例和实例，帮助读者更好地理解和应用所学知识。

4.《材料科学与工程：依法施工》- R. Balasubramaniam这本书主要介绍了材料科学与工程领域的基本原理和技术，包括材料的种类、制备方法、性质和应用等方面的内容。

书中结合现实案例和实验数据，讲解了材料的特性和行为，帮助读者深入理解材料科学与工程的相关知识。

5.《材料科学导论》- H. L. Geyer这本书介绍了材料科学与工程的基本概念、原理和应用。

书中讲解了各种材料的结构、性质、制备和处理等方面的内容，并提供了一些实际案例和实验数据，帮助读者更好地理解和应用所学知识。

总之，材料学在现代科学和工程中扮演着非常重要的角色，对各个领域的发展具有极大的影响力。

mit出版的材料科学书籍MIT出版的材料科学书籍材料科学是一门研究材料的性质、结构、制备和应用的学科，涉及到物理、化学、工程等多个领域。

而MIT出版的材料科学书籍则是这个领域中的重要参考资料，为学者和工程师提供了丰富的知识和实践经验。

本文将按照类别介绍几本值得推荐的MIT出版的材料科学书籍。

1. 金属材料《金属材料的力学性能》是一本由William F. Hosford编写的经典教材，介绍了金属材料的力学性能和变形机制。

该书详细讲解了金属材料的塑性、断裂和疲劳等方面的知识，对于研究金属材料的学者和工程师来说是一本不可或缺的参考书。

2. 半导体材料《半导体材料和器件》是一本由Donald A. Neamen编写的教材，介绍了半导体材料和器件的基本原理和应用。

该书详细讲解了半导体材料的物理性质、制备方法和器件结构等方面的知识，对于研究半导体材料和器件的学者和工程师来说是一本不可或缺的参考书。

3. 高分子材料《高分子材料的结构和性能》是一本由James E. Mark编写的教材，介绍了高分子材料的结构和性能。

该书详细讲解了高分子材料的分子结构、物理性质和化学性质等方面的知识，对于研究高分子材料的学者和工程师来说是一本不可或缺的参考书。

4. 纳米材料《纳米材料的制备和应用》是一本由Gleb Yushin和Yury Gogotsi编写的教材，介绍了纳米材料的制备和应用。

该书详细讲解了纳米材料的制备方法、物理性质和应用领域等方面的知识，对于研究纳米材料的学者和工程师来说是一本不可或缺的参考书。

5. 其他材料除了以上几个类别的材料，MIT出版的材料科学书籍还包括了其他类型的材料，如陶瓷材料、复合材料、玻璃材料等。

其中，《陶瓷材料的制备和性能》是一本由M. John Matthewson编写的教材，介绍了陶瓷材料的制备和性能。

该书详细讲解了陶瓷材料的制备方法、物理性质和化学性质等方面的知识，对于研究陶瓷材料的学者和工程师来说是一本不可或缺的参考书。

国外材料科学英文书籍1. "Materials Science and Engineering: An Introduction" by William D. Callister, Jr. 这本书是材料科学与工程领域的经典教材，涵盖了材料的结构、性能、制备和应用等方面。

2. "Principles of Materials Science and Engineering" by Donald R. Askeland and Pradeep P. Phulé. 本书提供了材料科学的全面概述，包括晶体结构、热力学、相图、材料的力学行为等内容。

3. "Introduction to Materials Science for Engineers" by James F. Shackelford. 这本书是为工程师编写的材料科学入门教材，强调了材料的工程应用和设计。

4. "Materials Science and Technology" by R.W. Cahn and P. Haasen. 该书是材料科学领域的权威著作，涵盖了材料的结构、性能、制备和应用等方面，内容深入且广泛。

5. "Materials Characterization: Introduction to Microscopic and Spectroscopic Methods" by Michael F. Ashby and David R. H. Jones. 这本书介绍了材料表征的各种技术和方法，包括显微镜、光谱学和衍射技术等。

这些书籍都是材料科学领域的经典著作，可以帮助读者深入了解材料的结构、性能、制备和应用等方面的知识。

你可以根据自己的需求和兴趣选择适合的书籍进行阅读。

参考书目第一章晶体学1.Б.К.因斯坦著吴自勤译《现代晶体学(第一卷)》中国科技大学出版社（1990）2. C.本斯 A.M.格莱泽著俞文海周贵恩译《固体科学中的空间群》高等教育出版社（1984）3.张克从著《近代晶体学基础(上册)》科学出版社(1987)4.晶王英华编著《晶体学导论》清华大学出版社(1989)5.Allen S M, Thomas E L. The Structure of Materials. New York: Wuley,1998.第二章固体材料中电子运动状态1．方俊鑫，陆栋编，《固体物理学》上海科学技术出版社（上海，1980年第一版，1993年第10次印刷）2．沈仲钧，冯茂仁，《量子力学》，上海科学技术出版社（上海，1988）3．陈端刚主编，《中华小百科全书――物理学卷》，四川教育出版社，四川辞书出版社，1994。

4．E. Arzt V orlesungsskript von Materialphysik III – Elektronische Eigenschaft, Universitaet Stuttgart第三章晶体结构1.T.B.Massalski，Structure of Solid Solutions in Physical Mwtallurgy，Third revised andenlarged，ed. by R.W.Cahn and P.Haasen. p.153. & p.219 North-Holland Physics Publishing, 1983.2.Kingery.W.D, Introduction to Ceramics, (chapter 2). John Wiley & Sons, Inc. New York.1976.3.张克从近代晶体学基础上册科学出版社19874.冯端冯步云放眼晶态之外湖南教育出版社19995.晶体化学6.M.V.斯温主编陶瓷的结构与性能第一章（材料科学与技术丛书第11卷）科学出版社1998第四章非晶态与半晶态1．Allen S M, Thomas E L. The Structure of Materials. New York: Wuley,1998.2．冯端，师昌绪刘治国材料科学导论化学工业出版社20023．余永宁强文江等译工程材料科学与设计机械出版社20034．张德庆张东兴刘立柱等编高分子材料科学导论哈尔滨工业大学出版社1999第五章相图1.张圣弼，李道子，相图−原理、计算及在冶金中的应用，冶金工业出版社，19862.Materials Science and Technology V ol 5, Phase Transformation in Materials, ed. by P.Haasen,New York, Basel, Cambridge:VCH, 19933. A.D. Pelton, Phase Diagrams in Physical Metallurgy，Third Ed. North-Holland PhysicsPublishing, 19834.M. Hansen, K. Anderko, Constitution of Binary Alloys, Mc-Graw-Hill, 1985第六章晶体中的点缺陷和线缺陷1.S.M.Allen, E.L.Thomas, The Structure of Materials. New York: Wuley, 19982.R.W.卡恩, P.哈森, E.J.克雷默, 主编；王佩璇等译, 材料科学与技术丛书, 第1卷, 第7章, 科学出版社, 19983.J.P.Hirth，Dislocation in Physical Metallurgy 3rd edition；North-Holland Physics Publishing，19834.冯顺华, 晶体位错理论基础, 第一卷, 科学出版社, 19885.W.F. Harris, Scientific American. 1977, 237; 666.余永宁, 毛卫民, 材料的结构, 冶金工业出版社, 2001第七章面缺陷和体缺陷1.S.M.Allen, E.L.Thomas. The Structure of Materials. New York: Wuley, 19982.James M. Howe. Interfaces in Materials. Wiley, 19993. F.J.Humphreys and M.Hatherly, Recrystallization and Related Annealing Phenomena，Galliard (Printers) Ltd. 19954.Materials Science and Technology, vol 1，Structure of Solids，ed. by V.Gerold，New York，Basel，Cambridge:VCH. 19935.李恒德，肖纪美主编，材料表面与界面，清华大学出版社，19906.R.E.Smallman, R.J.Bishop, Modern Physical Metallurgy and Materials Engineering, Sixth Ed,Oxford OX2 8DP 1999第八章材料中原子扩散1.J.L.Bocquet, J.Brebec and Y.Limoge Diffusion in Metals and Alloys, in Physical Metallurgy,Third Ed. North-Holland Physics Publishing, 19832.J. S. Kirkaldy, Diffusion in the Condensed State, The Universities Press (Belfast) Ltd. 19873.李长海，余永宁译, 金属和合金中的相变, 冶金工业出版社，19884.J. Crack, The Mathematics of Diffusion, Second Edition, Oxford University Press, 19755.Materials Science and Technology, V ol 5, Phase Transformation in Materials, ed. by P.Haasen,New York, Basel, Cambridge:VCH. 19936.W.D Kingery. Introduction to Ceramics, John Wiley & Sons, Inc. New York. 1976第九章材料的形变1. 哈宽富，金属力学性质的微观理论，科学出版社，19832. Richard W. Hertzberg, Deformation and Fracture Mechanics of Engineering Materials，JohnWiley & sons. 19833. B.Bay, N.Hansen, Acta Metall.Mater., 40, 1992, 205∼2194. F.J.Humphrey & M.Hatherly, Recrystallization and Related Annealing Phenomena，Pergamon，19955. R.W.卡恩，P.哈森，E.J.克雷默主编，颜鸣皋等译，材料科学与技术丛书，第6卷第1、2、3章，科学出版社，19986. 冯端，师昌绪，刘治国主编，材料科学导论，化学工业出版社，20027. R.W.Cahn and P.Haasen. Physical Metallurgy. Fourth, revised and enhanced edition. Elsevier Science BV. 1996, V ol.3, Chapter 32: A metallurgist’s guide to polymers. By A.H.Windel8．赵品，谢辅洲，孙文山，材料科学基础，哈尔滨工业大学出版社，19999. 刘智恩, 材料科学基础，西北工业大学出版社，200010. J.W.Christian and S.Mahajan. Deformation Twinning. Progress in materials science. 1995.V ol.39. 1-15711．余永宁，强文江等译，工程材料科学与设计，机械工业出版社，200312. J.Hirsch, K.Lücke and M.Hatherly. Mechanism of deformation and development of rollingtextures in polycrystalline F.C.C. metals---III. The influence of slip inhomogeneities and twinning. Acta metall. 36, 1988, No.11. 2905-2927第十章相变的基本原理1. D.A.Port, K.E.Eastering. Phase Transformations in metals and alloys. 19922.冯端，师昌绪，刘治国主编，材料科学导论，化学工业出版社，20023. R. W. Cahn，Physical Metallurgy，4rd，revised and enlarged ED.，ed. by R.W.Cahn andP.Haasen, Elsever Publishing, 19964. G.Gottstein. Physical foundations of materials science. Springer-V erlag, 2004第十一章凝固1.弗莱明斯著，关玉龙等译，凝固过程，冶金工业出版社, 19812.戴维斯著，舒震等译，凝固与铸造，机械工业出版社, 19813. D.A.Port, K.E.Eastering. Phase Transformations in metals and alloys. 19924.H.Boloni in Physical Metallury，Ed. by R.W.Cahn and P.Haasen, North-Holland Pub. 3rdedition, 477-579, 19835.I. Minkoff, Solidfication and Cast Structures, John Wiley and Sons Ltd., 19866.冯端，师昌绪，刘治国主编，材料科学导论，化学工业出版社，2002第十二章固态转变1. R. W. Cahn，Physical Metallurgy，4rd，revised and enlarged ED.，ed. by R.W.Cahn and P.Haasen,Elsever Publishing, 19962. Hsun Hu（胡郇），物理冶金进展评论，中国金属学会编译组译，冶金工业出版社,1985 ,pp.151-1873. R.K.Ray, J.J.Jonas and R.E.Hook. Cold rolling and annealing textures in low carbon and extralow carbon steels. International Materials Review. 1994. V ol.39. No.4. 129-1724. G.Gottstein. Rekristallisation metallischer Werkstoffe. Deutsche Gesellschaft fuer MetallkundeE.V. 19845. J.Hjelen, R.rsund and E.Nes. On the origin of recrystallization textures in Aluminium. Actametal. 39, 1991. No.7. 1377-14046. A.Berger, P.-J.Wilbrandt, F.Ernst, U.Klement and P.Haasen. On the generation of neworientations during recrystallization: Recent results on the recrystallization of tensile-deformed fcc single crystals. Progress in materials science. 32, 1988. 1-957．余永宁. 金属学原理. 冶金工业出版社，20008. D.A.Port, K.E.Eastering. Phase Transformations in metals and alloys. 19929．G.Gottstein. Physical foundations of materials science. 2004, Springer-V erlag 10. 徐恒均. 材料科学基础. 北京工业大学出版社，2001。

Problems - Chapter 5 1. FIND: Calculate the stress on a tensioned fiber.GIVEN: The fiber diameter is 25 micrometers. The elongational load is 25 g. ASSUMPTIONS: The engineering stress is requested.DATA: Acceleration due to gravity is 9.8 m/sec 2. A Newton is a kg-m/sec 2. A Pascal is a N/m 2. A MPa is 106 Pa.SOLUTION: Stress is force per unit area. The cross-sectional area is πR 2 = 1963.5 square micrometers. The force is 25 g (kg/1000g)(9.8 m/sec 2) = 0.245 N. Thus, the stress isσ = F/A =02451960100125262..N umum m MPa⨯☞☟☝✋ ☺=COMMENTS: You must learn to do these sorts of problems, including the conversions. 2. GIVEN: FCC Cu with a o = 0.362nmREQUIRED: A) Lowest energy Burgers vector, B) Length in terms of radius of Cu atom, C) Family of planesSOLUTION: We note that the Burgers vector is the shortest vector that connects crystallographically equivalent positions. A diagram of the structure is shown below:FCC structure with (111) shownWe note that atoms lying along face diagonals touch and are crystallographically equivalent. Therefore, the shortest vector connecting equivalent positions is ½ face diagonal. For example,one such vector is10]1[ 2a o as shown in (111).A. The length of this vector is0.256nm = 20.362= 2a= 4a+ 4a o 2o 2oB. By inspection, the size of the vector is 2 Cu atom radii.C. Slip occurs in the most densely packed plane which is of the type {111}. These are thesmoothest planes and contain the smallest Burgers vector. This means that the dislocations move easily and the energy is low.3. GIVEN:∣b∣ = 0.288nm in AgREQUIRED: Find lattice parameterSOLUTION: Recall the Ag is FCC. For FCC structures the Burgers vector is ½ a facediagonal as shown. We see that4. A. FCC structureThe (111) plane is shown in a unit cell with all atoms shown. Atoms touch along face diagonals. The (111) plane is the most closely packed, and the vectors shown connect equivalent atomic position. Thus 10]1[ 21= b etc. Then in general >110< 2a= bB. For NaC1We see that the shortest vector connecting equivalent positions is 10]1[ 2aas shown. Thisdirection lies in both the {100} and {110} planes and both are possible slip planes. However {110} are the planes most frequently observed as the slip planes. This is because repulsive interionic forces are minimized on these planes during dislocation motion. Thus we expect 1/2<110> Burgers vectors and {110} slip planes.5.GIVEN:Mo crystal0.272nm = ba o = 0.314nmREQUIRED: Determine the crystal structure.If Mo were FCC, then 0.222nm = 20.314= b __but |b| = 0.272 ⇒ Mo is not FCC.Assuming Mo is BCC, then 0.272nm.= 0.314 X 23 = b __Thus the Burgers vector isconsistent with Mo being BCC.6.FIND: Is the fracture surface in ionic solids rough or smooth?SOLUTION: Cleavages surfaces of ionic materials are generally smooth. Once a crack is started, it easily propagates in a straight line in a specific crystallographic direction on a specificcrystallographic plane. Ceramic fracture surfaces are rough when failure proceeds through the noncrystalline boundaries between small crystals.7. GIVEN: BCC Cr with |b| = 0.25nmREQUIRED: Find lattice parameter aASSUME: >111< 2a= b for BCC structureSOLUTION:2a 3= 4a+ 4a+ 4a= b 222__from the formula for the magnitude of a vector:8. GIVEN: Normal stress of 123 MPa applied to BCC Fe in [110] directionREQUIRED: Resolved shear in [101] on (010)SOLUTION: Recall that the resolved shear stress is given by:τ = σ cos θ cos φ (1)where θ = angle between slip direction and tensile axis; φ = angle between normal to slip plane and tensile axisThus MPa 43.5 = 2121 123 = ⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛τ9. GIVEN: Stress in [123] direction of BCC crystalREQUIRED: Find the stress needed to promote slip if τcR = 800 psi. The slip plane is (11_0) and slip direction is [111]. SOLUTION: Recall τ = σ cos θ cos φ (1)θ = [123] [111][123] ⋅ [111] = ∣[123]∣ ∣[111]∣cos θφ = [123] [11_0][123] ⋅ [11_0] = ∣[123]∣ ∣[11_0]∣cos φ10.Burgers vectors lie in the closest packed directions since the distance between equivalentcrystallographic positions is shortest in the close-packed directions. This means that the energy associated with the dislocation will be minimum for such dislocations since the energy is proportional to the square of the Burgers vector.11. Close packed planes are slip planes since these are the smoothest planes (on an atomic level) and would then be expected to have the lowest critical resolved shear stress.12.GIVEN: Dislocation lies on (11_1) parallel to intersection of (11_1) and (111) with Burgers vector parallel to [1_1_0]. Structure is FCC.REQUIRED: A) Burgers vector of dislocation and, B) Character of dislocation.SOLUTION: A) Since the structure is FCC, the Burgers vector is parallel to <110> and has magnitude. 2a For a Burgers vector parallel to [1_1_0] the scalar multiplier must be a/2. Thus b _ = a/2 [1_1_0]. B) We must determine the line direction of the dislocation. From the diagramwe see that the BV and line direction are at 60o which means the dislocation is mixed.13. GIVEN: Dislocation reaction below:REQUIRED: Show it is vectorially correct and energetically proper. SOLUTION: [100] a =] 111[ 2a+[111] 2aThe sum of the x, y & z components on the LHS must be equal to the corresponding component on the right hand side.x component (LHS) = x component (RHS)y component (LHS) = y component (RHS)z component (LHS) = z component (RHS)Energy: The reaction is energetically favorable if | b 1 | 2 + | b 2 | 2 > | b 3| 3Thus the reaction is favorable since a > a 43+ a 4322214. GIVEN: Dislocation in FCCParallel to [1_01] i.e. t_ = [1_01]REQUIRED: Character and slip planeSOLUTION: Character is found by angle between b_ and t_. Note b_ t_∙ = -1 + 0 + 1 = 0. Thus b__t_. Since b__t_the dislocation is pure edge.To find the slip plane we note that the cross produce of t_ & b_gives a vector that is normal to the plane in which t_ & b_lie. This vector so formed has the same indices as the plane since we have a fundamentally cubic structure.We see from the diagram that these vectors lie on (010).Thus, we have the plane (01_0) which is the same as the (010) plane. This does not move by glide since planes of the kind {100} are not slip planes for the FCC structure.15.FCC metals are more ductile than BCC or HCP because: 1) there is no easy mechanism for nucleation of microcracks in FCC as there is for BCC and HCP; 2) the stresses for plastic deformation are lower in FCC due to the (generally) smoother planes. This means that the microcracks that form in BCC & HCP will have high stresses tending to make them propagate. 16.For a simple cubic system, the lowest energy Burgers vectors are of the type <001> since this is the shortest distance connecting equivalent atomic positions. This means that the energy is lowest since the strain energy is proportional to the square of the Burgers vector. 17.GIVEN:At. wt. 0 = 16At. wt. Mg = 24.32 Same structure as NaCl ρ = 3.65 g/cm 3REQUIRED: Find length of Burgers Vector in MgOSOLUTION: The structure of MgO is shown schematically below along with the shortestBurgers vector. To solve the problem we first note that we require the lattice parameter a o . We can take a sub-section of the unit cell (cross-hatched cube) whose edge is 2a o unitslong.We can calculate the total mass of this cube and the volume and calculate the density. Since the mass is known and the density is known, the volume may be calculated from which a o may beextracted.and ½ Mg ++ ions in our cube. Thus a10 x 3.35 x 8= 8/ a 10 x 2.02) + (1.33= 3.653o-233o-2318.GIVEN: Critical resolved shear stress (0.34MPa), slip system (111)[1_10], and tensile axis [101]REQUIRED: Applied stress at which crystal begins to deform and crystal structure. SOLUTION: (A)The situation is shown belowτcrss = σ cos θ ⋅ cos φθ = angle between tensile axis and slip directionθ = angle between tensile axis and normal to slip planeφ = [111] [101] θ = [101] [1_01][111] ⋅ [101] = ∣[111]∣ ∣[101]∣cos φ [101] ⋅ [110] = ∣[101]∣ ∣[110]∣cos θ(B): To have a {111}<110> slip system, the material must have an FCC structure.19. GIVEN:τcrss = 55.2 MPa, (111)[1_01] slip system, [112] tensile axisREQUIRED: Find the highest normal stress that can be applied before dislocation motion in the [10 1_] direction.SOLUTION: The situation is shown below. Essentially the problem reduces to finding the value of the tensile stress when the critical resolved shear is reached.τcrss = σ cosθ⋅ cosφθ = [112] [1_01] φ = [112] {111}B. Would have exactly the same stress for a BCC metal (φ & θ would be interchanged).20. GIVEN:σ at yield = 3.5 MPa; (111) [11_0] slip system [11_1] tensile axisREQUIRED: Compute τcrssSOLUTION:τcrss = σcosθcosφθ = [11_1] [11_0] φ = [11_1] [111]21. Item Edge ScrewLinear defect? Yes YesElastic Distortion? Yes YesGlide? Yes YesClimb? Yes NoCross-slip? No YesBurgers Vector (BV) ⊥ to line // to lineUnique slip plane? Yes NoOffset // to BV // to BVMotion // to BV ⊥ to BV22. GIVEN: BCC metal with τcrss = 7MPa [001] tensile axis.REQUIRED: (a) Slip system that will be activated and (b) normal stress for plasticdeformation.SOLUTION: Recall that for BCC metals the usual slip system is <111> {110}. Deformation occurs on the plane and direction for which cos θ⋅cos φ is a maximum since this will have the maximum resolved shear stress. The situation is shown below.(Note that the slip directions are shown shortened in this view)Possible slip systems are listed below:sketch (also [11_1] on (011)) (also [1_11] on (01_1)) (also [1_1_1] on (101)) similar to planes shown in sketch. Also [111] on (1_01)We see by inspection that the resolved shear due to a tensile force in [001] will all be the same. The resolved shear on all other {110}<111> systems is zero.B. To compute the normal stress at the onset of plastic deformation we will consider (011) [1_1_1]τcrss = σcos θcos φ = 7θ = [001] [1_1_1]; cos φ = [001] [011]Note if we considered (101) [1_1_1] we would haveand we would obtain exactly the same answer.23. GIVEN: Yielding occurs at normal stress of σ = 170 MPa in [100] direction. Dislocationmoves on (101) in [111_] direction.REQUIRED:τcrss and crystal structuresSOLUTION: Assume an edge dislocation. τcrss = σcos⋅cosφθ = [100] [111_] φ = [100] [101]The - sign means that the slip direction is opposite to the motion of the dislocation. Essentially, we have a negative edge dislocation on (101) as shown below:The edge dislocation moves in [111_] direction but the offset is in [1_1_1] direction.The slip plane and slip direction are representative of BCC structures.24. GIVEN: (1_10)[111] slip system. [123] tensile axisτcrss = 800 psi for BCC crystal τcrss = 80 psi for FCC crystal withσFCC = 457 psi [123] tensile axis and (111)[11_0] system.REQUIRED: Normal stress at yield for BCC metalSOLUTION: The simplest way to solve this problem is to note cos θ⋅cos φ is the same for the BCC and FCC crystal with the meaning of φ and θ interchanged. Let M = cos θ⋅cos φ. (1) (2)(3)25.Here crystallographically equivalent positions join ions at cube corners (b v = a o ), face diagonals )a 2 = b (o v , cube diagonals )a 3 = b (o vThe most densely packed plane is the (110) in which we haveThe shortest vector that will reproduce all elements of the structure is a o . Thus b = a<100> COMMENT: We note that this is not sufficient for general deformation (e.g. a tensile axis ofthe type <100> produces zero shear on the 1<100> Burgers vectors. We expect then a<110> Burgers vectors as well.26. GIVEN:σ = 1.7 MPa [100] tensile axis (111)[101] slip systemsREQUIRED:τcrss, and crystal structure. Also find flaw in problem statement.SOLUTION: Since the slip system is of the type {111}<110> the structure is FCC. Theproblem is misstated since the Burgers vector must lie on the slip plane and [101] does not lieon (111). The slip direction would more appropriately be [101_]. Thus the slip system is(111)[101_] as shown below.27.⊥ = edge dislocation x = start of Burgers circuitb = Burgers vector y = end of Burgers circuit28. FIND: Show energy/area = force/length, that is, surface energy is surface tension in liquids.DATA: The units of energy are J = W/s or N-m. The units of force are N.SOLUTION: Energy/area = J/m2 =N-m/m2 = N/m = force/length29. GIVEN: Two grain sizes, 10μm and 40μmREQUIRED: A) ASTM GS# for both processes, B) Grain boundary area.SOLUTION: Assume that the grains are in the form of cubes for ease of calculation. TheASTM GS# is defined through the equation: n = 2N-1 where n = # grains/in2 at 100X.N=ASTM GS#To solve the problem we first convert the grain size to in. where D = length of cube edge in μm.At 100X linear magnification, the sides of the smaller grains will be:The area of each grain at 100X will beSimilarly the area of the 40μm grains at 100X isFor the 10μm dia grain, the # of grains per in 2 (at box) is645.16 = 10x 1.5501= n3-100X10μgrains/in 2at 100X Similarly 40.31 = 10x 24.811= n 3-100X 40μgrains/in 2at 100X For the 10μm grain size:B. In computing the total g.s. area we will assume 1 in 3 of materials. Since there are 6 facescube and the area of each face is shared by 2 cubes, each cube has an area of 3x Area of face. G.B. Area =d / 3 = d 3 x d 123⎥⎦⎤⎢⎣⎡GB Area (10μ gs) = 3/3.937 x 10-4 = 7620in 2/in 3 GB Area (40μ gs) = 3/15.75 x 10-4 = 1905in 2/in 330.GIVEN:σys = 200MPa at GS#4 = 300MPa at GS#6REQUIRED: σys at GS#9SOLUTION: Recall σys = σo + kd -1/2(1)for low carbon steel.If d = grain size (assume cubes) load = grain diameter at 100X(2)(3)16.82 = d11/24For ASTM GS# 4:For ASTM GS#6:23.78 = d11/24(5) Substituting (4) and (5) into (1) we have200 - σo + k(16.82) (6) 300 = σo + k(23.78) (7)Subtracting (6) from (7):100 = k(23.78 - 16.82)∴k = 14.37Substituting this value of κ into (6) yields 200 = σo + 14.37 x 16.82σo = -41.70 (this is not physically realistic since σo relates to the lattice friction stress which should not be negative)For ASTM GS#9Thus σys = σo + 14.37 x 40 = 41.70 + 574 = 533MPa31.GIVEN: ∣b ∣ = 0.25μm for BCC metal tilt boundary has angular difference of 2.5o REQUIRED: Dislocation density in tilt boundary wall SOLUTION: The physical situation is shown below:If b = Burgers vector, D = spacing between edge dislocation# of dislocations in boundary for a 1cm high boundary isD1(where D is in cm)32.33.FIND: Show D = b / θ.GIVEN: b is the magnitude of the Burger's vector; D is the spacing between dislocations, and θ is the tilt angle.SKETCH: See Fig. 5.3-4.SOLUTION: We can see the geometry more clearly using the following sketch:bFrom the Figure we can immediately write that tan/θ22=b D . Since the tan of a small angle is the angle itself:θ22=b D /, so that D = b / θ, as is written in the margin.34. FIND: How can you detect a cluster of voids or a cluster of precipitates in a material?SOLUTION: This can be a difficult challenge indeed. If the total void volume is large, then the density of the sample will be lower than that of dense material. The same is true for clusters ofprecipitate; however, usually the density difference between host and precipitate is not as greatas between host and air, so the technique does not work as well. Another possible technique is microscopy. Samples can be prepared for microscopy, perhaps by polishing and etching andthe defects observed using optical or electron microscopy. X-ray diffraction can also be used. With a random spacing of void or precipitate there is then an average spacing. SometimesBragg's law can be used to calculate the spacing if an intensity maximum is observed. Note that the angle of the maximum will be very small.COMMENTS: There are many other potential techniques that can potentially be used. Theyall rely on some property difference - magnetic, electrical, optical, or whatever.35. FIND: How can you ascertain whether a material contains both crystalline and noncrystallineregions?GIVEN: Recall that the density (and other properties) of crystalline material is greater than that of noncrystalline material of the same compositionSOLUTION: There are three methods in common usage to establish crystallinity polymers.These methods apply to all materials.1. Density. Measure the density of your sample and compare it to the density of noncrystallineand crystalline samples of the same composition.2. Differential Scanning Calorimetry. Heat your sample in a calorimeter. Samples that arecrystalline will absorb heat at the melting temperature and show a "melting endotherm". Somenoncrystalline samples (such as amorphous metals) will crystallize in the calorimeter and show a huge release of heat prior to melting. This is a "crystallization exotherm".3. X-ray diffraction. Crystalline materials show well-defined peaks.COMMENTS: Knowing whether a material is crystalline or noncrystalline is a commonchallenge to polymers scientists. We often need to quantify the fraction or percent crystallinity.Can you suggest a method for each of the 3 techniques outlined?36. FIND: State examples of materials' applications that require the material to behave in a purelyelastic manner.SOLUTION: There are many such possible examples. Since plastic deformation isnonrecoverable deformation, any application that requires repeated stressing and dimensional stability is a good example. Here are some examples:1. Springs in automobiles - leaf and coil springs2. A diving board3. Trusses in a bridge4. The walls in a building5. A bicycle frame6. Piano wire7. Airplane wings37. As the dislocation density ↑, there are more dislocation/dislocation interactions and the strengthgoes up. At the same time, the degree of “damage” also increases and the ductility decreases.38. If the point defect concentration ↑, the strength will go up as well. This is because the defectsmay migrate to edge dislocations where they cause jogs on the dislocations. A joggeddislocation is much harder to move and may itself require the generation of point defects tomove. In addition the point defects may collapse to form dislocation loops which also impede the motion of other dislocations making the materials stronger. If the defects are interstitials,they may migrate to areas around the dislocations in which the system energy is reduced. For the dislocation to move away from the interstitial an increase in the system energy is requiredwhich means the stress to move the dislocation must increase. If the point defect is asubstitutional atom, similar considerations apply. However, the magnitude of the energyreduction is less because of the less severe distortion. Thus the strength increase is not as high as for intersitital.39. As d↓σys↑ since this means the path over which a dislocation moves ↓. This means that thestress will have to increase to either nucleate or unlock dislocations in adjacent grains. Therelationship quantifying this behavior is the Hall-Petch equation: σys = σo + kd-1/240. The strength may increase as a result of:1. decreasing grain size - should not be too (see previous questions) temperature dependent.2. Adding impurities (e.g. C in Fe). The impurities “lock” the dislocation by associating withthe dislocation to lower the system energy. This will be very temperature dependent for dilute concentrations of impurities as the impurities will diffuse away at high temperatures.3. Adding precipitates - blocks the motion of dislocations through either having a differentcrystal structure or a large strain field. Since the precipitates are usually large compared to the atomistic dimension, strong temperature dependence is not expected.4. Cold work - increase quantity of dislocations.41. GIVEN: = 1012/cm 2 for low C steelREQUIRED: concentration of C atoms (at %) to lock all dislocationsSOLUTION: Recalling the At. weight of Fe is 55.85 and the density is about 7.8 gm/cm 3 we may write10 x 6.02 55.857.8 = N 23Fc(assume 1C atom for every Fe atom along dislocations)42. FIND: Why can you not bend the bar of tin?GIVEN: The bar has been well annealed, so the initial dislocation density is low. You are required to re-bend the bar after cold working.SOLUTION: The deformation has increased the dislocation density and the bar now requires much more stress, or force, to deform it. You are not necessarily a weakling, but you have been taken. Re-anneal the bar and bend it back or use brute force.COMMENTS: It is often difficult to bend a metal back to its original shape and this is just one of many possible reasons that depend on the metal and its thermo-mechanical。

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

4.3 Precision CastingMetal casting has a historic pedigree as the manufacturing process first used by man to produce intricate metal articles and art objects.It played a major role in the industrial revolution and remains the basis of current manufacturing equipment and manufactured goods.The process plays an important part in aerospace component production and,as such ,remains at the leading edge of technology development.Despite the competition from plastics and ceramics,metal still remain the dominant materials in the production of capital equipment and manufactured goods.Metal casting will continue to play a major role,as a manufacturing process of considerable versatility,for the foreseeable future金属铸造被人类作为生产工艺首次使用在复杂的金属制品和艺术品上有悠久的历史。

它在工业革命中发挥了重大作用，仍然是当今设备和产品制造的基本方法。

这种工艺在航空零部件生产中起着重要的作用，因此，仍然是科技发展的前沿技术。

Chapter 1 An Overview第一章概述1.1 Introduction1.1介绍Materials are the matter of the universe. These substances have properties that make them useful in structures, machines, devices, products, and systems. The term properties describe behavior of materials when subjected to some external force or condition. For example, the tensile strength of a metal is a measure of the material's resistance to a pulling force. The Family of Materials consists of four main groups of materials: Metals (e.g., steel), Polymers (e.g., plastics), Ceramics (e.g., porcelain), and Composites (e.g., glass-reinforced plastics). The materials in each group have similar properties and/or structures, as will be described later.材料是宇宙的物质。

这些物质的特性使其有用的结构、机器设备、产品和系统。

这个术语属性描述材料的行为时,受到一些外部力量或状态。

例如,抗拉强度的金属是测量的材料抵抗了拉力。

这个家庭的材料由四个主要群体的材料:金属(如钢)、高分子材料(例如:塑料)、陶瓷(如瓷),复合材料(例如,增强塑料)。

2023年材料科学与工程专业考研书目
1. 《材料科学基础》王义斌著
2. 《材料性能测试与表征技术》刘海峰著
3. 《材料工程学基础》严广学著
4. 《材料晶体学》徐光明著
5. 《材料加工工艺》郭锦坤著
6. 《高分子材料化学与物理》刘苏民著
7. 《功能材料学导论》王连兵著
8. 《材料科学与工程前沿问题研究》李群著
9. 《现代材料分析技术》张文华著
10. 《材料力学》陈兆康著
11. 《金属材料热力学基础》宋永平著
12. 《医用材料技术导论》田继球著
13. 《先进材料与先进制造技术》张伟壹著
14. 《材料科学与工程实验技术》赵建奇著
15. 《材料工程概论》刘磊著
16. 《复合材料力学与设计》马建宁著
17. 《高分子材料工程》鲁星著
18. 《表面与薄膜工程》梁福祥著
19. 《无机非金属材料工程》王飞著
20. 《材料科学概论》莫安民著。

关于材料专业的书籍以下是关于材料专业的一些推荐书籍：1. 《材料科学与工程概论（第二版）》：作者William D. Callister Jr.和David G. Rethwisch，介绍了材料科学与工程的基本概念、原理和应用，包括金属、陶瓷、聚合物和复合材料等。

2. 《材料科学与工程导论（第九版）》：作者William D. Callister Jr.和David G. Rethwisch，详细介绍了材料科学与工程领域中的原子结构、晶体结构、相变、机械性能等方面的知识。

3. 《材料科学基础（第三版）》：作者Robert W. Cahn、Peter Haasen和E. J. Kramer，涵盖了材料科学的基本概念、原理和应用，包括晶体学、相图、材料缺陷、热处理等内容。

4. 《工程材料及其应用（第五版）》：作者Florin Cîrciumaru，介绍了与工程材料相关的各种性能、结构和应用方面的知识，包括金属、陶瓷、高分子材料和复合材料等。

5. 《材料科学与工程（第八版）》：作者William D. Callister Jr.和David G. Rethwisch，包含了材料科学与工程的基本理论、方法和应用，涵盖了材料的结构、性能与加工等方面的内容。

6. 《材料选择与设计：材料性能图谱与数据手册》：作者Michael F. Ashby，介绍了材料选择和设计的理论和方法，提供了各种材料的性能图谱和数据，帮助工程师进行材料选择和设计。

7. 《高级材料科学与工程学》：作者G. Prabhakaran，涵盖了材料科学与工程领域的各个方面，包括材料的结构、性能、加工和应用等，同时还介绍了一些前沿的研究和应用领域。

这些书籍可以帮助学习者深入了解材料科学与工程的基础理论、原则和应用，适合材料专业学生和从事相关工作的专业人士阅读。