材料科学基础答案余永宁
材料科学工程基础_余永宁_第九章
9.1.4回复动力学
等温退火时电阻回复及机械性能回复都有相似的动力学特征， 共同点是：无孕育期，开始阶段变化率最快，然后逐渐变慢， 最后趋于零，性质回复到某一稳定的特征值。
I型动力学符合如下关系：
dr a r a ln t b dt t 回复速率和温度有Arrhenius关系
dr d t a t A exp( Q RT )
量热法
纯镍
再结晶
回复
回复
回复
再结晶
9
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纯铜经两种不同拉伸应变(真应变分别为0.28和0.16?) 后等温退火时能量释放率随退火时间变化的关系。
低形变试样再结晶 出现的时间比高应 变的迟。回复过程 只释放少量的能量， 且它不因形变量不 同而有很大差异， 而低形变量试样释 放的总能量比高形 变量试样低。
随着组织结构的变化，物理和机械性能也有变化，它们向 着未形变前的值变化。因回复过程在光学镜下难以直接观 察到，要用特殊方法测量：
研究方法
①量热法，测量回复时放出的储存能； ②电阻法，测量回复过程电阻的减小量； ③测量回复过程硬度或流变应力的降低量； ④测量回复过程位错密度的减小以及位错排列结构的变化； ⑤测量因形变而使X射线谱线的宽展和在回复过程中锋锐化程度；
多晶铁0C形变5%的回复动力学 (左)应变硬化回复程度r 与lnt 间 的 函 数 关 系 ； (下)回复激活能Q 与 回 复 分数间的关系(复杂 形变,激活能变化);
II型回复动力学符合如下关系：
dr c1r m dt
r (m1) r0(m1) (m 1)c1t
18
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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 and most materials contrast as they are cooled, the density of the glass is usually less than that of the liquid. The density ofthe 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 is frequently used by chemical physicists,who study noncrystalline materials.2.3. FIND:How does C p of an amorphous material change as the temperature isincreased through the glass transition temperature?SOLUTION:Heat capacity is an intensive property, one that does dependon the bulk properties of the material. Most thermodynamic intensiveproperties behave exactly the same as (molar) volume in the vicinity of theglass transition temperature. Hence, heat capacity changes 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 thermalexpansion coefficient of the glass is about 1/3 of that of the melt.SOLUTION:The linear thermal expansion coefficient is αth and thevolumetric thermal expansion 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 materials expands isotropically to length 1+δ1. We do the problem firstusing differentials.1 / 13Now 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, theglass transition occurs over a range of temperatures. T g is rate sensitive andstructure sensitive; it depends on the rate of heating or cooling and on thelocal structure which is statistically variable in a glass or melt.10. FIND:The temperature range for transitions that involve units smaller than amer.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 requiresless thermal input, kT, to excite smaller units, such as rotation of the ring sidegroup 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, sothis is not a design problem. That is fortunate, since flexible materials(polymers) do not stand up to radiation. (Look at what happens to your skinwhen you expose it to bright sunshine.) Temperature can also be a problem.High temperatures can soften, melt or degrade polymers. Low temperaturecan change a rubber to a glass, making a material that is flexible at roomtemperature to brittle at high temperature. Rubber seals that have becomeglassy may break or leak. Metal seals will decrease in volume as thetemperature is decreased. If your choice is a rubber gasket, then it needs toremain flexible at use temperature. Liquid oxygen is very cold and embrittles many materials.2 / 1312. 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 toidentify. 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 similarbut different properties - a material that flows under low stress but does notflow under its own weight (Bingham plastic). Plumbers putty is an example, but it also shows the characteristics of silly putty.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 decreasesexponentially. If motor oil does not behave this way, then it behaves in anunusual manner.COMMENTS: Motor oil, in fact, does behave in an unusual manner. It'sviscosity is essentially constant with temperature! Let me try to explain howthis is accomplished. Oil contains polymer molecules in a solvent. Themolecules do not like the solvent all that much, so they tend to ball up or coilsomewhat tightly on themselves. As the temperature is increased, theinteraction between solvent and polymer changes. The polymer begins to like the solvent, so it uncoils. The polymer molecules then become entangled inone another, raising the viscosity, which counteracts the normal decrease withincreasing temperature.14. FIND:Whether it is more difficult to obtain a GIVEN: shear strain rate witha high viscosity fluid 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 isproportional to velocity gradient. The constant of proportionality is viscosity:τ = η(dv/dx). Since the velocity gradient is invariant in this problem, the3 / 13shear stress varies with viscosity. The higher viscosity oil will require alarger stress to maintain the plate velocity. Skotch® tape is a polymer filmwith a thin coating of an oil-like material. It is simply the thinness of the oilfilm and its viscosity that prevents slippage between substrate and film.COMMENTS: When a large stress is required to shear a fluid, then muchwork 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 the viscosity 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, viscosity decreases manyorders 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 morelikely to be semicrystalline?GIVEN: The structure of PP is shown in Table 6.4-1. The structure ofatactic and isotactic polymer is shown in Fig. 6.4-5.SKETCH: i-PP has the methyl groups all on the same side. (The H sidegroups are not shown, but each carbon is bonded to 4 atoms. You shouldmentally visualize all the H atoms.)C C C C CCH3CH3CH333a-PP has the methyl groups appearing randomly on one side or the other.C CCC CCC33CH33CH3SOLUTION: 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 melting temperature, 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 is used in huge quantities.4 / 1317. FIND:Show the stereo isomers of PAN.GIVEN::PAN is poly(vinyl cyanide).5 / 13SKETCH: (H are not shown)isotacticsyndiotacticatacticCOMMENTS: Commercially available PAN is atactic.18. FIND: Which polymer is more likely to be semicrystalline, PVdF (which has 2F's) or PVF (a vinyl polymer, which has 1 F)?SOLUTION: Note that the PVdF [poly(vinylidene fluoride)] is symmetric.Symmetric molecules are easier to pack than nonsymmetric ones. PVdF issemicrystalline. 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 the melting 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 onthe specific polymer characteristics.20. FIND:Calculate the number of mers or degree of polymerization is a sampleof PP with a molecular 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 ofcalculations because the effect is negligible.21. FIND: Calculate the MW of a cellulose mer. If a molecule of cotton has aMW of 9,000 g/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 16.00 g/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 = 312.206 / 13g/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 formedby joining only about 29 mers.22. FIND:How will the addition of pentaerythritol affect the crystallinity andglass transition temperature of PET?GIVEN:Pentaerythritol is tetra functional.SOLUTION:Any branching agent will disrupt the ability of the moleculesto pack efficiently. Hence, crystallinity or the potential for crystallinity willbe reduced. Crosslinking makes molecular motion more difficult, so it raisesthe 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.Cleaving bonds in crystals destroys the balance of order. Some bonds arebroken, creating a difference in the bond arrangement than exists in the virgincrystal.COMMENTS:Organic molecules form crystals readily under appropriateconditions, but the structure of the crystal and the unit cell parameters do notresemble those of similar polymers.24. FIND: Does the fact that PP, crystallized under quiescent conditions, ischaracterized by Maltese cross patterned spherulites that fill the entire sampleimply 100% crystallinity?SOLUTION:Spherulites are aggregates of crystalline and noncrystallinematerial. Thus, a completely spherulitic sample is semicrystalline.25. FIND:Explain why amorphous PET that is hot stretched becomes opaqueand slowly cold drawn 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 notinduce crystallization in PET, whereas stretching above T g and orienting themore mobile molecules into a position similar to those occupied by themolecules in a crystal likely induces crystallization.COMMENTS:In making a very strong PET fiber it is necessary to orientthe molecules at first without inducing crystallization. Subsequent drawingsteps are typically carried out at progressively 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 therings; there are 3 as OH groups on each of the 2 rings; and there is one aspart of each of the 2 rings. Add them up: 2 + 2 x 3 + 1 x 2 = 10 O per mer.COMMENTS: They are extremely important in providing cellulose its properties!7 / 1327. 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. Theyloosen the silicate network, lowering the glass transition temperature significantly.Thus, they are added to silica to reduce the cost of raw materials and, moreimportantly, 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 acoordination number of 3 or 4. The valence of lead is such the PbO is theoxide predicted. Hence, PbO is not a good glass former. It is anintermediate.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 andbelow T g at room temperature. They are glasses. The crosslinks preventcrystallization.COMMENTS: On of the problems with some composite matrices is that theyare brittle and not tough. Thermoplastic matrices are being developed forvarious demanding applications. High molecular weight thermoplastics,which are semicrystalline polymers, have high viscosity. It is difficult to getthem 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 werelike a coin in your pocket, it might burn you. Use a calorimeter to quantifythe effect. X-ray diffraction will also show when crystallization occurs. Thedensity or specific volume of the material changes with crystallization.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:SiO2is a good glass former; PbO is not.SOLUTION:Since they form a solution, we expect the solution, which iseven more complex than either component, to be even slower to crystallize.Thus, PbO-SiO2mixtures should be excellent glass formers so long as thePbO 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.8 / 139 / 13 SOLUTION: The additives hinder crystallization, by increasing theviscosity of the melt, thereby reducing the diffusion coefficient, and byincreasing the size of the unit cell, thereby making 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 the fiber axis, forces are transmitted along covalent bonds only.SKETCH:M o d u lu sM o le c u la r M is a lig n m e n t a lo n g F ib e r A xisSOLUTION: 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 are packed together in a parallel fashion and stresses are carried by covalent bonds. The fiber requires 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 liquidnitrogen tank?GIVEN: Liquid nitrogen boils at a very low temperature.SOLUTION: You need a material that remains flexible even down to liquid nitrogen temperatures and one that does not react with nitrogen. Somesilicone rubbers (thermoset elastomers) are currently used for this application.36.37. FIND: Why is the modulus of Spectra ® PE roughly 30 times greater than thatof sandwich bag PE?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 molecularorientation in sandwich bag PE.38. FIND: Predict interesting properties of poly(dimethyl siloxane).SKETCH:SOLUTION:The polymer contain no carbon in the backbone. It isinorganic. The methyl side groups preclude crystallization, so the material is amorphous. The molecule is symmetric and its T g is well below roomtemperature. Hence, it is a rubber. Lightly crosslinked, it has excellentelastomeric properties. Because of its chemistry, it is chemically inert inmost environments, 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 toapproximate the end-to-end separation. There are 2 bonds, each 1 = 1.54Aper mer. The backbone is all carbon, so the factor (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 termsneglected in equation 6.6-2. A more sophisticated calculation shows thevalue 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 12.01g/mole + 5 x 1.01 g/mole = 101.13 g/mole. θin 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:10 / 1311 / 13 150,000 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 to estimate the end-to-end separation: (The molecule will look as is depicted in Fig. 6.6-8b.)end to-end separation = l m A 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 themechanical properties of 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 molecules cannot 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 crosslinkingis induced after fiber formation. Rubber bands, which are essentially thickfibers, are generally not round in cross-section. They are slit sheets, so thefibers are square or rectangular in cross-section. Thermoplastic fiber can bemelt- 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 on themselves. When they move to coil, the fiber shrinks.COMMENTS:Most of the shrinkage occurs during the first heat. Hence,the fiber can be heat set to minimize subsequent shrinkage.45. FIND:Explain why the modulus of rubber is much lower than thatcharacteristic of ceramic or oxide 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 glasses respond to stress by attempting to push or pullatoms out of their energy wells. Conformation changes are much easier toachieve. Rubber elasticity is entropy driven. Hookean elasticity is energydriven.46. FIND:Calculate the end-to-end separation of PP (a) coiled on itself and (b)completely stretched 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 moleculefor 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 twice12 / 13that 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 = 184A Note 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[文档可能无法思考全面，请浏览后下载，另外祝您生活愉快，工作顺利，万事如意!]13 / 13。
材料科学工程基础_余永宁第六章
1947年 Cottrell阐明溶质原子和位错的交互作用并用以解释低碳纲
的屈服现象，第一次成功地利用位错理论解决金属机械性能的具 体问题。 同年 Shockley描绘了面心立方形成扩展位错的过程 1950年 。 Frank和Read共同提出了位错的增殖机制。 上面所列出的是早期位错理论的发展的重要过程，到那时，对于 单个位错的运动规律，位错的交互作用等理论基本已经解决。 1953年Nye和1954年Bilby以及以后的krö ner提出的无限小位错连续分 布模型，为研究更复杂位错组态提供方法。 在解决任意形状的位错线的性质方面，由Burgers在1939年提 出的位移公式、Peach和krö ner在1950年提出的应力场公式和位错受 力公式及Blin在1955年提出的交互作用能公式等基本上能得到解 决 。1956年 Menter直接在电镜观察了铂钛花青晶体中位错的存在， 同年，Hirsch等应用相衬法在电镜观察到位错的运动，位错理论就 在更坚实的基础上发展了。 近几十年，随着实验设备和计算机的发展，研究位错核心的 组态以及在复杂结构中的位错方面取得很多很有成效的结果。
第六章
位错 （Dislocations）
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引言
•学习本章的意义：
strength
晶体非完整性/ 力学性能与缺陷密度的关系；
theoretic strength strength of crystal beard strength of metallic glass
6.2 位错基本概念
理论切变强度的估算 假设能量曲线是正弦形式。这 样，要使原子面相对切开所需要 的切应力为： 2nx
ı ı C sin
材料科学基础2版余永宁 (5)
区域内的相数之间有如下关系:
R R D D
式中，R’是邻接两个相区的边界维数；R是相图的维数；D、D+ 分别是从一个相区越过边界进入邻接另一相区后消失和出现的 相数。
相图的局限性： 相图是描述体系平衡状态的 ，不能说明达到平衡过程的动力学， 不能知道转变后的组织，也不能判断体系中可能出现的亚稳相 。 由于固态材料往往难达到整体稳定的平衡 ，实际测得的相图多数 都或多或少地偏离真正平衡，甚至有些相实际上是亚稳相。
因为①大多数聚合物熔化过程永远达不到平衡，只有稀溶液中 的聚合物才有可能达到平衡；②聚合物中含有很多大分子，每一 个分子的相对分子质量都不同，每个分子的行为也不同，熔融的 聚合物系统比较复杂；③实验观察和计算表明，除非聚合物的化 学结构非常相似，否则它们通常是不相容的（即它们不会溶合）。 所以，聚合物相图的实际意义不大，在本章不讨论聚合物相图。
● 2相平衡：F=1。两相平衡点的集合构成一对共轭曲线，若要维 持两相平衡，当温度改变时，两个平衡相的成分随之而变。
● 单相平衡： F=2，即相的温度和成分可以独立改变而不改变平 衡状态，即单相平衡点的集合构成任意形状的区域。
5.3.1 单相平衡和两相平衡
单相平衡在恒压的T-x（或w）相图上相区表现为一平面区域。 两相平衡时，每一个参与两相平衡的相成分是温度的函数，两 个平衡相的这一函数关系在相图上表现为一对共轭的温度-成分 线，这种平衡又称两变量平衡。这对共轭线所包围的区域称为 两相区。恒温连接线与共轭线两个交点的成分是两个相的平衡 成分，它不因体系成分改变而改变。 注意区分两相区及两相平衡的区别。
F 2 1 3
一般选取的2个强度变量是T、x（或w），x和w分别是摩尔分数和 质量分数。A-B二元系任一成分的2个组元摩尔分数之和为1，即 xA+xB=1；2个组元的质量分数w之和亦为1，即wA+wB=1。
材料科学基础2版余永宁 (8)
一定的参考坐标架而言的。第i组元在该处相对于所选参考坐标架 的流量Ji为：
Ji Civi
vi , Ci(质量浓度 i，kgL3)分别是第i组元的移动速度和体积浓度。
两种参考系：
实验参考系：参考系是相对于观察者不动的。一般把坐标架固 定在所研究试样的端部，我们忽略试样尺寸的变化，这种参考 坐标架也可以固定在离试样端部任一固定的距离处。相对这种 参考系的扩散流量记为J0。 点阵参考系：参考系是把坐标固定在晶体点阵的原子面上。相 对于这种参考系的扩散流量记为J。可以在晶体内插入惰性标志 物来获得这种参考系。
若在某处点阵参考系相对于实验参考系的运动速度为v，则在该 处第i组元（“组元”应包括空位）相对于两种参考坐标架的流量 之间的关系为：
J
0 i
Ji
Ci v
Ci (vi
v)
系统中含有n个组元，则系统的体积浓度C应为：
n
C Ci i1
因为空位浓度非常低，在计算其它组元浓度时可以忽略它，故上 式的i也不包括空位。相对于点阵坐标架，由于点阵固定，所以通 过点阵坐标架的流量总和为零，即
Ji DikCk
k 1
Dik是比例系数，称扩散系数，它的单位是cm2/s 。
i j k x y z
对于实验坐标架，因J
0 i
Ji
Ci v
Ci (vi
v)
n1
J
0 i
DikCk Civ
k1
，得
可以看到，一个组元的扩散流量不单受自身的浓度梯度控制，
也受其它组元的浓度梯度的控制。i=k时，Dik即组元扩散流量和 自身浓度梯度的比例系数，当ik时，Dik即扩散流量和其它组元 浓度梯度的比例系数。
材料科学基础2版余永宁 (4)
在熔点时的 数值对不同材料是不同：一般的bcc 金属，大约为
0.11，一般的fcc金属，大约为0.07。
非晶态是一种过冷液态 不具有长程有序，但因为它相对于液体熔点有很大的过冷，所 以原子是定域化的。
材料科学基础
北京科技大学 材料科学与工程学院
编制
第4章 非晶态与半晶态
固态
物体可分为
液态
气态
从结构看可分为 原子的迁移性接近
晶态
非晶态
气态
刚性固体
周期性长程有序
缺少长程有序 具有短程序
原子的迁移性接近
非晶态材料
• 大多数热固性塑料 • 氧化物及非氧化物（硫属化合物及氟化物）玻璃态 • 非晶态聚合物、干凝胶 • 非晶态半导体 • 非晶态金属或合金 • 非晶态电介质 • 非晶态离子导体 • 非晶态超导体
非晶态材料具有其他状态物质所没有的特性和优异性能。 在很多新材料应用领域如：光通信材料、激光材料、光集成电 路、新型太阳能电池、高效磁性材料、输电和输能材料等其都 是研究和开发的热点。
聚合物
• 聚合物是一类由长分子组成的有机材料，通常由碳骨架与 其他元素或结构单位连接作为侧基构成。
• 聚合物不容易晶化。即使在缓慢冷却条件下，聚合物通常 只是部分结晶或完全不结晶，而形成半晶态或非晶态。
• 在20世纪已经能够人工合成聚合物，并且可通过人工调整分 子的结构获得所希望的性能，这使得聚合物成为一大类重 要的材料，在建筑、汽车、通信甚至航空工业等领域都获 得重要的应用。
本章主要讨论非晶态和半晶态的结构、它们的基本特 征和一些简单材料。
4.1 非晶态
新版新疆大学材料与化工专业考研经验考研参考书考研真题
在决定考研的那一刻，我已预料到这一年将是怎样的一年，我做好了全身心地准备和精力来应对这一年枯燥、乏味、重复、单调的机械式生活。
可是虽然如此，我实在是一个有血有肉的人呐，面对诱惑和惰性，甚至几次妥协，妥协之后又陷入对自己深深的自责愧疚当中。
这种情绪反反复复，曾几度崩溃。
所以在此想要跟各位讲，心态方面要调整好，不要像我一样使自己陷入极端的情绪当中，这样无论是对自己正常生活还是考研复习都是非常不利的。
所以我想把这一年的经历写下来，用以告慰我在去年饱受折磨的心脏和躯体。
告诉它们今年我终于拿到了心仪学校的录取通知书，你们的付出和忍耐也终于可以扬眉了。
知道自己成功上岸的那一刻心情是极度开心的，所有心酸泪水，一扫而空，只剩下满心欢喜和对未来的向往。
首先非常想对大家讲的是，大家选择考研的这个决定实在是太正确了。
非常鼓励大家做这个决定，手握通知书，对未来充满着信念的现在的我尤其这样认为。
当然不是说除了考研就没有了别的出路。
只不过个人感觉考研这条路走的比较方便，流程也比较清晰。
没有太大的不稳定性，顶多是考上，考不上的问题。
而考得上考不上这个主观能动性太强了，就是说，自己决定自己的前途。
所以下面便是我这一年来积攒的所有干货，希望可以对大家有一点点小小的帮助。
由于想讲的实在比较多，所以篇幅较长，希望大家可以耐心看完。
文章结尾会附上我自己的学习资料，大家可以自取。
新疆大学材料与化工专考研初试科目：(101)思想政治理论(204)英语二(302)数学二(815)材料科学基础新疆大学材料与化工专考研参考书目：《材料科学基础》（第三版），胡赓祥、蔡珣、戎咏华，上海交通大学出版社，2010年5月。
《材料科学基础》（第一版），余永宁，高等教育出版社，2006年5月。
《材料科学基础》（第一版），潘金生主编，清华大学出版社，1998年1月。
先谈谈英语吧其实英语每什么诀窍，就是把真题读透彻，具体方法我总结如下：第一，扫描提干，划关键项。
北科大余永宁金属学原理课后解答10解
3. 导出二元合金中母相α和析出相β均为理想溶体以及规则溶体的相变总驱动力和形核驱动 力（以 J/mol 表示）。设原始成分为 x0，在脱溶温度α相平衡成分为 xα，脱溶物核心成分 和β相平衡成分近似相等为 xβ，交互作用系数为Ω。 解：(1)设 A-B 二元合金（如下图 a 所示），x0 成分的α相在 T1 下脱溶，相变总驱动力为
=
(1 −
xβ
)(
µ
α A
(
xα
)
−
µ ) α A( x0 )
+
xβ
µ( α B( xα )
−
µ ) α B( x0 )
因
µ
Φ i
= GiΦ
+ RT ln aΦi ，故
10-3
∆G I
=
(1 −
xβ
)µ
α A(
xα
)
+
µ x α β B( xα )
− [(1 −
xβ
µ) α A(x0 )
+
x
β
µ
α B(
10-1
对于立方体状核心，设 a 为边长，形核时的自由能变化∆G 为
∆G = a 3∆GV + 6a 2γ
上式对 a 的导数等于 0 时，求出临界核心的边长 a＊，
d∆G da
= 3a 2∆GV
+ 12aγ
=0
得
a∗ = − 4γ = 4 × 0.6 m = 1.43 ×10-9 m
∆GV 7 ×108
∆G
α →α x0
'+
β
为
∆Gmα→α ' +β
=
RT
⎡ ⎢(1 ⎣
北京科技大学材料科学基础A第2章-固态结构(3)
由两种或多种元素组成、以金属元素为主体、大多通 过冶炼或熔合而成、并在宏观上具有一般金属元素所 具有的共同特征，这一类材料或物质通称为合金。
纯金属 + 异类原子
怎样进入主组元中
代位固溶体结构 间隙固溶体结构 金属间化合物类型
第三节 合金相结构
一. 影响合金相结构的基本因素 1. 异类原子间的相互作用 异类原子间的相互作用与同类原子间的相互作用相比： ≈ 不使能量变化 均匀混合 > 使能量降低 有序固溶体 金属间化合物 正常价化合物 < 使能量升高 各自集聚、各自固有的晶体结构 2. 原子相对尺寸因素 原子尺寸差异，使得点阵发生畸变，引起能量升高， 量 变 结构变得不稳定，结构类型发生变化。 质 变
第三节 合金相结构
元素周期表中 同一周期 原子序数 负电性 同一族 原子序数 负电性
组元间负电性相差越大，
有利于形成化合物，不利于形成固溶体。
形成的化合物越稳定，固溶体的固溶度越小。 当电负性差小于0.4～0.5时，才有可能获得较大的固溶度。 金属元素 + VIA S、Se、Te, VA P、As、Sb、Bi, IVA Si、Ge、Sn、P
合金元素原子百分数
给定金属晶体结构，电子浓度有限，超过限度， 结构不稳定，发生改组 一价fcc, C电子<1.4（1.36） 一价bcc，C电子<1.5（1.48）
第三节 合金相结构
电子浓度确定困难： ① 每个原子不一定都贡献价电子； ② 变价金属的原子价随条件而变； ③ 过渡金属的 ns 层电子既可以为价电子， 又可以进入(n-1)d层和(n-2)f层， 而 (n-1)d层和(n-2)f层的电子又可能贡献价电子。 对于过渡金属的价电子数，通常有两种处理方法： 按零处理； 确定为除去与惰性气体相当的满壳层电子以外的所有电子。
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2版余永宁 (2)
自由电子的波函数
Ψ r,t Aexp it r k Aexp i t r p
A为振幅，t为时间，r为空间 位置矢量，i为虚数单位。
依据量子理论微观粒子的状态完全由波函数Ψ r,t 描述。
一般波函数 Ψ r,t的物理意义：粒子出现的几率密度
Hund法则通过光谱试验规律总结，得出基态下未满的亚电 子层中电子的排布情况，有关内容涉及了电子的角动量，故 此在介绍角动量之后，介绍Hund法则
电子能级的试验测试手段
设法使电子在不同能级之间跃迁， 尤其在基态与激发态之间变化
能量守恒：电子能级跃迁释放 光子的能量波长关系
Eph
h ph
hc
ph
~
E2
载流子 电子对 自由电子 电子、空穴 电子和/或离子
电导率 ： 1 1m-1
例(室温下)： • 金属 Ag —6.3x107 ，Cu —6.0x107 ，Al —4.3x107，Fe —1x107； • 纯 Si — 掺杂1ppm的As、P、B、Al后(代位)，提高100 000倍； • PE — <10-14, PP : <10-13 , 酚醛树脂(电木)—10-11 ， • SiO2 — <10-12 ， 石英 —10-18
第二章 固体电子态
课程要求 •具体内容要点——第二章总结
1. 理解、掌握电子的波动性及其应用 2. 全面理解并掌握原子的内层电子状态 3. 了解并掌握固体原子的外层电子状态、特征和描述方法 4. 了解固体材料电子状态的一些实际应用 5.了解电子态的基本理论依据
课程内容框架
固体电子态——在材料学科中的应用
有关材料的研究工作，有大量的关于
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
第二章 晶体缺陷1.0
第二章 晶体缺陷问题2.1.1根据最近邻假设，估算简单立方晶体中一个空位的形成能。
（假定每一个体内原子间的结合能为U 0）答：首先，简单立方晶体中最近邻原子数为6个，形成一个空位断6根键，空位处原子移至晶体表面成键3根，故空位形成能为021U ；而形成空位后周围原子向空位处偏移导致应变能增加，该移动同时导致结合能的增加。
由于该偏移是自发过程，所以能量降低，综上，空位形成能小于021U 。
（参考P184 图6-9）？问题2.1.2自间隙原子的形成能远大于空位形成能。
请从应变能和结合能的角度给予分析。
答：自间隙原子的半径远大于间隙尺寸，因此会引起应变能很大的上升；（结合能？）？问题2.1.3以室温为参考点，去测量纯铜的点阵常数随温度变化率，然后再测量纯铜的线膨胀系数随温度的变化，将两条随温度变化的曲线画在一张图上，你认为200℃一下会怎样？900℃以上的高温有会怎样？答：问题2.1.4 图2－2中的置换原子(黑色)的尺寸画得有些随意。
假定(b)图中黑原子半径比白的小10％，而(c)图中大10％，问哪种情况下基体内的应变能更大些？为什么？如果数据由10%变为0.1%，上述结论会变化吗？为什么？ 答：（b ）图中应变能更大。
①应变能是由附近白原子点阵常数的变化引起的结合能的改变量。
②由结合能的图像可知，在平衡位置r0左右，曲线并非对称(形变大时非弹性成分的存在)。
产生相同且较大(与0.01%对比着看)的形变时，压缩引起的应变能更大。
若数据仅为0.1%，则看不出差别。
问题2.1.5对于置换固溶体，溶质加入对点阵常数有影响吗？请对溶质原子直径大于溶剂直径的情况予以分析。
答：有影响。
溶质原子直径大于溶剂直径时，溶质原子溶入对周围原子施加压应变，使周围原子远离溶质原子，导致晶格常数增大。
问题2.1.6 Al 2O 3溶入MgO(具有NaCl 结构)中，形成的非禀性点缺陷在正离子的位置，还是相反？答：Al 2O 3溶入MgO 晶体，由于Al 离子是+3价，，而Mg 离子是+2价，所以当两个铝离子取代两个镁离子的位置后，附近的一个镁离子必须空出，形成的非禀性点缺陷在正离子的位置。
北科大余永宁金属学原理课后解答4解
同距离 x1 和 x2 所对应的时间 t1 和 t2 有如下关系：即
x1 = x2
Dt1
Dt2
即
t2
=
( x2 x1
)2 t1
故在距表面 0.1cm 处获得同样的浓度（0.45%）所需时间 t2 为
t2
=
( 0.1 )2 0.05
×1.822 ×104 s
=
7.288 ×104 s
=
20.24h
③根据②的解释，同一温度下渗入距离和时间关系的一般表达式为
②因扩散系数随浓度线性变化，设
D=a+bC
因 D1 = a + bC1 D0 = a + bC0
求得
a
=
D1
−
D1 C1
− −
D0 C0
C1
b = D1 − D0 C1 − C0
扩散流量 J = −(a + bC) dC dx
上式积分得 − Jx = aC + b C2 + d 2
边界条件：x=l，C=C0；代入上式得：
把数据转换成 lnα和 x2，得
x2×10−2=z
0.01 0.04 0.09 0.16 0.25
lnα=y
8.52 8.29 7.83 7.25 6.26
用线性回归，方程 y=a+bz 得
a=8.659 b=−935.82
4-4
D=− 1 =
1
mm2 ⋅ s−1 = 3.71 × 10−9 mm2 ⋅ s−1
距顶端距离 x/cm 0.1 α(任意单位) 5012
0.2 3981
0.3 2512
0.4 1413
0.5 524.8
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。
材料科学基础课后习题第1-第4章
材料科学基础课后习题第1-第4章《材料科学基础》课后习题答案第一章材料结构的基本知识4. 简述一次键和二次键区别答：根据结合力的强弱可把结合键分成一次键和二次键两大类。
其中一次键的结合力较强，包括离子键、共价键和金属键。
一次键的三种结合方式都是依靠外壳层电子转移或共享以形成稳定的电子壳层，从而使原子间相互结合起来。
二次键的结合力较弱，包括范德瓦耳斯键和氢键。
二次键是一种在原子和分子之间，由诱导或永久电偶相互作用而产生的一种副键。
6. 为什么金属键结合的固体材料的密度比离子键或共价键固体为高？答：材料的密度与结合键类型有关。
一般金属键结合的固体材料的高密度有两个原因：（1）金属元素有较高的相对原子质量；（2）金属键的结合方式没有方向性，因此金属原子总是趋于密集排列。
相反，对于离子键或共价键结合的材料，原子排列不可能很致密。
共价键结合时，相邻原子的个数要受到共价键数目的限制；离子键结合时，则要满足正、负离子间电荷平衡的要求，它们的相邻原子数都不如金属多，因此离子键或共价键结合的材料密度较低。
9. 什么是单相组织？什么是两相组织？以它们为例说明显微组织的含义以及显微组织对性能的影响。
答：单相组织，顾名思义是具有单一相的组织。
即所有晶粒的化学组成相同，晶体结构也相同。
两相组织是指具有两相的组织。
单相组织特征的主要有晶粒尺寸及形状。
晶粒尺寸对材料性能有重要的影响，细化晶粒可以明显地提高材料的强度，改善材料的塑性和韧性。
单相组织中，根据各方向生长条件的不同，会生成等轴晶和柱状晶。
等轴晶的材料各方向上性能接近，而柱状晶则在各个方向上表现出性能的差异。
对于两相组织，如果两个相的晶粒尺度相当，两者均匀地交替分布，此时合金的力学性能取决于两个相或者两种相或两种组织组成物的相对量及各自的性能。
如果两个相的晶粒尺度相差甚远，其中尺寸较细的相以球状、点状、片状或针状等形态弥散地分布于另一相晶粒的基体内。
如果弥散相的硬度明显高于基体相，则将显著提高材料的强度，同时降低材料的塑韧性。