材料科学基础(武汉理工大学,张联盟版)课后习题及答案 第七章

《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。

SOLUTION FOR CHAPTER 71. FIND: Label the phase fields in Figure HP7-1.SOLUTION: There are two regions of single phase equilibrium separated by a region of two-phase equilibrium. The line separating the single phase liquid from the two phaseliquid and solid region is called the liquidus, and the line separating the two phase liquid and solid from the single phase solid is called the solidus.SKETCH:2. FIND: The name given to the type of equilibrium diagram shown in Fig. HP7-1.SOLUTION: The diagram is termed isomorphous. Above the liquidus there is only one phase, namely the liquid, and it has the same structure regardless of composition. Below the solidus is the solid phase, and as well it has the same structure regardless ofcomposition.3. FIND: What is the crystal structure of component B, why?SOLUTION: There are no phase boundaries below the solidus consequently the phase and therefore the structure must be the same. If A is FCC, then B and all compositionsbetween A and B must be the same phase and hence same structure. If A is FCC, then B must also be FCC.4. FIND: Sketch equilibrium cooling curves for alloy X o and pure component B. Explainwhy they have different shapes.SOLUTION: The slope of the temperature versus time behavior for the alloy in thesingle phase region is controlled by the cooling rate and the heat transferred from theliquid. At the liquidus temperature a small amount of solid is formed, releasing anamount of latent heat of fusion related to the volume of solid transformed. Hence, at the liquidus temperature there will be a change in slope. Since heat is being generated theslope will be less than that above the liquidus. As the temperature is reduced through the two-phase region a small amount of solid is formed and a corresponding heat released;once the solidus temperature is reached no additional transformation takes place and there is again a change in slope. The slope is greater as the solid cools than as the liquid andsolid cooled. For the pure component, solid and liquid are in equilibrium at T B, and hence,a horizontal line when liquid transforms to solid at T B.SKETCH:5. FIND: The liquidus temperature and the solidus temperature of alloy X oSOLUTION: From figure HP7-1 the liquidus temperature is approximately 1110o C and the solidus temperature is approximately 1070o C.6. FIND: Determine compositions and phase fractions of each phase in equilibrium at1100o C for alloy X o.SOLUTION: At equilibrium, the temperature of two phases must be the same, and thecomposition of the solid is found where the tie line intersects the solidus and the liquid is at the intersection of the tie line with the liquidus. From the phase diagram thecomposition of solid is 0.35B and that of the liquid is 0.55B. Using the lever rule thefraction of liquid, f L, and fraction of solid, f s is determined.7. FIND: Sketch f L and f S for alloy X o as the alloy is cooled under equilibrium conditionsfrom 1200o C to room temperature.SOLUTION: At the liquidus temperature the amount of liquid is almost 100% with onlya very small amount of solid. Conversely at the solidus temperature, the amount of solidis almost 100% with only a very small amount of liquid. In a two-phase field:f L + f s = 1.8. FIND: Changes in the compositions of the liquid and solid phases during quilibriumcooling of alloy X o through the two-phase field.SOLUTION: At the liquidus temperature, the composition of the solid is approximately0.30B. As equilibrium solidification progresses, the composition of the solid increases to0.5 B, the maximum value it can reach. The liquid on the other hand is initially thecomposition of the alloy X o, and as equilibrium solidification progresses the composition increases in B to the maximum composition in the liquid of approximately 0.7 B.9. FIND:From the following data construct a plausible equilibrium phase diagram.Component A melts at 800︒C and B melts at 1000︒C; A and B are completely soluble in one another at room temperature; and if solid α containing 0.3 B is heated underequilibrium conditions, the solid transforms to liquid having the same composition at500︒C.GIVEN: Melting temperatures for the two components, congruent melting temperature and composition.SOLUTION: The sketch shown below satisfies all of the requirements stated in theproblem. The melting temperature of pure A is 800︒C and that for pure B is 1000︒C. The congruent melting temperature is at 500︒C at a composition containing 0.3 B. Below the congruent melting temperature there is a continuous α solid solution.10. FIND: For alloys containing 10, 22, 25, 27, and 40 wt% Ni, determine the number ofphases present and the composition of the phases in equilibrium at 1200o C.11. FIND: Beginning with a statement of mass balance, derive the lever rule in a two-phasesystem.SOLUTION: In a two phase field for an alloy of some over all composition X o, thesolute is distributed in the two phases: x o = fα xα + fβ xβ where the compositions areexpressed in terms of one of the components, fα + fβ = 1. Then, fα= 1 - fβ. Substituting:x o = (1 - fβ) xα + fβ xβx o = xα - xα fβ + xβfβx o - xα = (xβ - xα) fβ12. FIND: Discuss each of the factors that permit the Cu-Ni system to be isomorphous overthe temperature range 350-1000o C.SOLUTION: The empirical rules of Hume-Rothery identify the characteristics that two elements must have in common for extensive solubility. This should require thata. the two components must have the same crystal structureb. the atomic radii of the two atoms must be similarc. the two components have the comparable electro negativities, andd. the two components have the similar valence.13. FIND: What does the temperature 322O C represent in Figure HP7-2?SOLUTION: 322o C is called the critical temperature. At the critical temperature there is a corresponding critical composition. For an alloy of this composition, cooling under equilibrium conditions from above the critical temperature to below this temperatureresults in the formation of two phases from one phase of the critical composition. Ascooling continues the two phases that form have different compositions.14. FIND: In Figure HP7-2, are α1 and α2 different crystal structures?SOLUTION: α1 and α2 are two phases having the same structure but differentcompositions. The composition of the two phases are determined as in any two phasesystem by using the tie-line. Where the tie-line at the equilibrium temperature intersects the phase boundaries determines the composition of the two phase in equilibrium.15. FIND: Location of the equilibrium phase boundary at temperature T.GIVEN: Alloy 1 containing 30%B and alloy 2 containing 50% B, when equilibrated at temperature T are in the same two-phase (L + S) region. The fraction of liquid in alloy 1 is 0.8 and the fraction of liquid in alloy 2 is 0.4.SKETCH:SOLUTION: Since the fraction of liquid in alloy 1 is greater than that of solid, theliquidus is to the left of 0.3, and since the fraction of solid in alloy 2 is greater than that of liquid, the solidus must lie to the right of alloy 2. Using the lever rule:for alloy 1for alloy 20.8X L - 0.8X s = X s - 0.30.4X L - 0.4X s = X s - 0.5Solve for X s and X L,0.8X s - 0.8X L = X s - 0.30.8X s - 0.8X L = 2X s - 1.0____________________0 = -X s + 0.7, orX s = 0.7BTo find X L, substitute X s into one of the equations:0.4 - 0.56 = -0.8X L-0.16 = -0.8X L, orX L = 0.2BAs a check, use the other equation to calculate the fraction of liquid from thecompositions.16. FIND: Label all regions of the phase diagram and the boundaries of monovariantequilibrium for the diagram shown in Figure HP7-3.SKETCH/SOLUTION:17. FIND: Sketch an equilibrium cooling curve from above the eutectic to roomtemperature for an alloy of eutectic composition.SKETCH/SOLUTION:18. FIND: Explain why the equilibrium cooling curves for alloys on either side of theeutectic composition will be different than the equilibrium cooling curve for a eutecticalloy.SOLUTION: At the eutectic temperature, liquid of composition X E is in equilibriumwith two solids, Xα and Xβ . For a hypoeutectic alloy (composition to the left of theeutectic), the first phase to form at the liquidus temperature of the alloy is α. When the eutectic temperature is reached the liquid of the eutectic composition is in equilibriumwith the two solids, one of composition Xα and the other Xβ. Consequently, depending upon composition, the closer the overall composition of the alloy is to X E, the lessproeutectic α and the more eutectic. Similarly, for alloys to the right, proeutectic β will form. In terms of the phase rule at constant pressure, for the eutectic reactionF = C - P + 1 = 2 - 3 + 1 = 0. The eutectic is invariant, and solidifies under equilibriumconditions at one temperature, T E. For alloys on either side of the eutectic composition but between Xα and Xβ, the proeutectic phase forms and cooling occurs as it does in any two-phase s-l region. Once the eutectic isotherm is reached, the remaining liquid ofeutectic composition solidifies isothermally.19. FIND: The maximum solid solubility of B in A and of A in B in Figure HP7-3.SOLUTION: The maximum solubility of B in A occurs at 0.15 B and the maximumsolubility of A in B is 0.1 A.20. FIND: For an alloy of eutectic composition in Figure HP7-3, determine the compositionof the solid phases in equilibrium with the liquid.SOLUTION: The composition of the two solid phase α and β in equilibrium with liquid at T_ occurs at 0.15 B and 0.9 B.21. FIND: Plot f L, fα, and fβ as a function of temperature for the equilibrium cooling of analloy of eutectic composition.SOLUTION: At just above the eutectic temperature, the microstructure consists of allliquid of composition X E. (Determined from the phase diagram to be approximately0.63B.) Just below the eutectic temperature, the microstructure is a mixture of twophases, α and β. From the phase diagram, Xα ~ 0.15B and Xβ ~ 0.9B. To determine the fractions of α and β at temperatures below T E, use the lever-rule. Tabulated below arecompositions of the phases in equilibrium at several temperatures estimated from thephase diagram. These compositions are then used to calculate the amount of each phase present.Fraction of phases:At 600: f L = 1.0, fα = fβ = 0At 500: f L = 0Instal l Equa tion E ditor a nd do uble-click h ere to view equat ion.Instal l Equa tion E ditor a nd do uble-click h ere to view equat ion.At 400: f L = 0Instal l Equa tion E ditor a nd do uble-click h ere to view equat ion.Instal l Equa tion E ditor a nd do uble-click h ere to view equat ion.At 300: f L = 0Instal l Equa tion E ditor a nd do uble-click h ere to view equat ion.Instal l Equa tion E ditor a nd do uble-click h ere to view equat ion.22. FIND: For alloy X o (in Figure HP7-3), calculate the fraction of α that forms as primary αand the fraction of α that forms by eutectic decomposition when the alloy is cooled from 850o C to room temperature under equilibrium conditions.SOLUTIONS: The fraction of proeutectic α that forms under equilibrium conditionspresent in alloy X o cooled from 850o C to just above the eutectic is:Then the fraction of liquid which is of eutectic composition is:f L = 1 - fα = 1. - 0.69 = 0.31.The fraction of α that forms from liquid of eutectic composition, fα,E is :fα,E = (fraction of α formed from eutectic) (0.31)To check these results, the sum of proeutectic α plus that which forms from the eutectic =0.69 + 0.11 = 0.8. That amount should be the same as if we calculated directly fα that would be present just below the eutectic for alloy of composition X o.23. FIND: The total fraction of α at room temperature for alloy X o in Figure HP7-3.SOLUTION: Solubility of B in α at room temperature is approximately 0.02 A and the solubility of A in β is approximately 0.04 A (in terms of B 0.96 B).24. FIND: Equilibrium phase diagram given the information below.GIVEN: Component A melts at 900o C; component B melts at 1000o C; and there is aninvariant reaction at 600o C. The solubility of B in α is known to increase from almost nil at room temperature to a maximum of 10%. When an alloy containing 30% B is cooled under equilibrium conditions just above 600o C, a two-phase mixture is present, 50% αand 50% liquid. When the alloy is cooled just below 600o C, the alloy contains two solid phases, α and β. The fraction of α is 0.75. After cooling under equilibrium conditions to room temperature, the amount of α in the α + β mixture decreases to 68%.SOLUTION: At just above the invariant temperature there is a region of two phaseequilibrium, a mixture of 50% α and 50% L. Consequently if the overall composition is 50%β, and α is located at 10%β, then the liquid must be at ˜50% B. Just below theinvariant reaction, two solid phases are present, αat ˜10%B and β at some composition, Xβ. If fα = 0.75, then:At room temperature, if fα = 0.68, and the solubility of B in α is nil, thenThe invariant reaction is L _ α + β, a eutectic, hence the diagram shown below fits allconditions.25. FIND: A binary phase diagram consistent with the information given below.GIVEN: The binary system A-B with T B > T A is known to contain two invariantreactions of the type: L _ α + β at T1, and L _ β + γ at T2 where T1 < T A and T2 > T A ,and β (0.5 B) is a congruently melting phase at a temperature higher than T B.SKETCH/SOLUTION: Given T B > T A, and T1 < T2 and T2 > T A, then at the congruent temperature, solid transforms to liquid at the same composition, hence for the threephases, α, β and γ with some solubility, is shown below along with a similar phasediagram in which there is limited solubility of B in A and A in B and the congruent phase appears as a line compound.26. FIND: Which phase diagram in Figure HP7-4 is MgO-NiO and which is NiO-CaO?Label the regions on the diagram and identify the invariant reaction.SOLUTION: To determine whether the isomorphous system is NiO-MgO or NiO-CaO, one needs to know the ionic radii of the three cations. r(Ni2+) = 0.078 nm, r(Mg2+) =0.078 nm, and r(Ca2+) = 0.106 nm. Therefore, based upon size of the ions occupying thecation sublattice we would expect the NiO-MgO system to be the isomorphous system.27. FIND: What is the maximum solid solubility of Ag in Pt?GIVEN: Phase diagram in Figure HP7-5.SOLUTION: Maximum solid solubility of Ag in Pt is 10.5 wt% Ag.28. FIND: For an alloy of peritectic composition, what is the composition of the last liquidto solidify at 1186o C?SOLUTION: The composition of the last liquid to solidify is 66.3% Ag (the peritiectic composition is 42.4% Ag).29. FIND: What is the range of alloy compositions that will peritectically transform duringequilibrium cooling?SOLUTION: Compositions between 10.5 and 66.3 wt % Ag.30. FIND: Plot the fraction of liquid, f L, the fraction of α, fα, and the fraction of β, fβ, as afunction of temperature during equilibrium cooling from 1800 to 400o C. For an alloy of peritectic composition.SOLUTION: First, you must determine the compositions of the phase(s) that is (are) in equilibrium. If, at a particular temperature the alloy is in a single phase then thecomposition of the phase is the composition of the alloy and the microstructure contains 100% of that phase. When the alloy is in a two phase field, the compositions aredetermined in using the tie-lines, and the phase fractions are determined using the lever.If, however, the microstructure is equilibrated at an invariant temperature and thecomposition lies along the invariant line, then only the composition of the phases can be determined, not their relative amounts. For an alloy of 42.4% Ag cooled underequilibrium conditions the approximate compositions of the phases in equilibrium atseveral temperatures are summarized in the following table.Fractions of phases at 1400:At 1300:At 1200At 1100At 1000At 900At 800At 700 At 600At 500At 400Fraction of phases as a function of temperature are plotted below.31. FIND: Sketch a possible diagram given the information below.GIVEN: The two component system A-B with T A > T B contains two invariant reactions of the type L + α _ β at T1, and L _ β + γ at T2 (<T1).SKETCH/SOLUTION: Reaction at T1 is a peritectic, reaction at T2 a eutectic, T A > T B and T1 > T2.In the above figure some B is soluble in A and likewise some A is soluble in B. Thephase that melts incongruently at T1 has a change in solubility with temperature.Alternatively, we can construct a diagram where the solubility of the end members are nil and the incongruently melting phase, X, is a line compound. That diagram wouldresemble the diagram shown below.32. FIND: Label all phase fields and identify the invariant reactions in the Ag-Al Phasediagram shown in figure HP7-6.SKETCH/SOLUTION:33. FIND: Label the phase fields and identify the invariant reactions in the V2O5 - NiOphase diagram.SKETCH/SOLUTION:34. FIND: Apply the 1-2-1...rule to the V2O5 - NiO diagram at 600o C.SOLUTION: At 600o C the single-phase regions are V2O5, N-V, 2N-V, 3N-V and NiO.These narrow single-phase fields define the regions of two-phase equilibrium. Also, itshould be recalled that at a fixed temperature the composition of each phase in a two-phase field is constant. All that varies is the relative amount of each phase (i.e. the lever-rule).35. FIND: Label all phase fields and identify the invariant reactions in the V2O5 - Cr2O3system.SKETCH/SOLUTION:36. FIND: For compositions X and Y, plot the fraction of phases present as a function oftemperature.SOLUTION: From the phase diagram in Figure HP7-8 the relevant we find:Composition of alloy X ~ 0.47BComposition of alloy Y ~ 0.55BComposition of the liquid at the peritectic isotherm ~ 0.44BComposition of the eutectic ~ 0.21BLiquidus temperature of X ~ 1100o CLiquidus temperature of Y ~ 1400o CPeritectic temperature ~ 900o CEutectic temperature ~ 700o CFor alloy X, in the temperature range from the liquidus (1100o C) to just above theperitectic isotherm, f L , decreases from 1.0 to:while the fraction of the oxide increases from 0 to (1 - 0.95) = 0.05. In the temperature range just below the peritectic isotherm to just above the eutectic isotherm, the phase fractions change from:andtoandjust above the eutectic temperature.In the temperature range below the eutectic isotherm, the Instal l Equa tion E ditor a nd do uble -click h ere to view equat ion.areconstant and from the lever rule are:Similarly for alloy Y:In the temperature range from the liquidus temperature (1400o C) to just above theperitectic isotherm the fraction of liquid decreases from f L = 1 towhile the fraction of Cr 2O 3 increasing from 0 toBelow the eutectic isotherm the amount of Cr 2O 3⋅V 2O 5 and Cr 2O 3 are constant.37.FIND: Identify the invariant reactions occurring in the Cu-Pb system.SOLUTION: There are two invariant reactions in the Cu - Pb system, a monotectic at 955o C and a eutectic at 326o C. For a complete illustration examine how the reduction in solubility of the components in each other effect the monotectic systems shown in Figure 7.6-1.38.FIND: Phase diagram HP7-9 indicates no mutual solid solubility of Cu in Pb or Pb in Cu. Explain why this is not strictly true.SOLUTION: Although the phase diagram appears to indicate no solubility of Cu in Pb or Pb in Cu, as pointed out on page 274 of the text, the free energy of a pure component can always be reduced by small additions of a second component since the presence of the second component increases the randomness of the system. It then is a competition between the enthalpy and the entropy terms as to how much is added, since:∆G = ∆H - T ∆S39.FIND: If an alloy containing 63 wt% Pb is cooled under equilibrium conditions from 1100o C to room temperature, plot the fraction of phases present as a function of temperature.SKETCH/SOLUTION: When an alloy containing 63% Pb, is cooled from 1100o C to room temperature under equilibrium conditions the following phases form. Below 991o C and just above the monotectic isotherm, liquid phase separation occurs. The composition of the two liquids follow the liquidus boundary, and the amount is calculated by the lever-rule. Below the monotectic isotherm to just above the eutectic isotherm two phases are present, Cu with essentially no soluble Pb and liquid rich in Pb, the composition of which follows the liquidus. Below the eutectic isotherm there is a two-phase mixture of essentially pure Cu and pure Pb. At 991o C the microstructure is essentially all liquid containing 63% Pb. At 975o C the microstructure has separated into two liquids, one (L I ) containing approximately 80% Pb and the second (L II ) containing approximately 43% Pb.Just above the monotectic isotherm.Below the monotectic isotherm, the fraction of liquid decreases, while the fraction ofsolid Cu increases. Just below 955o CAt 900o CAt 800o CThere is so little solubility of Cu in the liquid that once the temperature drops below 600o C, the system is essentially solid Cu with liquid essentially composed of Pb. Below 326o C the microstructure is a mixture of solid copper and lead. The phase diagramsuggests that there is no measurable change in solubility of the components. Then,40. FIND: Most alloying elements used in commercial titanium alloys can be classified aseither alpha stabilizers or beta stabilizers. Figure HP7.10 contains the Ti-Al and Ti-Vequilibrium phase diagrams. Which alloying element would most likely be consideredthe alpha stabilizer? Explain.GIVEN: The Ti-Al and Ti-V phase diagrams.SOLUTION: Solid titanium exists in two crystal forms. β, the high temperatureallotrope which is bcc, and α, the low temperature form which is hcp. At 882 ︒C thesetwo phases are in equilibrium at 1 atm pressure for pure titanium. When alloyingelements are added the relative stability of the phases may change. For the Ti-Al system, adding aluminum to pure titanium increases the temperature at which α begins to form.Thus, aluminum is said toIn the Ti-V system, thetemperature at which the αphases formsdecrease with increasing V,said to be a βthus V is41. FIND: In the Ti-Al system, identify a phase and its composition that melts congruently.Estimate the congruent melting temperature from the phase diagram.GIVEN: The Ti-Al phase diagram and the definition of a congruently melting phase. A phase melts congruently, when the composition of the solid and the composition of theliquid for an alloy are the same.SOLUTION: Examination of the Ti-Al phase diagram shows that the composition near5 weight % ( ~ 10 atomic %) melts congruently at approximately 1730 ︒C.42.FIND: Label all the two-phase fields on the Ti-Al phase diagram.GIVEN: The Ti-Al phase diagram with all the single phase fields labeled.SOLUTION: To complete this problem, we need only recognize that in a binary system, phases appear across the diagram alternating 1-2-1-2-...-1. Included below is the Al-Ti phase diagram with all phase fields labeled.α + Ti Al 33Ti Al + TiAl TiAl + TiAl 2α + Ti Al 3β + TiAlβ + Lα + TiAlTiAl + LTiAl + δTiAl + δ2α700900110013001500T e m p e r a t u r e , °C23TiAl + βTiAl 23TiAl + αTiAl 3αTiAl + Al 3βTiAl + Al Allotropic phaseL + δα + β3δ + βTiAl 3βTiAl + L 170043. FIND: Label all the invariant reactions occurring in the Ti-Al system.GIVEN: The Ti-Al phase diagram.SOLUTION:See the sketch of the Ti-Al diagram below.44. FIND: From the phase diagram, estimate the temperature at which Ti3Al congruentlytransforms to α Ti.GIVEN: The Ti-Al phase diagram.SOLUTION: The phase Ti3Al transforms the α Ti of the same composition atapproximately 1200︒C and a composition of approximately 21 wt % Al (33 atomic % Al).45. FIND: When the line compound β TiAl3 is heated, what is the composition of the firstliquid that forms in equilibrium with β TiAl3? Does this compound melt congruently?Explain.GIVEN: The Ti-Al phase diagram.SOLUTION: TiAl3 appears as a line compound on the phase diagram. Atapproximately 1375︒C TiAl3 undergoes a peritectic reaction on heating underequilibrium conditions. The composition of the liquid in equilibrium with TiAl3 is notthe composition of TiAl3 and hence the phase does not melt congruently, but meltsincongruently. The composition of the liquid is approximately 63 wt % Al. Theperitectic reaction written on heating is:TiAl3→ liquid (63 wt % Al) + δ Ti.46. FIND: The presence of Ti3Al in Ti-Al alloys has a detrimental effect on the ductility. Tocontrol this problem, the amount of aluminum needs to be less than 6 wt. %.Consequently, all of the products that you are producing contain a maximum of 6 wt. %of aluminum. You have been told that additions of tin, zirconium, and oxygen (oftenpresent as an impurity) are all known to be alpha stabilizers. If there is a possibility thatsmall additions of oxygen may enter the system as your alloy is melted, should youreduce the amount of aluminum in your alloy or not worry about it if ductility is animportant property for your product? Explain your reasoning.GIVEN: The Ti-Al phase diagram.SOLUTION: Since both aluminum and oxygen are α stabilizers, a judicious choicewould be to reduce the amount of aluminum in your alloy. This would be a particularlywise choice since the critical amount is just at 6 % Al, the addition of any alloyingelement that might affect the amount of Ti3Al should be avoided. Therefore reducing the amount of Al should be reduced.47. FIND: Label all the two-phase fields in the Ti-V system.GIVEN: The Ti-V phase diagram with all the single phase fields labeled.SOLUTION: To complete this problem, we need only recognize that in a binary system, phase fields appear across the diagram alternating 1-2-1-2-...-1. Included below is asketch of the Al-V phase diagram with all phase fields labeled.48. FIND: Identify the invariant reaction that is occurring at 675︒C in the Ti-V system.GIVEN: The Ti-V phase diagram.SOLUTION: The invariant reaction that is occurring at 675︒C is a monotectoid reaction and can be written symbolically as:β1→α + β2.from 900︒C down to 500︒C, plot the fraction of phases present as a function of temperature.GIVEN: The Ti-V phase diagram.SOLUTION: The compositions and the fraction of each phase from the phase diagram are shown in the table below. From the table the sketch given below illustrates the relative fraction of phases present as a function of temperature.50.FIND: Consider an alloy containing 52 wt. % V. Describe the phases present and their compositions as the alloy is cooled under equilibrium conditions from 900︒C to 500︒C. GIVEN: The Al-V phase diagram.SOLUTION: An alloy containing 52 wt. % V at 900︒C is in the β Ti,V single phase field. When the alloy is cooled it passes through the critical temperature of a monotectoid at 850︒C. Cooling below this temperature results in the formation of a two phase mixture of V-poor and V-rich solids that have the β Ti,V structure. Decreasing the temperature results in two solid phases whose compositions continuously change. The composition of the two solids follow the solid-solid miscibility boundary as shown in the sketch below. Cooling to 675︒C results in the invariant monotectoid reaction at that temperature. Symbolically that reaction can be written as β1 → α + β2. Cooling below themonotectoid temperature results in the disappearance of the β1 phase. Since the relative solubilities of V in Ti and Ti in V decrease with temperature, cooling to 500︒C changes the compositions of the α and β2 phases following the solvus boundaries as shown in the sketch.51.FIND: If a titanium alloy containing 5 wt. % V is cooled under equilibrium conditions from 900︒C down to 500︒C, plot the fraction of phases present as a function of temperature. Which phase is richer in vanadium, α or β? GIVEN: The Ti-V phase diagram.SOLUTION: The β phase is richer in vanadium than the α phase.L + β ( Ti, V )β ( Ti, V ) + α ( Ti )β1β2+Comp. of β decreases in V along phase boundary12Comp. of β increases in V along phase boundaryMonotectoid isotherm。

第七章答案7-1略7-2浓度差会引起扩散，扩散是否总是从高浓度处向低浓度处进行?为什么?解：扩散是由于梯度差所引起的，而浓度差只是梯度差的一种。

当另外一种梯度差，比如应力差的影响大于浓度差，扩散则会从低浓度向高浓度进行。

7-3欲使Ca2＋在CaO中的扩散直至CaO的熔点（2600℃）时都是非本质扩散，要求三价离子有什么样的浓度？试对你在计算中所做的各种特性值的估计作充分说明。

已知CaO肖特基缺陷形成能为6eV。

解：掺杂M3+引起V’’Ca的缺陷反应如下：当CaO在熔点时，肖特基缺陷的浓度为：所以欲使Ca2＋在CaO中的扩散直至CaO的熔点（2600℃）时都是非本质扩散，M3+的浓度为，即7-4试根据图 7-32查取：（1）CaO在1145℃和1650℃的扩散系数值；（2）Al2O3在1393℃和1716℃的扩散系数值；并计算CaO和Al2O3中Ca2＋和Al3＋的扩散活化能和D0值。

解：由图可知CaO在1145℃和1650℃的扩散系数值分别为，Al2O3在1393℃和1716℃的扩散系数值分别为根据可得到CaO在1145℃和1650℃的扩散系数的比值为：，将值代入后可得，Al2O3的计算类推。

7-5已知氢和镍在面心立方铁中的扩散数据为cm2/s和cm2/s，试计算1000℃的扩散系数，并对其差别进行解释。

解：将T=1000℃代入上述方程中可得，同理可知。

原因：与镍原子相比氢原子小得多，更容易在面心立方的铁中通过空隙扩散。

7-6在制造硅半导体器体中，常使硼扩散到硅单晶中，若在1600K温度下，保持硼在硅单晶表面的浓度恒定（恒定源半无限扩散），要求距表面10－3cm深度处硼的浓度是表面浓度的一半，问需要多长时间（已知D1600℃＝8×10－12cm2/s；当时，）？解：此模型可以看作是半无限棒的一维扩散问题，可用高斯误差函数求解。

其中=0，，所以有0.5=，即=0.5，把=10－3cm，D1600℃＝8×10－12cm2/s代入得t=s。

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

6、描述不同材料常用的加工方法。

7、叙述金属材料的类型及其分类依据。

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

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

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

6、计算体心立方、面心立方、密排六方晶胞中的原子数、配位数、堆积系数。

7、从理论计算公式计算NaC1与MgO的晶格能。

MgO的熔点为2800℃，NaC1为80l℃, 请说明这种差别的原因。

8、根据最密堆积原理，空间利用率越高，结构越稳定，金钢石结构的空间利用率很低(只有34.01％)，为什么它也很稳定?9、证明等径圆球面心立方最密堆积的空隙率为25．9％；10、金属镁原子作六方密堆积，测得它的密度为1.74克/厘米3，求它的晶胞体积。

第一章8．计算下列晶体的离于键与共价键的相对比例(1)NaF (2)CaO (3)ZnS解：1、查表得：X Na =0.93,X F =3.98根据鲍林公式可得NaF 中离子键比例为：21(0.93 3.98)4[1]100%90.2%e ---⨯=共价键比例为：1-90.2%=9.8% 2、同理，CaO 中离子键比例为：21(1.00 3.44)4[1]100%77.4%e---⨯=共价键比例为：1-77.4%=22.6%3、ZnS 中离子键比例为：21/4(2.581.65)[1]100%19.44%ZnS e --=-⨯=中离子键含量共价键比例为：1-19.44%=80.56%10说明结构转变的热力学条件与动力学条件的意义．说明稳态结构与亚稳态结构之间的关系。

答：结构转变的热力学条件决定转变是否可行，是结构转变的推动力，是转变的必要条件；动力学条件决定转变速度的大小，反映转变过程中阻力的大小。

稳态结构与亚稳态结构之间的关系：两种状态都是物质存在的状态，材料得到的结构是稳态或亚稳态，取决于转交过程的推动力和阻力(即热力学条件和动力学条件)，阻力小时得到稳态结构，阻力很大时则得到亚稳态结构。

稳态结构能量最低，热力学上最稳定，亚稳态结构能量高，热力学上不稳定，但向稳定结构转变速度慢，能保持相对稳定甚至长期存在。

但在一定条件下，亚稳态结构向稳态结构转变。

第二章1．回答下列问题：(1)在立方晶系的晶胞内画出具有下列密勒指数的晶面和晶向：(001）与[210]，(111）与[112]，(110)与 [111],(132)与[123],(322)与[236］(2)在立方晶系的一个晶胞中画出（111)和 （112)晶面，并写出两晶面交线的晶向指数。

(3)在立方晶系的一个晶胞中画出同时位于（101). (011)和（112)晶面上的[111]晶向。

解：1、2．有一正交点阵的 a=b, c=a/2。

某晶面在三个晶轴上的截距分别为 6个、2个和4个原子间距，求该晶面的密勒指数。

材料科学基础（张联盟版）课后部分习题材料基部分题目1.晶体结构1、解释下列概念晶系、晶胞、晶胞参数、空间点阵、米勒指数（晶面指数）、离子晶体的晶格能、原子半径与离子半径、配位数、离子极化、同质多晶与类质同晶、正尖晶石与反正尖晶石、反萤石结构、铁电效应、压电效应.2、（1）一晶面在x、y、z轴上的截距分别为2a、3b、6c，求出该晶面的米勒指数；（2）一晶面在x、y、z轴上的截距分别为a/3、b/2、c，求出该晶面的米勒指数。

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

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

6、计算体心立方、面心立方、密排六方晶胞中的原子数、配位数、堆积系数。

7、从理论计算公式计算NaC1与MgO的晶格能。

MgO的熔点为2800℃，NaC1为80l℃, 请说明这种差别的原因。

8、根据最密堆积原理，空间利用率越高，结构越稳定，金钢石结构的空间利用率很低(只有34.01％)，为什么它也很稳定?9、证明等径圆球面心立方最密堆积的空隙率为25．9％；10、金属镁原子作六方密堆积，测得它的密度为1.74克/厘米3，求它的晶胞体积。

11、根据半径比关系，说明下列离子与O2—配位时的配位数各是多?r o2-=0.132nm r Si4+=0.039nm r K+=0.133nm r Al3+=0.057nm r Mg2+=0.078n m12、为什么石英不同系列变体之间的转化温度比同系列变体之间的转化温度高得多？13、有效离子半径可通过晶体结构测定算出。

材料科学与基础习题集和答案第七章回复再结晶，还有相图的内容。

第一章1.作图表示立方晶体的()()()421,210,123晶面及[][][]346,112,021晶向。

2.在六方晶体中，绘出以下常见晶向[][][][][]0121,0211,0110,0112,0001等。

3.写出立方晶体中晶面族{100}，{110}，{111}，{112}等所包括的等价晶面。

4.镁的原子堆积密度和所有hcp 金属一样，为0.74。

试求镁单位晶胞的体积。

已知Mg 的密度3Mg/m 74.1=mg ρ，相对原子质量为24.31，原子半径r=0.161nm 。

5.当CN=6时+Na 离子半径为0.097nm ，试问：1) 当CN=4时，其半径为多少？2) 当CN=8时，其半径为多少？6. 试问：在铜（fcc,a=0.361nm ）的<100>方向及铁(bcc,a=0.286nm)的<100>方向,原子的线密度为多少？7.镍为面心立方结构，其原子半径为nm 1246.0=Ni r 。

试确定在镍的（100），（110）及（111）平面上12mm 中各有多少个原子。

8. 石英()2SiO 的密度为2.653Mg/m 。

试问： 1) 13m 中有多少个硅原子（与氧原子）？2) 当硅与氧的半径分别为0.038nm 与0.114nm 时，其堆积密度为多少（假设原子是球形的）？9.在800℃时1010个原子中有一个原子具有足够能量可在固体内移动，而在900℃时910个原子中则只有一个原子，试求其激活能（J/原子）。

10.若将一块铁加热至850℃，然后快速冷却到20℃。

试计算处理前后空位数应增加多少倍（设铁中形成一摩尔空位所需要的能量为104600J ）。

11.设图1-18所示的立方晶体的滑移面ABCD 平行于晶体的上、下底面。

若该滑移面上有一正方形位错环，如果位错环的各段分别与滑移面各边平行，其柏氏矢量b ∥AB 。

材料科学基础（武汉理工大学，张联盟版）课后习题及答案第七章第七章答案7-1略7-2浓度差会引起扩散，扩散是否总是从高浓度处向低浓度处进行?为什么?解：扩散是由于梯度差所引起的，而浓度差只是梯度差的一种。

当另外一种梯度差，比如应力差的影响大于浓度差，扩散则会从低浓度向高浓度进行。

7-3欲使Ca在CaO中的扩散直至CaO的熔点（2600℃）时都是非本质扩散，要求三价离子有什么样的浓度？试对你在计算中所做的各种特性值的估计作充分说明。

已知CaO肖特基缺陷形成能为6eV。

解：掺杂M引起V’’Ca的缺陷反应如下：当CaO在熔点时，肖特基缺陷的浓度为：3+2＋所以欲使Ca在CaO中的扩散直至CaO的熔点（2600℃）时都是非本质扩散，M的浓度为，即2＋3+7-4试根据图7-32查取：（1）CaO在1145℃和1650℃的扩散系数值；（2）Al2O3在1393℃2＋3＋和1716℃的扩散系数值；并计算CaO和Al2O3中Ca和Al的扩散活化能和D0值。

解：由图可知CaO在1145℃和1650℃的扩散系数值分别为，Al2O3在1393℃和1716℃的扩散系数值分别为根据可得到CaO在1145℃和1650℃的扩散系数的比值为：，将值代入后可得，Al2O3的计算类推。

7-5已知氢和镍在面心立方铁中的扩散数据为2cm/s和2cm/s，试计算1000℃的扩散系数，并对其差别进行解释。

解：将T=1000℃代入上述方程中可得。

，同理可知原因：与镍原子相比氢原子小得多，更容易在面心立方的铁中通过空隙扩散。

7-6在制造硅半导体器体中，常使硼扩散到硅单晶中，若在1600K温度下，保持硼在硅单晶－3表面的浓度恒定（恒定源半无限扩散），要求距表面10cm深度处硼的浓度是表面浓度的一半，问需要多长时间（已知D1600℃＝8×10－12cm/s；当2时，）？解：此模型可以看作是半无限棒的一维扩散问题，可用高斯误差函数求解。

其中－122=0，，所以有0.5=s。