中南大学 材料科学基础 课后习题答案

材料科学基础习题与参考答案（doc14页）完美版第⼀章材料的结构⼀、解释以下基本概念空间点阵、晶格、晶胞、配位数、致密度、共价键、离⼦键、⾦属键、组元、合⾦、相、固溶体、中间相、间隙固溶体、置换固溶体、固溶强化、第⼆相强化。

⼆、填空题1、材料的键合⽅式有四类，分别是（），（），（），（）。

2、⾦属原⼦的特点是最外层电⼦数（），且与原⼦核引⼒（），因此这些电⼦极容易脱离原⼦核的束缚⽽变成（）。

3、我们把原⼦在物质内部呈（）排列的固体物质称为晶体，晶体物质具有以下三个特点，分别是（），（），（）。

4、三种常见的⾦属晶格分别为（），（）和（）。

5、体⼼⽴⽅晶格中，晶胞原⼦数为（），原⼦半径与晶格常数的关系为（），配位数是（），致密度是（），密排晶向为（），密排晶⾯为（），晶胞中⼋⾯体间隙个数为（），四⾯体间隙个数为（），具有体⼼⽴⽅晶格的常见⾦属有（）。

6、⾯⼼⽴⽅晶格中，晶胞原⼦数为（），原⼦半径与晶格常数的关系为（），配位数是（），致密度是（），密排晶向为（），密排晶⾯为（），晶胞中⼋⾯体间隙个数为（），四⾯体间隙个数为（），具有⾯⼼⽴⽅晶格的常见⾦属有（）。

7、密排六⽅晶格中，晶胞原⼦数为（），原⼦半径与晶格常数的关系为（），配位数是（），致密度是（），密排晶向为（），密排晶⾯为（），具有密排六⽅晶格的常见⾦属有（）。

8、合⾦的相结构分为两⼤类，分别是（）和（）。

9、固溶体按照溶质原⼦在晶格中所占的位置分为（）和（），按照固溶度分为（）和（），按照溶质原⼦与溶剂原⼦相对分布分为（）和（）。

10、影响固溶体结构形式和溶解度的因素主要有（）、（）、（）、（）。

11、⾦属化合物（中间相）分为以下四类，分别是（），（），（），（）。

12、⾦属化合物（中间相）的性能特点是：熔点（）、硬度（）、脆性（），因此在合⾦中不作为（）相，⽽是少量存在起到第⼆相（）作⽤。

13、CuZn、Cu5Zn8、Cu3Sn的电⼦浓度分别为（），（），（）。

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 = MPa m um um N 125.010*******.0262=⎪⎪⎭⎫⎝⎛⨯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 facediagonal. For example, one such vector is 10]1[ 2ao as shown in (111).A. The length of this vector is 0.256nm = 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 the smoothest 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 that 4.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 connectequivalent 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.This direction 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 crystal 0.272nm = b a 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 is consistent 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 specific crystallographic 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 = 21 21 123 = ⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛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 equivalent crystallographic 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> andhas magnitude . 2aFor 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 diagram we 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]2a The 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 shortest Burgers 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 units long.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 be extracted.O= and ½ Mg++ ions in our cube.Thus10x3.35x88/a10x2.02)+(1.33=3.65-233o-2318. GIVEN: Critical resolved shear stress (0.34MPa), slip system (111)[1_10], and tensileaxis [101]REQUIRED: Applied stress at which crystal begins to deform and crystal structure.SOLUTION: (A)The situation is shown belowτcrss = σ cosθ⋅ cosφθ = 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]∣cosφ [101] ⋅ [110] = ∣[101]∣∣[110]∣cosθ(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 dislocationmotion 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 plastic deformation.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.Dislocation moves 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 crystalτcrss = 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 cosθ⋅cosφ is the same for the BCC and FCC crystal with the meaning of φ and θ interchanged. Let M = cosθ⋅cosφ.(1)(2)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 of the 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. The problem is misstated since the Burgers vector must lie on the slip plane and [101] does not lie on (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 inliquids.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.The ASTM 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 = 10 x 1.5501 = n 3-100X10μgrains/in 2 at 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 faces cube and the area of each face is shared by 2 cubes, each cube has an area of 3xArea 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 3 30.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 100X4For 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 = 533MPa 31.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 betweendislocations, and θ is the tilt angle.SKETCH: See Fig. 5.3-4.SOLUTION: We can see the geometry more clearly using the following sketch:From the Figure we can immediately write that tan/θ22=bD. Since the tan of a small angle isthe angle itself:θ22=bD/, 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 islarge, then the density of the sample will be lower than that of dense material. Thesame is true for clusters of precipitate; however, usually the density difference between host and precipitate is not as great as between host and air, so the technique does notwork as well. Another possible technique is microscopy. Samples can be prepared for microscopy, perhaps by polishing and etching and the defects observed using opticalor electron microscopy. X-ray diffraction can also be used. With a random spacing of void or precipitate there is then an average spacing. Sometimes Bragg's law can beused 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.They all rely on some property difference - magnetic, electrical, optical, or whatever.35. FIND: How can you ascertain whether a material contains both crystalline andnoncrystalline regions?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 crystallinitypolymers. These methods apply to all materials.1. Density. Measure the density of your sample and compare it to the density ofnoncrystalline and crystalline samples of the same composition.2. Differential Scanning Calorimetry. Heat your sample in a calorimeter. Samplesthat are crystalline will absorb heat at the melting temperature and show a "meltingendotherm". Some noncrystalline 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 acommon challenge to polymers scientists. We often need to quantify the fraction orpercent 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 ina purely elastic manner.SOLUTION: There are many such possible examples. Since plastic deformation isnonrecoverable deformation, any application that requires repeated stressing anddimensional 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 andthe strength goes 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 thedefects may migrate to edge dislocations where they cause jogs on the dislocations. A jogged dislocation is much harder to move and may itself require the generation ofpoint defects to move. In addition the point defects may collapse to form dislocationloops which also impede the motion of other dislocations making the materialsstronger. If the defects are interstitials, they may migrate to areas around thedislocations in which the system energy is reduced. For the dislocation to move away from the interstitial an increase in the system energy is required which means the stress to move the dislocation must increase. If the point defect is a substitutional atom,similar considerations apply. However, the magnitude of the energy reduction is less because of the less severe distortion. Thus the strength increase is not as high as forintersitital.39. As d↓σys↑ since this means the path over which a dislocation moves ↓. This meansthat the stress will have to increase to either nucleate or unlock dislocations in adjacent grains. The relationship 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) temperaturedependent.2. Adding impurities (e.g. C in Fe). The impurities “lock” the dislocation byassociating with the dislocation to lower the system energy. This will be verytemperature dependent for dilute concentrations of impurities as the impurities willdiffuse away at high temperatures.3. Adding precipitates - blocks the motion of dislocations through either having adifferent crystal 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 7.8gm/cm 3 we may write 10 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 nowrequires 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%at10x4.76=10x8.4110x4_100x10x4+10x8.4110x4=%Cat10x4=10x2.510=10x2.510=Ncm10x2.5_ddcm10=N2-222119221919198-12c8-FeFe12c。

《材料科学基础》习题附答案第⼆章思考题与例题1. 离⼦键、共价键、分⼦键和⾦属键的特点，并解释⾦属键结合的固体材料的密度⽐离⼦键或共价键固体⾼的原因？2. 从结构、性能等⽅⾯描述晶体与⾮晶体的区别。

3. 何谓理想晶体？何谓单晶、多晶、晶粒及亚晶？为什么单晶体成各向异性⽽多晶体⼀般情况下不显⽰各向异性？何谓空间点阵、晶体结构及晶胞？晶胞有哪些重要的特征参数？4. ⽐较三种典型晶体结构的特征。

（Al 、α-Fe 、Mg 三种材料属何种晶体结构？描述它们的晶体结构特征并⽐较它们塑性的好坏并解释。

）何谓配位数？何谓致密度？⾦属中常见的三种晶体结构从原⼦排列紧密程度等⽅⾯⽐较有何异同？5. 固溶体和中间相的类型、特点和性能。

何谓间隙固溶体？它与间隙相、间隙化合物之间有何区别？（以⾦属为基的）固溶体与中间相的主要差异（如结构、键性、性能）是什么？6. 已知Cu 的原⼦直径为2.56A ，求Cu 的晶格常数，并计算1mm 3Cu 的原⼦数。

7. 已知Al 相对原⼦质量Ar （Al ）=26.97，原⼦半径γ=0.143nm ，求Al 晶体的密度。

8 bcc 铁的单位晶胞体积，在912℃时是0.02464nm 3；fcc 铁在相同温度时其单位晶胞体积是0.0486nm 3。

当铁由bcc 转变为fcc 时，其密度改变的百分⽐为多少？9. 何谓⾦属化合物？常见⾦属化合物有⼏类？影响它们形成和结构的主要因素是什么？其性能如何？10. 在⾯⼼⽴⽅晶胞中画出[012]和[123]晶向。

在⾯⼼⽴⽅晶胞中画出（012）和（123）晶⾯。

11. 设晶⾯（152）和（034）属六⽅晶系的正交坐标表述，试给出其四轴坐标的表⽰。

反之，求（3121）及（2112）的正交坐标的表⽰。

（练习），上题中均改为相应晶向指数，求相互转换后结果。

12.在⼀个⽴⽅晶胞中确定6个表⾯⾯⼼位置的坐标，6个⾯⼼构成⼀个正⼋⾯体，指出这个⼋⾯体各个表⾯的晶⾯指数，各个棱边和对⾓线的晶向指数。

第1章 习题
1-10 纯铁点阵常数0.286nm ，体心立方结构，求1cm 3中有多少铁原子。

解：体心立方结构单胞拥有两个原子，单胞的体积为
V ＝(0.286×10-8)3 cm 3，所以1cm 3中铁原子的数目为
2283128.5510(2.8610)Fe n -=
⨯=⨯⨯
1-11 一个位错环能否各部分都是螺型位错，能否各部分都是刃型位错？为什么？
解：螺型位错的柏氏矢量与位错线平行，一根位错只有一个柏氏矢量，而一个位错环不可能与一个方向处处平行，所以一个位错环不能各部分都是螺型位错。

刃位错的柏氏矢量与位错线垂直，如果柏氏矢量垂直位错环所在的平面，则位错环处处都是刃型位错。

这种位错的滑移面是位错环与柏氏矢量方向组成的棱柱面，这种位错又称棱柱位错。

1-15 有一正方形位错线，其柏氏矢量及位错线的方向如图1-51所示。

试指出图中各段位错线的性质，并指出刃型位错额外串原子面所处的位置。

解：由柏氏矢量与位错线的关系可以知道，DC 是右螺型位错，BA 是左螺型位错。

由右手法则，CB 为正刃型位错，多余半原子面在纸面上方。

AD 为负刃型位错，多余半原子面在纸面下方。

A B
C
D。