工程材料科学与设计 答案
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。
材料科学与工程专业英语第三版翻译以及答案
UNIT 1一、材料根深蒂固于我们生活的程度可能远远的超过了我们的想象，交通、装修、制衣、通信、娱乐（recreation）和食品生产，事实上（virtually），我们生活中的方方面面或多或少受到了材料的影响。
历史上，社会的发展和进步和生产材料的能力以及操纵材料来实现他们的需求密切（intimately）相关，事实上，早期的文明就是通过材料发展的能力来命名的（石器时代、青铜时代、铁器时代）。
二、早期的人类仅仅使用（access）了非常有限数量的材料，比如自然的石头、木头、粘土（clay）、兽皮等等。
随着时间的发展，通过使用技术来生产获得的材料比自然的材料具有更加优秀的性能。
这些性材料包括了陶瓷（pottery）以及各种各样的金属，而且他们还发现通过添加其他物质和改变加热温度可以改变材料的性能。
此时，材料的应用（utilization）完全就是一个选择的过程，也就是说，在一系列有限的材料中，根据材料的优点来选择最合适的材料，直到最近的时间内，科学家才理解了材料的基本结构以及它们的性能的关系。
在过去的100年间对这些知识的获得，使对材料性质的研究变得非常时髦起来。
因此，为了满足我们现代而且复杂的社会，成千上万具有不同性质的材料被研发出来，包括了金属、塑料、玻璃和纤维。
三、由于很多新的技术的发展，使我们获得了合适的材料并且使得我们的存在变得更为舒适。
对一种材料性质的理解的进步往往是技术的发展的先兆，例如：如果没有合适并且没有不昂贵的钢材，或者没有其他可以替代（substitute）的东西，汽车就不可能被生产，在现代、复杂的（sophisticated）电子设备依赖于半导体（semiconducting）材料四、有时，将材料科学与工程划分为材料科学和材料工程这两个副学科（subdiscipline）是非常有用的，严格的来说，材料科学是研究材料的性能以及结构的关系，与此相反，材料工程则是基于材料结构和性能的关系，来设计和生产具有预定性能的材料，基于预期的性能。
材料工程复习思考题部分答案
材料工程基础》复习思考题第一章绪论1、材料科学与材料工程研究的对象有何异同？答：材料科学侧重于发现和揭示组成与结构，性能，使用效能，合成与加工等四要素之间的关系，提出新概念，新理论。
而材料工程指研究材料在制备过程中的工艺和工程技术问题，侧重于寻求新手段实现新材料的设计思想并使之投入使用，两者相辅相成。
6、进行材料设计时应考虑哪些因素？答：. 材料设计的最终目标是根据最终需求，设计出合理成分，制订最佳生产流程，而后生产出符合要求的材料。
材料设计十分复杂，如模型的建立往往是基于平衡态，而实际材料多处于非平衡态，如凝固过程的偏析和相变等。
材料的力学性质往往对结构十分敏感，因此，结构的任何细小变化，性能都会发生明显变化。
相图也是材料设计不可或缺的组成部分。
7、在材料选择和应用时，应考虑哪些因素？答：一，材料的规格要符合使用的需求：选择材料最基本的考虑，就在满足产品的特性及要求，例如：抗拉强度、切削性、耐蚀性等；二，材料的价格要合理；三，材料的品质要一致。
8、简述金属、陶瓷和高分子材料的主要加工方法。
答：金属：铸造（砂型铸造、特种铸造、熔模铸造、金属型铸造、压力铸造、低压铸造、离心铸造、连续铸造、消失模铸造）、塑性加工（锻造、板料冲压、轧制和挤压、拉拨）、热处理、焊接（熔化焊、压力焊、钎焊）；橡胶：塑炼、混炼、压延、压出、硫化五部分；高分子：挤制成型、干压成型、热压铸成型、注浆成型、轧膜成型、等静压成型、热压成型和流延成型。
10、如何区分传统材料与先进材料？答：传统材料指已经成熟且已经在工业批量生产的材料，如水泥、钢铁，这些材料量大、产值高、涉及面广，是很多支柱产业的基础。
先进材料是正在发展，具有优异性能和应用前景的一类材料。
二者没有明显界限，传统材料采用新技术，提高技术含量、性能，大幅增加附加值成为先进材料；先进材料长期生产应用后成为传统材料，传统材料是发展先进材料和高技术基础，先进材料推到传统材料进一步发展。
材料制备科学与技术答案
材料制备科学与技术答案【篇一：化学与人类_尔雅慕课_课后答案】.1化学研究的对象和内容1.天然气中主要成分是什么？（）a、丁烷b、丙烷c、乙烷d、甲烷我的答案：d2.（）研究除碳元素以外的所有元素的单质及其化合物的性质。
a、生物化学b、有机化学c、无机化学d、物理化学我的答案：c3.分析化学是研究（）成分的测定方法和原理。
a、原子b、量子c、物质d、电子我的答案：c4.甲烷的泄露会引起人的中毒。
（）我的答案：x5.有机化学研究的是碳氢化合物及其衍生物的性能。
（）我的答案：√6.化学是一门研究物质变化的科学。
（）我的答案：√1.2煤气和爆炸极限1煤气中的臭味是因为在其中人为加入了什么？（）a、甲醇b、乙醇c、乙醚d、硫醇我的答案：d2在常温常压下，氢会呈现什么状态？（）a、固态b、气态c、液态d、混合态我的答案：b31.3关于广告的问题1上个世纪80年代谁表演了“水变油”的荒唐骗局。
（）a、王洪志b、李洪成c、王洪成d、李洪志我的答案：c2．1卡路里是等于多少焦耳？（）a、4.19b、5.19c、6.19d、7.19我的答案：a34二恶烷别名又叫做二氧六环。
（）我的答案：√1.4化学与新闻专业的关系1白金是哪种化学元素的俗称？（）a、金b、银c、铂d、钋我的答案：c2铹元素的名称是为纪念回旋加速器的发明者（）。
a、洛伦兹b、伦琴c、斯特恩d、劳伦斯我的答案：d3氯是一种非金属元素。
（）我的答案：√2.1石油是一种碳氢化合物的混合物1石油实际上是以（）为主的烃类化合物。
a、烯烃b、炔烃c、烷烃d、环烃我的答案：c2以下哪项不是当今人类使用的四大能源之一？（）a、石油b、煤c、原子能d、风能我的答案：d3煤是地球上分布最广、储量最大的能源资源。
（）我的答案：√ 2.2汽油的制备1以下哪一项不是内燃机气缸的工作原理？（）a、进气b、排气c、点火d、爆缸我的答案：d2所谓93号汽油就是指（）为93的汽油。
材料科学与工程专业英语第三版-翻译以及答案
UNIT 1一、材料根深蒂固于我们生活的程度可能远远的超过了我们的想象，交通、装修、制衣、通信、娱乐（recreation）和食品生产，事实上（virtually），我们生活中的方方面面或多或少受到了材料的影响。
历史上，社会的发展和进步和生产材料的能力以及操纵材料来实现他们的需求密切（intimately）相关，事实上，早期的文明就是通过材料发展的能力来命名的（石器时代、青铜时代、铁器时代）。
二、早期的人类仅仅使用（access）了非常有限数量的材料，比如自然的石头、木头、粘土（clay）、兽皮等等。
随着时间的发展，通过使用技术来生产获得的材料比自然的材料具有更加优秀的性能。
这些性材料包括了陶瓷（pottery）以及各种各样的金属，而且他们还发现通过添加其他物质和改变加热温度可以改变材料的性能。
此时，材料的应用（utilization）完全就是一个选择的过程，也就是说，在一系列有限的材料中，根据材料的优点来选择最合适的材料，直到最近的时间内，科学家才理解了材料的基本结构以及它们的性能的关系。
在过去的100年间对这些知识的获得，使对材料性质的研究变得非常时髦起来。
因此，为了满足我们现代而且复杂的社会，成千上万具有不同性质的材料被研发出来，包括了金属、塑料、玻璃和纤维。
三、由于很多新的技术的发展，使我们获得了合适的材料并且使得我们的存在变得更为舒适。
对一种材料性质的理解的进步往往是技术的发展的先兆，例如：如果没有合适并且没有不昂贵的钢材，或者没有其他可以替代（substitute）的东西，汽车就不可能被生产，在现代、复杂的（sophisticated）电子设备依赖于半导体（semiconducting）材料四、有时，将材料科学与工程划分为材料科学和材料工程这两个副学科（subdiscipline）是非常有用的，严格的来说，材料科学是研究材料的性能以及结构的关系，与此相反，材料工程则是基于材料结构和性能的关系，来设计和生产具有预定性能的材料，基于预期的性能。
材料科学与工程基础 (1)
• 课程建设
• 1996年将原教材改编为《高分子材料导论》上、下册 • 1998年，正式开设《材料科学与工程基础》课程 • 编写《材料科学与工程基础》教材—高分子为主线 1998年，四川大学“九五”重点资助教材。
1999年，教育部“面向21世纪课程教材”（我校
牵 头、三校合编，顾宜任主编）---2002年4月出版
材料科学与工程学科教学指导委员会
四个专业教学指导分委员会 材料化学与物理 金属材料及冶金工程
无机非金属材料工程
高分子材料与工程
பைடு நூலகம்
国家级精品课程（九门）
• 2003年，材料科学基础， • 2004年，材料科学基础， 上海交通大学， 西北工业大学， 蔡珣 陈铮 顾宜
• 2004年，材料科学与工程基础，四川大学，
四大要素
Performance
Properties
Synthesis and Processing Compositions and Structures
研究内容：
• 科学性
a. 从化学角度出发，研究材料的化学组 成、键性、 结构与性能的关系 b. 从物理角度，阐述材料的组成原子、 分子及其运 动状态与各物性之间的关系 • 技术性 c. 材料的制备工艺 d. 材料的性能表征 e. 材料的应用 • 工程性
(3) 材料工程 Materials Engineering
对于工程技术人员：如何选择特定 应用环境下需要的材料，来满足使用要 求，如何按实际要求设计新材料，须弄 清以下三个关系 （材质）——材料内部结构与性能 （内部形态）——加工工艺与性能 （使用环境，耐久性）——材料的性 能与使用过程
材料高等教育
材料在人类社会发展中的地位和作用
材料科学与工程专业英语第三版-翻译以及答案
UNIT 1一、材料根深蒂固于我们生活的程度可能远远的超过了我们的想象，交通、装修、制衣、通信、娱乐（recreation）和食品生产，事实上（virtually），我们生活中的方方面面或多或少受到了材料的影响。
历史上，社会的发展和进步和生产材料的能力以及操纵材料来实现他们的需求密切（intimately）相关，事实上，早期的文明就是通过材料发展的能力来命名的（石器时代、青铜时代、铁器时代）。
二、早期的人类仅仅使用（access）了非常有限数量的材料，比如自然的石头、木头、粘土（clay）、兽皮等等。
随着时间的发展，通过使用技术来生产获得的材料比自然的材料具有更加优秀的性能。
这些性材料包括了陶瓷（pottery）以及各种各样的金属，而且他们还发现通过添加其他物质和改变加热温度可以改变材料的性能。
此时，材料的应用（utilization）完全就是一个选择的过程，也就是说，在一系列有限的材料中，根据材料的优点来选择最合适的材料，直到最近的时间内，科学家才理解了材料的基本结构以及它们的性能的关系。
在过去的100年间对这些知识的获得，使对材料性质的研究变得非常时髦起来。
因此，为了满足我们现代而且复杂的社会，成千上万具有不同性质的材料被研发出来，包括了金属、塑料、玻璃和纤维。
三、由于很多新的技术的发展，使我们获得了合适的材料并且使得我们的存在变得更为舒适。
对一种材料性质的理解的进步往往是技术的发展的先兆，例如：如果没有合适并且没有不昂贵的钢材，或者没有其他可以替代（substitute）的东西，汽车就不可能被生产，在现代、复杂的（sophisticated）电子设备依赖于半导体（semiconducting）材料四、有时，将材料科学与工程划分为材料科学和材料工程这两个副学科（subdiscipline）是非常有用的，严格的来说，材料科学是研究材料的性能以及结构的关系，与此相反，材料工程则是基于材料结构和性能的关系，来设计和生产具有预定性能的材料，基于预期的性能。
材料科学：建筑工程材料考试卷及答案模拟考试练习.doc
材料科学：建筑工程材料考试卷及答案模拟考试练习 考试时间：120分钟 考试总分：100分遵守考场纪律，维护知识尊严，杜绝违纪行为，确保考试结果公正。
1、单项选择题 石油沥青防腐中，当气温低于（ ）时，应按冬季施工处理。
A.15℃ B.10℃ C.5℃ D.0℃ 本题答案： 2、问答题 金属复合材料的基体性能与应用？ 本题答案： 3、名词解释 标准稠度用水量 本题答案： 4、填空题 水玻璃常用的促硬剂为（ ）。
本题答案： 5、判断题 以塑料为基体、玻璃纤维为增强材料的复合材料，通常称为玻璃钢。
本题答案： 6、单项选择题 已知甲种钢筋的伸长率δ5与乙种钢筋的伸长率δ10相等，则说明（ ）。
A.甲种钢筋塑性好姓名：________________ 班级：________________ 学号：________________--------------------密----------------------------------封 ----------------------------------------------线----------------------B.甲种钢筋弹性好C.乙种钢筋塑性好D.乙种钢筋弹性好本题答案：7、单项选择题水泥混凝土配合比设计时需进行水灰比和水泥用量的校核，其目的是考虑（）。
A.经济性B.耐久性C.设计强度D.环境因素本题答案：8、多项选择题水玻璃可用来（）。
A.配制水玻璃混凝土B.加固地基C.涂刷材料D.紧急抢修本题答案：9、填空题低合金高强度钢是在（）的基础上加入了少量的合金元素而形成的钢。
本题答案：10、问答题影响金属基复合材料性能的关键因素？损伤及失效机制？本题答案：11、填空题材料的抗渗性是指材料（）的性质。
混凝土等的抗渗性常用（）表示。
S6（P6）表示该材料最大能抵抗的水压力为（）。
本题答案：12、判断题钢材防锈的根本方法是防止潮湿和隔绝空气。
材料工程基础答案
材料工程基础答案1. 简介材料工程是研究材料的特性和应用的学科。
它涉及材料的制备、性能评价、结构分析、应用设计等方面。
本文将回答一些关于材料工程基础的问题，以帮助读者更好地了解这个领域。
2. 问题回答2.1 什么是材料工程？材料工程是研究材料的性能、结构和应用的学科。
它使用物理、化学、力学等原理来研究材料的特性和行为，并将这些知识应用于材料的设计、制备和应用过程中。
2.2 材料工程的研究内容包括哪些方面？材料工程的研究内容包括：•材料的结构与组织•材料的性能评价•材料的制备与加工•材料的性能优化与改善•材料的应用设计与开发2.3 材料工程的重要性是什么？材料工程在现代工业生产中起着至关重要的作用。
其重要性主要体现在以下几个方面：•材料工程可以为产品的设计和制造提供材料的选择和优化建议，从而提高产品性能和质量。
•材料工程研究可以推动材料科学的发展，推动工业技术的进步。
•材料工程可以解决社会发展中的资源短缺和环境污染等问题，促进可持续发展。
2.4 材料工程的研究方法有哪些？材料工程的研究方法包括：•实验方法：通过实验，研究材料的结构、性能和行为。
•数值模拟方法：利用计算机和数学模型，模拟材料的性能和行为，并给出预测。
•理论分析方法：从物理学、化学等基本原理出发，分析材料的行为和性能。
•综合方法：结合实验、数值模拟和理论分析等多种方法，综合研究材料的性能和行为。
2.5 材料工程的发展趋势是什么？材料工程的发展趋势主要体现在以下几个方面：•多功能材料：研究并开发具有多种功能的材料，如具有光、电、磁等多种性能的复合材料。
•纳米材料：研究纳米级的材料，利用纳米技术改变材料的结构和性能。
•可持续发展材料：研究使用环保、可再生资源制备的材料，以解决资源短缺和环境污染等问题。
•新型制备技术：研究新型的材料制备方法，如3D 打印、纳米制备技术等。
3. 结论本文回答了一些关于材料工程基础的问题，从定义和研究内容到发展趋势都进行了简要介绍。
材料科学与工程专业英语匡少平课后翻译答案精编WORD版
材料科学与工程专业英语匡少平课后翻译答案精编W O R D版IBM system office room 【A0816H-A0912AAAHH-GX8Q8-GNTHHJ8】Alloy合金applied force作用力amorphous materials不定形材料artificial materials人工材料biomaterials生物材料biological synthesis生物合成biocompatibility生物相容性brittle failure脆性破坏carbon nanotub e碳纳米管carboxylic acid羟酸critical stress临近应力dielectric constant介电常数clay minera l粘土矿物cross-sectional area横截面积critical shear stress临界剪切应力critical length临界长度curing agent固化剂dynamic or cyclic loading动态循环负载linear coefficient of themal expansio n性膨胀系数electromagnetic radiation电磁辐射electrodeposition电极沉积nonlocalizedelectrons游离电子electron beam lithography电子束光刻elasticity 弹性系数electrostation adsorption静电吸附elastic modulus弹性模量elastic deformation弹性形变elastomer弹性体engineering strain工程应变crystallization 结晶fiber-optic光纤维Ethylene oxide环氧乙烷fabrication process制造过程glass fiber玻璃纤维glass transition temperature 玻璃化转变温度heat capacity热熔Hearing aids助听器integrated circuit集成电路Interdisplinary交叉学科intimate contact密切接触inert substance惰性材料implant移植individual application个体应用deformation局部形变mechanical strength机械强度mechanical attrition机械磨损Mechanical properties力学性Materials processing材料加工质mechanical behavior力学行为magnetic permeability磁导率magnetic hybrid technique混合技术induction磁感应mass per unit of volume单位体积质量monomer identity单体种类molecular mass分子量microsphere encapsulation technique微球胶囊技术macroscopical宏观的naked eye 肉眼nonlocalized nanoengineered materials纳米材料nanostructured materials纳米结构材料nonferrous metal有色金属线nucleic acid核酸nanoscale纳米尺度Nanotechnology纳米技术nanobiotechnology纳米生物技术nanocontact printing纳米接触印刷optical property光学性质optoelectronic device光电设备oxidation degradation 氧化降解piezoelectric ceramics压电陶瓷Relative density相对密度stiffnesses刚度sensor传感材料semiconductors半导体specific gravity比重shear 剪切Surface tention表面张力self-organization自组装static loading静载荷stress area应力面积stress-strain curves应力应变曲线sphere radius球半径submicron technique亚微米技术substrate衬底supramolecalar超分子sol-gel method溶胶凝胶法thermal/electrical conductivity 热/点导率thermoplastic materials热塑性材料Thermosetting plastic热固性塑料thermal motion热运动toughness test韧性试验tension张力torsion扭曲Tensile Properties拉伸性能Two-dimentional nanostructure二维纳米结构Tissue engineering组织工程transplantation of organs器官移植the service life使用寿命the longitudinal direction纵向the initial length of the materials初始长度the acceleration gravity重力加速度the normal vertical axis垂直轴the surface to volume ratio 比表面密度the burgers vector伯格丝矢量the mechanics and dynamics of tissues 组织力学和动力学phase transformation temperature相转变温度plastic deformation塑性形变Pottery陶瓷persistence length余晖长度polymer synthesis聚合物合成Polar monomer记性单体polyelectrolyte高分子电解质pinning point钉扎点plasma etching 等离子腐蚀pharmacological acceptability药理接受性pyrolysis高温分解ultrasonic treatment超射波处理yield strength屈服强度vulcanization硫化1-1:直到最近，科学家才终于了解材料的结构要素与其特性之间的关系。
材料科学与工程专业英语1-19单元课后翻译答案
材料科学与工程专业英语1-19单元课后翻译答案Unit 11.“材料科学”涉及到研究材料的结构与性能的关系。
相反，材料工程是根据材料的结构与性质的关系来涉及或操控材料的结构以求制造出一系列可预定的性质。
2.实际上，所有固体材料的重要性质可以分为六类：机械、电学、热学、磁学、光学、腐蚀性。
3.除了结构与性质，材料科学与工程还有其他两个重要的组成部分，即加工与性能。
4.工程师或科学家越熟悉材料的各种性质、结构、性能之间的关系以及材料的加工技术，根据以上的原则，他或她就会越自信与熟练地对材料进行更明智的选择。
5.只有在少数情况下，材料才具有最优或最理想的综合性质。
因此，有时候有必要为某一性质而牺牲另一性能。
6.Interdisciplinary dielectric constant Solid material(s)heat capacity Mechanical property electromagnetic radiationMaterial processing elastic modulus 7.It was not until relativcalproperties relate deformation to an applied load or force.Unit 21. 金属是电和热很好的导体，在可见光下不透明；擦亮的金属表面有金属光泽。
2. 陶瓷是典型的导热导电的绝缘体，并且比金属和聚合物具有更高的耐热温度和耐恶劣环境性能。
3. 用于高科技领域的材料有时也被称为先进材料。
4.压电陶瓷在电场作用下膨胀和收缩；反之，当它们膨胀和收缩时，他们也能产生一个电场。
5. 随着能够观察单个原子或者分子的扫描探针显微镜的出现，操控和移动原子和分子以形成新结构成为可能，因此，我们能通过一些简单的原子水平的构建就可以设计出新的材料。
6. advanced materials ceramic materials high-performance materials clay mineralsalloy implant glass fibre carbon nanotube 7. Metallic materials have large numbers ofnonlocalized electrons and many properties of metals are directlyattributable to these electrons. 8. Many of polymeric materials areorganic compounds with very large molecular structures. 9. Semiconductors hace electrical properties that are intermediate betweenthe electrical conductors(viz. metals and metal alloys) andinsulators(viz. ceramics and polymers). 10. Biomaterials must notproduce toxic substances and must be compatible with body tissues.Unit 31．金属的行为（性质）不同于陶瓷的行为（性质），陶瓷的行为（性质）不同于聚合物的行为（性质）。