機(jī)械制造關(guān)鍵工程開卷習(xí)題答案

1、 QUESTIONS1.1Explain the difference between a unit cell and a single crystal. Answer (from page # 40): The Smallest group of atoms showing the characteristic lattice structure of a particular metal is known as a Unit Cell. It is the building block of a crystal, and a single crystal can have many uni

2、t cells.1.2In tables on crystal structures, iron is listed as having both a bcc and a fee structure. Why? Answer (from page # 41): Same metal may form different structures, because of a lower energy requirement at that temperature. For example, iron forms a bcc structure (alpha iron) below 912 C (16

3、74 F) and above 1394 C (2541 F), but if forms an fcc structure (gamma iron) between 912 C and 1394 C.1.3a. Define anisotropy, and what materials can you think of other than metals that exhibit anisotropic behavior? Answer (from page # 42): Because the b/a ratio is different for different directions

4、within the crystal, a single crystal has different properties when tested in different directions. We say that a single crystal is anisotropic. Where “a” is the spacing of the atomic planes and “b” is inversely proportional to the atomic density in the atomic planes.A common example of anisotropy is

5、 woven cloth, which stretches differently when we pull it in different directions, or plywood, which is much stronger in the planar direction than along its thickness direction (it splits easily). 1.4What effects does recrystallization have on properties of metals? Answer (from page # 50): Recrystal

6、lization decreases the density of dislocations, lowers the strength, and raises the ductility of the metal.1.5 What is strain hardening? And what effects does it have on the properties of metals? Answer (from page # 45): The effect o9 fan increase in shear stress that causes an increase in the overa

7、ll strength of the metal is known as the work hardening or strain hardening. The effect on the properties of metals is the greater the deformation, the greater the number of entanglements, hence an increase in the metals strength.1.6 Explain what is meant by structure sensitive and structurein sensi

8、tive properties of metals. Answer (from page # 43, 44): Mechanical and electrical properties of metals, such as yield, fracture strength, and electrical conductivity, are adversely affected by these defects; these are known as structure-sensitive properties. On the other hand, their physical and che

9、mical properties such as melting point, specific heat, coefficient of thermal expansion, and elastic constants (e.g., modulus of elasticity and modulus of rigidity) are not sensitive to these defects; these are known as structure-insensitive properties.1.7 Make a list of each of the major kinds of i

10、mperfection in the crystal structure of metals, and describe them. Answer (from page # 4): Line defects, called dislocationsPoint defects, such as vacancy (missing atom), and interstitial atom (extra atom in the lattice), or an impurity (foreign atom) that has replaced the atom of the pure metal.Vol

11、ume or bulk imperfections, such as voids or inclusions (nonmetallic elements such as oxides, sulfides, and silicates);Planar imperfection, such as grain boundaries.1.8 What influence does grain size have on the mechanical properties of metals? Answer (from page # 46): Grain size significantly influe

12、nces the mechanical properties of metals. At room temperature, a large grain size is generally associated with low strength, low hardness, and low ductility. Large grains, particularly in sheet metals, also cause a rough surface appearance after the material has been stretched. 1.9 What is the relat

13、ionship between the nucleation rate and the number of grains per unit volume of a metal? Answer (from page # 45): The number and size of the grains developed in a unit volume of the metal depends on the rate at which nucleation (the initial stage of formation of crystals) take place.1.10 What is a s

14、lip system, and what is its significance? Answer (from page # 42): The combination of a slip plane and its direction of slip are known as a slip system. In general, metals with slip systems of 5 or above are ductile, whereas those with slip systems below 5 are not. It also determines the probability

15、 of Shear stress.1.11 Explain the difference between recovery and recrystallization. Answer (from page # 49): Recovery, which occurs at a certain temperature range below the recrystallization temperature of the metal, the stresses in the highly deformed regions are relived. Subgrain boundaries begin

16、 to form (a process called polygonization), with no appreciable change in mechanical properties such as hardness and strength.Recrystallization is the process in which, at a certain temperature range, new equiaxed and strain-free grains are formed, replacing the older grains, in called recrystalliza

17、tion.1.12 What is hot shortness, and what is its significance? Answer (from page # 47): Hot shortness is caused bye local melting of a constituent or an impurity in the grain boundary at a temperature below the melting point of the metal itself. When subjected to plastic deformation at elevated temp

18、eratures (hot-working), the piece of metal crumbles and disintegrates along the grain boundaries. Examples are antimony in copper, leaded steels, and leaded brass. 1.13 Explain the differences between cold, warm, and hotworking of metals. Answer (from page # 51): Cold-working refers to plastic defor

19、mation that is usually, but not necessarily, carried out at room temperature. When the deformation is carried out above the recrystallization temperature, it is called hot-working. As the name implies, warm-working is carried out at intermediate temperatures. Thus warm working is a compromise betwee

20、n cold-and hot-working.1.14 Describe what the orange peel effect is. Answer (from page # 50, 51): Large grains produces a rough surface appearance on the sheet metals, called orange peel, when they are stretched to form a part, or when a piece of metal is subjected to compression (such as in forging

21、 operations).1.15 Why cant some metals such as lead become stronger when cold worked? Answer (from page # 50):Recrystallization decreases the density of dislocations, lowers the strength, and raises the ductility of the metal. Lead recrystallization at about room temperature; as a result, when cold-

22、worked, they do not work harden. Recrystallization depends on the degree of prior cold work (work hardening): the more cold work, the lower the temperature required for recrystallization to occur. The reason is that, as the amount of cold work increases, the number of dislocations and the amount of

23、energy stored in dislocations (stored energy) also increase. This energy supplies the work required for recrystallization.1.16 Describe the difference between preferred orientation and mechanical fibering. Answer (from page # 48, 49): Preferred Orientation is also called crystallographic anisotropy.

24、 When a metal crystal is subjected to tension, the sliding blocks rotate toward the direction of the pulling force. As a result, slip planes and slip bands tend to align themselves with direction of deformation. Mechanical Fibering results form the alignment of impurities, inclusions (stringers), an

25、d voids in the metal during deformation. 1.17 In microscopy, it is common practice to apply a chemical etchant to a metal in order to highlight grain boundaries and microstructures. Explain why. Answer (from page # 46): Grain boundaries are more reactive than the grains themselves, because the atoms

26、 along the grain boundaries are packed less efficiently and are more disordered. As a result, they have higher energy than the atoms in orderly lattice within the grains. For this reason, a polished and etched surface can become rougher 1.18 What is twinning? How does it differ from slip? Answer (fr

27、om page # 42): The second mechanism of plastic deformation is twinning, in which a portion of the crystal forms a mirror image of itself across the plane of twinning. Whereas in slip the plastic deformation can only take place along planes of maximum atomic density or, in other words, that slip take

28、s place in closely packed direction.= End of Homework Assignment QUESTIONS2.1 Distinguish between engineering stress and true stress. Answer (from page #_57, 60_): The engineering stress or nominal stress is defined as the ratio of the applied load p to the original cross-sectional area A of the spe

29、cimen:Engineering Stress, = P/ATrue stress is defined as the ratio of the local p to the actual (instantaneous, hence true) cross-sectional area A of the specimen:True stress, = P/A2.2 Describe the events that occur when a specimen undergoes a tension test. Sketch a plausible stressstrain curve, and

30、 identify all significant regions and points between them. Assume that loading continues up to fracture. Answer (from page #_56,57_): When the load is first applied, the specimen elongates in proportion to the load. This effect is called linear elastic behavior. If the load is removed, the specimen

31、returns to its original length and shape, in an elastic process similar to stretching a rubber band and releasing it. As the load is increased, the specimen begins, at some level of stress, to undergo permanent (plastic) deformation. Beyond that level, the stress and strain are no longer proportiona

32、l, as they were in the elastic region. The stress at which this phenomenon occurs is known as they yield stress, y, of the material. If the specimen is loaded beyond its ultimate tensile strength, it begins to neck or neck down. Finally specimen fractures at the necked region or know as breaking or

33、fractures stress.2.3 What is ductility, and how is it measured? Answer (from page #_59_): Ductility is the extent of plastic deformation that the material undergoes before fracture. There are two common measures of ductility. The first is the total elongation of the specimen:Elongation = (l f l o)/(

34、l o) * 100The second measure of ductility is the reduction of area.Reduction of area = (A o A f)/(A o) * 1002.5 What is strainrate sensitivity, and how is it measured? Answer (from page #_64,65_): When increasing the strain rate increases the strength of the material (strain-rate hardening). The slo

35、pe of these curves is called the strain-rate sensitivity exponent, m. The value of m is obtained from log-log plots, provided that the vertical and horizontal scales are the same. A slope of 45 would, therefore, indicate a value of m = 1. The relationship is given by the equation. = C mWhere C is a

36、strength coefficient. The constant C has the units of stress; is the true strain rate, defined as the strain that the material undergoes per unit time. The sensitivity of strength to strain rate increase with temperature.2.6 What test can measure the properties of a material undergoing shear strain?

37、 Answer (from page #_67_): Torsion test2.7 What testing procedures can be used to measure the properties of brittle materials, such as ceramics and carbides? Answer (from page #_67_): Disk Test2.8 Describe the differences between brittle and ductile fracture. Answer (from page #_76,78_): Ductile fra

38、cture is characterized by plastic deformation which precedes failure of the part. Brittle fracture occurs with little or no gross plastic deformation.2.9 Explain the difference between stress relaxation and creep. Answer (from page #_73,75_): Creep is the permanent elongation of a component under a

39、static load maintained for a period of time. It is a phenomenon of metals and of certain nonmetallic materials, such as thermoplastic and rubbers, and it can occur at any temperature. Stress Relaxation is closely related to creep. In stress relaxation the stresses resulting from a loading of a struc

40、tural component decrease in magnitude over a period of time, even though the dimensions of the component remain constant. 2.10 Describe the difference between elastic and plastic behavior. Answer (from page #_57_): An elastic behavior is when the load is first applied, the specimen elongates in prop

41、ortion to the load, and if the load is removed, the specimen returns to its original length and shape. A Plastic behavior is when the load is increase, the specimen begins, at some level of stress, to undergo permanent(plastic) deformation. Beyond that level, the stress and strain are no longer prop

42、ortional, as they were in the elastic region. 2.11 Explain what uniform elongation is in tension testing. Answer (from page #_58_): Under increasing load, to elongate beyond Y, its cross-sectional area decreases permanently and uniformly throughout its gage length. If the specimen is unloaded from a

43、 stress level higher than the yield stress, the curve follows a straight line downward and parallel to the original slope. The stress-strain curve is offset by a 0.002, or 0.2% elongation.2.12 Describe the difference between deformation rate and strain rate. What unit does each one have? Answer (fro

44、m page #_64_): Deformation rate is defined as the speed at which a tension test is being carried out, in units of, say m/s or ft/min. The strain rate, on the other hand, is a function of the specimen length. A short specimen elongates proportionately more during the same time period than does a long

45、 specimen. The strain rates are usually stated in terms of orders of magnitude, such as 102 s-1, 104 S-1, and so on. 2.13 Describe the difficulties involved in making a compression test. Answer (from page #_66_): The compression test, in which the specimen is subjected to a compressive load, gives i

46、nformation useful for these processes. This test is usually carried out by the compressing a solid cylindrical specimen between two flat dies (platens). Because of friction between the specimen and the platens, the specimens cylindrical surface bulges; this effect is called barreling. Friction preve

47、nts the top and bottom surfaces from expanding freely.Because the cross-sectional area of the specimen now changes along its height, being maximum in the middle, obtaining stress-strain curves in compression is difficult. Furthermore, friction dissipates energy, so the compressive force is higher th

48、an it otherwise would be, in order to supply the work required to overcome friction. With effective lubrication, friction can be minimized, and a reasonably constant cross-sectional area can be maintained during the test.2.14 What is Hookes Law? Youngs Modulus? Poissons ratio? Answer (from page #_58

49、,59_): The ratio of stress to strain in the elastic region is known as the modulus of elasticity, E or Youngs modulus.Modulus of elasticity, E = / eThis linear relationship is known as Hookes law. The modulus of elasticity is essentially a measure of the slope of the elastic portion of the curve and

50、 hence the stiffness of the material. That because engineering strain is dimensionless, E has the same unit as stress. The higher the E value, the higher the load required to stretch the specimen to the same extent, and thus the stiffer the material.The absolute value of the ratio in the specimen of

51、 the lateral strain to the longitudinal strain is known as Poissons ratio. 2.15 What is necking? Answer (from page #_58_): If the specimen is loaded beyond its ultimate tensile strength, it begins to neck, or neck down. The cross-sectional area of the specimen is no longer uniform along the gage len

52、gth and is smaller in the necked region. 2.16 What is the reason that “a yield” strength is defined as a 0.2% offset strength? Answer (from page #_58_): For soft and ductile materials, it may not be easy to determine the exact location on the stress-strain curve at which yielding occurs, because the

53、 slope is of the straight (elastic) portion of the curve begins to decrease slowly. Therefore we usually define Y as the point of the stress-strain curve that is offset by a strain of 0.002, or 0.2% elongation. 2.17 Why does the fatigue strength of a specimen or part depend on its surface finish? An

54、swer (from page #_80,81_): The Surface finish can help reduce decarburization surface pits (due to corrosion) that act as stress raisers hydrogen embrittlement galvanizing, and electroplating. In simply way to put it, it reduces Stress-Corrosion and stress cracking.2.18 If striations are observed un

55、der microscopic examination of a fracture surface, what do they suggest regarding the mode of fracture? Answer (from page #_): 802.19 Explain the difference between transgranular and intergranular fracture. Answer (from page #_79_): In polycrystalline metals, the fracture paths most commonly observe

56、d are transgranular (transcrystalline or intragranular) that is, the crack propagates through that grain. Intergranular fracture, where the crack propagates along the grain boundaries, generally occurs when the grain boundaries are soft, contain a brittle phase, or have been weakened by liquid-or so

57、lid-metal embrittlement.= End of Homework Assignment QUESTIONS3.1 List reasons why density is an important material property in manufacturing. Answer (from page # 91,92_): A significant role that Density plays is in the specific strength strength-to-weight ratio) and specific stiffness (stiffness-to

58、-weight ratio) of materials and structure. Density is an important factor in the selection of the materials for high-speed equipment. For example, the use of magnesium in printing and textile machinery, many components of which usually operate at very high speeds. In order to avoid vibrations and pa

59、rt failure because of their low density, Density is considered to be as an important factor.3.2 Why is the melting point of a material an important factor in manufacturing processes? Answer (from page #_94_): The melting point of the materials has a number of indirect effects on manufacturing operat

60、ions. The choice of the material for the high-temperature applications is the most obvious effect. Because of the recrystallization temperatures of a metal is related to its melting point operations such as annealing, heat treating, and hot working required a knowledge of the melting point of the ma