材料科学基础课后习题答案

1、Problems - Chapter 51. 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/sec2. A Newton is a kg-m/sec2. A Pascal is a N/m2. A MPa is

2、106 Pa. SOLUTION: Stress is force per unit area. The cross-sectional area is R2 = 1963.5 square micrometers. The force is 25 g (kg/1000g)(9.8 m/sec2) = 0.245 N. Thus, the stress is = F/A = COMMENTS: You must learn to do these sorts of problems, including the conversions. 2.GIVEN: FCC Cu with ao = 0.

3、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) sho

4、wnWe note that atoms lying along face diagonals touch and are crystallographically equivalent. Therefore, the shortest vector connecting equivalent positions is face diagonal. For example, one such vector is as shown in (111).A. The length of this vector is B. By inspection, the size of the vector i

5、s 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: R

6、ecall the Ag is FCC. For FCC structures the Burgers vector is a face diagonal 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 at

7、omic position. Thus etc. Then in general B. For NaC1We see that the shortest vector connecting equivalent positions is as 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 becau

8、se repulsive interionic forces are minimized on these planes during dislocation motion. Thus we expect 1/2 Burgers vectors and 110 slip planes.5.GIVEN:Mo crystalao = 0.314nmREQUIRED: Determine the crystal structure.If Mo were FCC, then but |b| = 0.272 Mo is not FCC.Assuming Mo is BCC, then Thus the

9、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 specifi

10、c 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: for BCC structureSOLUTION:from the formula for the magnitude of a vector:8. GIV

11、EN: 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:t = s cosq cosf(1)where q = angle between slip direction and tensile axis; f = angle between normal to slip plane and tensile axisThus 9.

12、 GIVEN: Stress in 123 direction of BCC crystalREQUIRED: Find the stress needed to promote slip if tcR = 800 psi. The slip plane is (10) and slip direction is 111.SOLUTION: Recall t = s cosq cosf(1)q = 123 111123 111 = 123 111cosqf = 123 10123 10 = 123 10cosf10.Burgers vectors lie in the closest pack

13、ed 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 pa

14、cked 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) parallel to intersection of (11) and (111) with Burgers vector parallel to 0. Structure is FCC.REQUIRE

15、D: A) Burgers vector of dislocation and, B) Character of dislocation.SOLUTION: A) Since the structure is FCC, the Burgers vector is parallel to and has magnitude For a Burgers vector parallel to 0 the scalar multiplier must be a/2. Thus = a/2 0. B) We must determine the line direction of the disloca

16、tion. 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: The sum of the x, y & z components on the LHS must be equal to the correspondi

17、ng 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 | b1 | 2 + | b2 | 2 | b3| 3 Thus the reaction is favorable since 14.GIVEN: Dislocation in FCCParallel

18、to 01i.e. = 01REQUIRED: Character and slip planeSOLUTION: Character is found by angle between . Note = -1 + 0 + 1 = 0. Thus. Sincethe dislocation is pure edge.To find the slip plane we note that the cross produce ofgives a vector that is normal to the plane in whichlie. This vector so formed has the

19、 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 (00) 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.FC

20、C 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 w

21、ill have high stresses tending to make them propagate.16.For a simple cubic system, the lowest energy Burgers vectors are of the type 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

22、 of the Burgers vector.17.GIVEN:At. wt. 0 = 16At. wt. Mg = 24.32 Same structure as NaClr = 3.65 g/cm3REQUIRED: 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 th

23、e lattice parameter ao. We can take a sub-section of the unit cell (cross-hatched cube) whose edge is 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 ao may be

24、extracted. We note that there are 40= ions and 4Mg+ ions located at the corners. However, an ion at a corner is shared by 8 such cubes. Thus, we have O= and Mg+ ions in our cube.Thus18.GIVEN: Critical resolved shear stress (0.34MPa), slip system (111)10, and tensile axis 101REQUIRED: Applied stress

25、at which crystal begins to deform and crystal structure.SOLUTION: (A)The situation is shown belowtcrss = s cosq cosfq = angle between tensile axis and slip directionq = angle between tensile axis and normal to slip planef = 111 101 q = 101 01111 101 = 111 101cosf 101 110 = 101 110cosq(B): To have a

26、111 slip system, the material must have an FCC structure.19.GIVEN: tcrss = 55.2 MPa, (111)01 slip system, 112 tensile axisREQUIRED: Find the highest normal stress that can be applied before dislocation motion in the 10 direction.SOLUTION: The situation is shown below. Essentially the problem reduces

27、 to finding the value of the tensile stress when the critical resolved shear is reached.tcrss = s cosq cosfq = 112 01 f = 112 111B. Would have exactly the same stress for a BCC metal (f & q would be interchanged).20.GIVEN: s at yield = 3.5 MPa; (111) 10 slip system 11 tensile axisREQUIRED: Compute t

28、crssSOLUTION: tcrss = scosqcosfq = 11 10 f = 11 11121.ItemEdgeScrewLinear defect?YesYesElastic Distortion?YesYesGlide?YesYesClimb?YesNoCross-slip?NoYesBurgers Vector (BV) to line/ to lineUnique slip plane?YesNoOffset/ to BV/ to BVMotion/ to BV to BV22.GIVEN: BCC metal with tcrss = 7MPa 001 tensile a

29、xis.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 110. Deformation occurs on the plane and direction for which cosqcosf is a maximum since this will have the maximum resolved shear stress.

30、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 on (011) (also 11 on (01) (also 1 on (101)similar to planes shown in sketch. Also 111 on (01)We see by inspection that the resolved shear due to a tensi

31、le force in 001 will all be the same. The resolved shear on all other 110 systems is zero.B. To compute the normal stress at the onset of plastic deformation we will consider (011) 1tcrss = scosqcosf = 7q = 001 1; cosf = 001 011Note if we considered (101) 1 we would haveand we would obtain exactly t

32、he same answer.23.GIVEN: Yielding occurs at normal stress of s = 170 MPa in 100 direction. Dislocation moves on (101) in 11 direction.REQUIRED: tcrss and crystal structuresSOLUTION: Assume an edge dislocation. tcrss = scoscosfq = 100 11 f = 100 101The - sign means that the slip direction is opposite

33、 to the motion of the dislocation. Essentially, we have a negative edge dislocation on (101) as shown below:The edge dislocation moves in 11 direction but the offset is in 1 direction.The slip plane and slip direction are representative of BCC structures.24.GIVEN: (10)111 slip system. 123 tensile ax

34、istcrss = 800 psi for BCC crystaltcrss = 80 psi for FCC crystal withsFCC = 457 psi 123 tensile axis and (111)10 system.REQUIRED: Normal stress at yield for BCC metalSOLUTION: The simplest way to solve this problem is to note cosqcosf is the same for the BCC and FCC crystal with the meaning of f and

35、q interchanged. Let M = cosqcosf.(1)(2)(3)25.Here crystallographically equivalent positions join ions at cube corners (bv = ao), face diagonals , cube diagonals The most densely packed plane is the (110) in which we haveThe shortest vector that will reproduce all elements of the structure is ao. Thu

36、s b = aCOMMENT: We note that this is not sufficient for general deformation (e.g. a tensile axis of the type produces zero shear on the 1 Burgers vectors. We expect then a Burgers vectors as well.26.GIVEN: s = 1.7 MPa 100 tensile axis (111)101 slip systemsREQUIRED: tcrss, and crystal structure. Also

37、 find flaw in problem statement.SOLUTION: Since the slip system is of the type 111 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 10. Thus the slip system is (111)10 as s

38、hown below.27. = edge dislocationx = start of Burgers circuitb = Burgers vectory = 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

39、/m2 =N-m/m2 = N/m = force/length29.GIVEN: Two grain sizes, 10mm and 40mmREQUIRED: 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

40、 100X. N=ASTM GS#To solve the problem we first convert the grain size to in. where D = length of cube edge in mm.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 40mm grains at 100X isFor the 10mm dia grain, the # o

41、f grains per in2 (at box) isgrains/in2 at 100XSimilarlygrains/in2 at 100XFor the 10mm grain size:B. In computing the total g.s. area we will assume 1 in3 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 3x Area of face. G.B. Area = G

42、B Area (10m gs) = 3/3.937 x 10-4 = 7620in2/in3GB Area (40m gs) = 3/15.75 x 10-4 = 1905in2/in330.GIVEN:sys = 200MPa at GS#4= 300MPa at GS#6REQUIRED: sys at GS#9SOLUTION: Recall sys = so + kd-1/2(1)for low carbon steel. If d = grain size (assume cubes) load = grain diameter at 100X(2)(3)For ASTM GS# 4

43、: (4)For ASTM GS#6:(5)Substituting (4) and (5) into (1) we have200 - so + k(16.82)(6)300 = so + k(23.78)(7)Subtracting (6) from (7):100 = k(23.78 - 16.82)k = 14.37Substituting this value of k into (6) yields 200 = so + 14.37 x 16.82so = -41.70 (this is not physically realistic since so relates to th

44、e lattice friction stress which should not be negative)For ASTM GS#9Thus sys = so + 14.37 x 40 = 41.70 + 574 = 533MPa31.GIVEN: b = 0.25mm for BCC metal tilt boundary has angular difference of 2.5oREQUIRED: Dislocation density in tilt boundary wallSOLUTION: The physical situation is shown below:If b

45、= Burgers vector, D = spacing between edge dislocation# of dislocations in boundary for a 1cm high boundary is (where D is in cm)32. 33. FIND: Show D = b / .GIVEN: b is the magnitude of the Burgers vector; D is the spacing between dislocations, and is the tilt angle. SKETCH: See Fig. 5.3-4. SOLUTION

46、: We can see the geometry more clearly using the following sketch:From the Figure we can immediately write that . Since the tan of a small angle is the angle itself:, so that D = b / , as is written in the margin.34. FIND: How can you detect a cluster of voids or a cluster of precipitates in a mater

47、ial?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 of precipitate; however, usually the density difference between host and precipitate is not as great as b

48、etween 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 and the defects observed using optical or electron microscopy. X-ray diffraction can also be used. With a random spacing of

49、void or precipitate there is then an average spacing. Sometimes Braggs 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. They all rely o

50、n some property difference - magnetic, electrical, optical, or whatever.35. FIND: How can you ascertain whether a material contains both crystalline and noncrystalline regions?GIVEN: Recall that the density (and other properties) of crystalline material is greater than that of noncrystalline materia

51、l 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 noncrystalline and crystalline samples of the same composition.2. Diff

52、erential Scanning Calorimetry. Heat your sample in a calorimeter. Samples that are crystalline will absorb heat at the melting temperature and show a melting endotherm. Some noncrystalline samples (such as amorphous metals) will crystallize in the calorimeter and show a huge release of heat prior to

53、 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 common challenge to polymers scientists. We often need to quantify the fraction or percent crystallinity. Can y

54、ou 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 purely elastic manner.SOLUTION: There are many such possible examples. Since plastic deformation is nonrecoverable deformation, any application that re

55、quires 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 wings 37.As the dislocation density , there are more dislocation/dislocation interactions and the