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。