工程材料科学与设计(扫描版本)习题答案chapter5(bak)

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/sec2. A Newton is a kg-m/sec2. A Pascal is a N/m2. A MPa is 106 Pa. SOLUTION: Stress is force per unit area. The cross-sectional area is SYMBOL 112 \f "Symbol"R2 = 1963.5 square micrometers. The force is 25 g (kg/1000g)(9.8 m/sec2) = 0.245 N. Thus, the stress is SYMBOL 115 \f "Symbol" = F/A = COMMENTS: You must learn to do these sorts of problems, including the conversions. 2. GIVEN: FCC Cu with ao = 0.362nm REQUIRED: A) Lowest energy Burgers vector, B) Length in terms of radius of Cu atom, C) Family of planes SOLUTION: 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) shown We 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 is as shown in (111). A. The length of this vector is B. 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 Ag REQUIRED: Find lattice parameter SOLUTION: Recall the Ag is FCC. For FCC structures the Burgers vector is ½ a face diagonal as shown. We see that 4. A. FCC structure The (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 etc. Then in general B. For NaC1 We 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 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 ao = 0.314nm REQUIRED: Determine the crystal structure. If Mo were FCC, then but |b| = 0.272  Mo is not FCC. Assuming Mo is BCC, then 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| = 0.25nm REQUIRED: Find lattice parameter a ASSUME: for BCC structure SOLUTION: from the formula for the magnitude of a vector: 8. GIVEN: Normal stress of 123 MPa applied to BCC Fe in [110] direction REQUIRED: Resolved shear in [101] on (010) SOLUTION: Recall that the resolved shear stress is given by:  =  cos cos (1) where  = angle between slip direction and tensile axis;  = angle between normal to slip plane and tensile axis Thus 9. GIVEN: Stress in [123] direction of BCC crystal REQUIRED: Find the stress needed to promote slip if cR = 800 psi. The slip plane is (1EQ \O(1,_)0) and slip direction is [111]. SOLUTION: Recall  =  cos cos (1)  = [123] [111] [123]  [111] = [123] [111]cos  = [123] [1EQ \O(1,_)0] [123]  [1EQ \O(1,_)0] = [123] [1EQ \O(1,_)0]cos 10. 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 (1EQ \O(1,_)1) parallel to intersection of (1EQ \O(1,_)1) and (111) with Burgers vector parallel to [EQ \O(1,_) EQ \O(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 For a Burgers vector parallel to [EQ \O(1,_) EQ \O(1,_)0] the scalar multiplier must be a/2. Thus EQ \O(b,_) = a/2 [EQ \O(1,_) EQ \O(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: 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 | b1 | 2 + | b2 | 2 > | b3| 3 Thus the reaction is favorable since 14. GIVEN: Dislocation in FCC Parallel to [EQ \O(1,_)01] i.e. ADVANCE \l0 ADVANCE \u0 = [EQ \O(1,_)01] REQUIRED: Character and slip plane SOLUTION: Character is found by angle between . Note = -1 + 0 + 1 = 0. Thus . Since the dislocation is pure edge. To find the slip plane we note that the cross produce of gives a vector that is normal to the plane in which 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 (0EQ \O(1,_)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 = 16 At. wt. Mg = 24.32 Same structure as NaCl  = 3.65 g/cm3 REQUIRED: Find length of Burgers Vector in MgO SOLUTION: 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 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 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 ½ O= and ½ Mg++ ions in our cube. Thus 18. GIVEN: Critical resolved shear stress (0.34MPa), slip system (111)[EQ \O(1,_)10], and tensile axis [101] REQUIRED: Applied stress at which crystal begins to deform and crystal structure. SOLUTION: (A)The situation is shown below crss =  cos  cos  = angle between tensile axis and slip direction  = angle between tensile axis and normal to slip plane  = [111] [101]  = [101] [EQ \O(1,_)01] [111]  [101] = [111] [101]cos [101]  [110] = [101] [110]cos (B): To have a {111}<110> slip system, the material must have an FCC structure. 19. GIVEN: crss = 55.2 MPa, (111)[EQ \O(1,_)01] slip system, [112] tensile axis REQUIRED: Find the highest normal stress that can be applied before dislocation motion in the [10 EQ \O(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. crss =  cos  cos  = [112] [EQ \O(1,_)01]  = [112] {111} B. Would have exactly the same stress for a BCC metal ( &  would be interchanged). 20. GIVEN:  at yield = 3.5 MPa; (111) [1EQ \O(1,_)0] slip system [1EQ \O(1,_)1] tensile axis REQUIRED: Compute crss SOLUTION: crss = coscos  = [1EQ \O(1,_)1] [1EQ \O(1,_)0]  = [1EQ \O(1,_)1] [111] 21. Item Edge Screw Linear defect? Yes Yes Elastic Distortion? Yes Yes Glide? Yes Yes Climb? Yes No Cross-slip? No Yes Burgers Vector (BV)  to line // to line Unique slip plane? Yes No Offset // to BV // to BV Motion // to BV  to BV 22. GIVEN: BCC metal with crss = 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cos 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 [1EQ \O(1,_)1] on (011)) (also [EQ \O(1,_)11] on (0EQ \O(1,_)1)) (also [EQ \O(1,_) EQ \O(1,_)1] on (101)) similar to planes shown in sketch. Also [111] on (EQ \O(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) [EQ \O(1,_) EQ \O(1,_)1] crss = coscos = 7  = [001] [EQ \O(1,_) EQ \O(1,_)1]; cos = [001] [011] Note if we considered (101) [EQ \O(1,_) EQ \O(1,_)1] we would have and we would obtain exactly the same answer. 23. GIVEN: Yielding occurs at normal stress of  = 170 MPa in [100] direction. Dislocation moves on (101) in [11EQ \O(1,_)] direction. REQUIRED: crss and crystal structures SOLUTION: Assume an edge dislocation. crss = coscos  = [100] [11EQ \O(1,_)]  = [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 [11EQ \O(1,_)] direction but the offset is in [EQ \O(1,_) EQ \O(1,_)1] direction. The slip plane and slip direction are representative of BCC structures. 24. GIVEN: (EQ \O(1,_)10)[111] slip system. [123] tensile axis crss = 800 psi for BCC crystal crss = 80 psi for FCC crystal with FCC = 457 psi [123] tensile axis and (111)[1EQ \O(1,_)0] system. REQUIRED: Normal stress at yield for BCC metal SOLUTION: The simplest way to solve this problem is to note coscos is the same for the BCC and FCC crystal with the meaning of  and  interchanged. Let M = coscos. (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 have The shortest vector that will reproduce all elements of the structure is ao. 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:  = 1.7 MPa [100] tensile axis (111)[101] slip systems REQUIRED: crss, 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 [10EQ \O(1,_)]. Thus the slip system is (111)[10EQ \O(1,_)] as shown below. 27.  = edge dislocation x = start of Burgers circuit b = Burgers vector y = end of Burgers circuit 28. 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/length 29. GIVEN: Two grain sizes, 10m and 40m REQUIRED: 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 m. At 100X linear magnification, the sides of the smaller grains will be: The area of each grain at 100X will be Similarly the area of the 40m grains at 100X is For the 10m dia grain, the # of grains per in2 (at box) is grains/in2 at 100X Similarly grains/in2 at 100X For the 10m 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 = GB Area (10 gs) = 3/3.937 x 10-4 = 7620in2/in3 GB Area (40 gs) = 3/15.75 x 10-4 = 1905in2/in3 30. GIVEN: ys = 200MPa at GS#4 = 300MPa at GS#6 REQUIRED: ys at GS#9 SOLUTION: Recall ys = o + 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: (4) For ASTM GS#6: (5) Substituting (4) and (5) into (1) we have 200 - o + k(16.82) (6) 300 = o + k(23.78) (7) Subtracting (6) from (7): 100 = k(23.78 - 16.82) k = 14.37 Substituting this value of  into (6) yields 200 = o + 14.37 x 16.82 o = -41.70 (this is not physically realistic since o relates to the lattice friction stress which should not be negative) For ASTM GS#9 Thus ys = o + 14.37 x 40 = 41.70 + 574 = 533MPa 31. GIVEN: b = 0.25m 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 is (where D is in cm) 32. 33. FIND: Show D = b / SYMBOL 113 \f "Symbol". GIVEN: b is the magnitude of the Burger's vector; D is the spacing between dislocations, and SYMBOL 113 \f "Symbol" 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 . Since the tan of a small angle is the angle itself: , so that D = b / SYMBOL 113 \f "Symbol", 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 of precipitate; however, usually the density difference between host and precipitate is not as great as 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 and the 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. Sometimes Bragg'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. They all rely on 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 material of the same composition SOLUTION: 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. Differential 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 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 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 purely elastic manner. SOLUTION: There are many such possible examples. Since plastic deformation is nonrecoverable 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 springs 2. A diving board 3. Trusses in a bridge 4. The walls in a building 5. A bicycle frame 6. Piano wire 7. Airplane wings 37. As the dislocation density , there are more dislocation/dislocation interactions and the 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 the defects 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 of point defects to move. 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 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 for intersitital. 39. As dys since this means the path over which a dislocation moves . This means that 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: ys = o + kd-1/2 40. 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 with the 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 different 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/cm2 for low C steel REQUIRED: concentration of C atoms (at %) to lock all dislocations SOLUTION: Recalling the At. weight of Fe is 55.85 and the density is about 7.8 gm/cm3 we may write (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