材料科学基础作业

Fundamentals of Materials Science 1. Determine the Miller indices for the planes shown in the following unit cell:A:(2 1 -1) B:(0 2 -1)2. Show that the atomic packing factor for HCP is 0.74.Solution:This problem calls for a demonstration that the APF for HCP is 0.74. Again, the APF is just the total sphere-unit cell volume ratio. For HCP, there are the equivalent of six spheres per unit cell, and thusNow, the unit cell volume is just the product of the base area times the cell height, c. This base area is just three times the area of the parallelepiped ACDE shown below.The area of ACDE is just the length of CD times the height BC. But CD is just a or 2R, and3. For both FCC and BCC crystal structures, the Burgers vector b may be expressed aswhere a is the unit cell edge length and [hkl] is the crystallographic direction having the greatest linear atomic density.(a) Compute and compare the linear densities of the [100], [110], and [111] directions for FCC and BCC.(b) What are the Burgers vector representations for FCC, BCC?(c) If the magnitude of the Burgers vector b isdetermine the values of b for aluminum, For Al which has an FCC crystal structure, R=0.1431nm.solutions:(a) In this problem we are to compute the linear densities of several crystallographic planes for the face-centered cubic crystal structure. For FCC the linear density of the [100] direction is computed as follows:The linear density, LD, is defined by the ratiowhere L l is the line length within the unit cell along the [100] direction, and Lc is line length passing through intersection circles. Now, L l is just the unit cell edge length, a which, for FCC is related to the atomic radius R according to a = 2R√2 [Equation (3.1)]. Also for this situation, Lc = 2R and therefore4. Show that the minimum cation-to-anion radius ratio for a coordination number of 6 is 0.414.solutions:In this problem we are asked to show that the minimum cation-to-anion radius ratio for a coordination number of six is 0.414. Below is shown one of the faces of the rock salt crystal structure in which anions and cations just touch along the edges, and also the face diagonals.5. The number-average molecular weight of a polypropylene is 1,000,000 g/mol. Compute the number-average degree of polymerization. Solution:We are asked to compute the number-average degree of polymerization for polypropylene, given that the number-average molecular weight is 1,000,000 g/mol.The mer molecular weight of polypropylene is justm = 3(A C) + 6(A H)= (3)(12.01 g/mol) + (6)(1.008 g/mol) = 42.08 g/molIf we let n n represent the number-average degree of polymerization, then from Equation (4.4a)6. A sheet of BCC iron 1mm thick was exposed to a carburizing gas atmosphere on one side and a decarburizing atmosphere on the other side at 725o C. After having reached steady state, the iron was quickly cooled to room temperature. The carbon concentrations at the two surfaces of the sheet were determined to be 0.012 and 0.0075 wt%. Compute the diffusion coefficient if the diffusion flux is 1.4 ×10-8 kg/m2-s. Hint: Use Equation 5.9 to convert the concentrations from weight percent to kilograms of carbon per cubic meter of iron.7. For a steel alloy it has been determined that a carburizing heat treatment of 10 h duration will raise the carbon concentration to 0.45 wt% at a point 2.5 mm from the surface. Estimate the time necessary to achieve the same concentration at a 5.0-mm position for an identical steel and at the same carburizing temperature.8. The diffusion coefficients for iron in nickel are given at two temperatures:(a) Determine the values of D0 and the activation energy Qd .(b) What is the magnitude of D at 1100 o C (1373 K)?9. A load of 44,500 N (10,000 lbf ) is applied to a cylindrical specimen of steel (displaying the stress–strain behavior shown in Figure 7.33) that has a cross-sectional diameter of 10 mm (0.40 in.).(a) Will the specimen experience elastic or plastic deformation? Why?(b) If the original specimen length is 500 mm (20 in.), how much will it increase in length when this load is applied?Solution:This problem asks us to determine the deformation characteristics of a steel specimen, the stress-strain behavior of which is shown in Figure7.33.(a)In order to ascertain whether the deformation is elastic or plastic, wemust first compute the stress, then locate it on the stress-strain curve, and, finally, note whether this point is on the elastic or plastic region.Thus,10. Consider a single crystal of silver oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a (111) plane and in a[101] direction, and is initiated at an applied tensile stress of 1.1 MPa (160 psi), compute the critical resolved shear stress.Solution: λφστcos cos =11. For alloys of two hypothetical metals A and B, there exist an α, A-rich phase and a β, B-rich phase. From the mass fractions of both phases for two different alloys, which are at the same temperature,determine the composition of the phase boundary (or solubility limit) for both α and βphases at this temperature.12. What is the carbon concentration of an iron–carbon alloy for which the fraction of total ferrite is 0.94?Solution:This problem asks that we compute the carbon concentration of an iron-carbon alloy for which the fraction of total ferrite is 0.94. Application of the lever rule [of the form of Equation (10.12)] yields13. Is it possible to have an iron–carbon alloy for which the mass fractions of total ferrite and proeutectoid cementite are 0.846 and 0.049, respectively? Why or why not?Solution:This problem asks if it is possible to have an iron-carbon alloy for which Wα= 0.846 and W Fe3C = 0.049. In order to make this determination, it isnecessary to set up lever rule expressions for these two mass fractions in terms of the alloy composition, then to solve for the alloy composition of each; if both alloy composition values are equal, then such an alloy is possible. The expression for the mass fraction of total ferrite is14. The fatigue data for a brass alloy are given as follows:(a) Make an S–N plot (stress amplitude versus logarithm cycles to failure) using these data.(b) Determine the fatigue strength at 5×105 cycles.(c) Determine the fatigue life for 200 MPa.Solution:This problem first provides a tabulation of fatigue data (i.e., stress amplitude and cycles to failure) for a brass alloy.(a)These fatigue data are plotted below.(b) As indicated by one set of dashed lines on the plot, the fatigue strength at 5×105 cycles [log (5×105) = 5.7] is about 250 MPa.(c) As noted by the other set of dashed lines, the fatigue life for 200 MPa is about 2×106 cycles (i.e., the log of the lifetime is about 6.3).15. Is it possible to produce a continuous and oriented aramid fiber(芳香尼龙纤维)-epoxy matrix composite having longitudinal and transverse moduli of elasticity of 57.1 GPa and 4.12 GPa, respectively? Why or why not? Assume that the modulus of elasticity of the epoxy is 2.4 GPa. Solution:This problem asks for us to determine if it is possible to produce a continuous and oriented aramid fiber-epoxy matrix composite having longitudinal and transverse moduli of elasticity of 57.1 GPa and 4.12 GPa,respectively, given that the modulus of elasticity for the epoxy is 2.4 GPa. Also, from Table 15.4 the value of E for aramid fibers is 131 GPa. The approach to solving this problem is to calculate two values of Vf using the data and Equations (15.10b) and (15.16); if they are the same then this composite is possible.16. Below, weight gain-time data for the oxidation of copper at an elevated temperature are tabulated.(a) Determine whether the oxidation kinetics obey a linear, parabolic, or logarithmic rate expression.(b) Now compute W after a total time of 450 min.Solution:For this problem we are given weight gain-time data for the oxidation ofCu at an elevated temperature.(a)We are first asked to determine whether the oxidation kinetics obey aparabolic, linear, or logarithmic rate expression, expressions which are described by Equations (16.34), (16.35), and (16.36), respectively.One way to make this determination is by trial and error. Let us assume that the parabolic relationship is valid; that is, from Equation(16.34)which means that we may establish three simultaneous equations using the three sets of given W and t values, then using two combinations of two pairs of equations, solve for K1 and K2; if K1 and K2 have the same values for both solutions, then the kinetics are parabolic. If the values are not identical then the other kinetic relationships need to be explored. Thus, the three equations are。