The atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or APF = V S V C. = 2(sphere volume) = 2 = V C = 4R

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1 3.5 Show that the atomic packing factor for BCC is The atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or APF = V S V C Since there are two spheres associated with each unit cell for BCC V S 4πR 8πR = 2(sphere volume) = 2 = Also, the unit cell has cubic symmetry, that is V C = a 3. But a depends on R according to Equation 3.3, and Thus, V C = 4R 3 3 = 64 R VS 8 π R /3 APF = = = V 64 R / 3 3 C

2 3.61 The metal iridium has an FCC crystal structure. If the angle of diffraction for the (220) set of planes occurs at (first-order reflection) when monochromatic x-radiation having a wavelength of nm is used, compute (a) the interplanar spacing for this set of planes, and (b) the atomic radius for an iridium atom. (a) From the data given in the problem, and realizing that = 2θ, the interplanar spacing for the (220) set of planes for iridium may be computed using Equation 3.13 as d 220 nλ (1)(0.154 nm) = = = 2sinθ (2) sin nm (b) In order to compute the atomic radius we must first determine the lattice parameter, a, using Equation 3.14, and then R from Equation 3.1 since Ir has an FCC crystal structure. Therefore, ( ) a = d + + = = (2) (2) (0) ( nm) nm And, from Equation 3.1 a nm R = = = nm

3 4.2 Calculate the number of vacancies per cubic meter in iron at 855 C. The energy for vacancy formation is 1.08 ev/atom. Furthermore, the density and atomic weight for Fe are 7.65 g/cm 3 and g/mol, respectively. Determination of the number of vacancies per cubic meter in iron at 855 C (1128 K) requires the utilization of Equations 4.1 and 4.2 as follows: N Q N ρ Qv exp exp kt A kt v A Fe v = N = Fe And incorporation of values of the parameters provided in the problem statement into the above equation leads to Nv ( )( ) atoms / mol 7.65 g / cm 1.08 ev / atom = exp g / mol ( ev / atom K )(1128 K) = cm 3 = m 3

4 5.7 A sheet of steel 1.8 mm thick has nitrogen atmospheres on both sides at 1200 C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is m 2 /s, and the diffusion flux is found to be kg/m 2 s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m 3. How far into the sheet from this high-pressure side will the concentration be 2.0 kg/m 3? Assume a linear concentration profile. This problem is solved by using Equation 5.3 in the form J = D C A C B x A x B If we take C A to be the point at which the concentration of nitrogen is 4 kg/m 3, then it becomes necessary to solve for x B, as x B = x A + D C A C B J Assume x A is zero at the surface, in which case x 4 kg / m 2 kg / m kg / m s B = 0 + ( 6 10 m /s) = m = 1 mm

5 5.22 The diffusion coefficients for silver in copper are given at two temperatures: T ( C) D (m 2 /s) (a) Determine the values of D 0 and Q d. (b) What is the magnitude of D at 875 C? (a) Using Equation 5.9a, we set up two simultaneous equations with Q d and D 0 as unknowns as follows: ln D 1 = lnd 0 Q d 1 R T 1 ln D 2 = lnd 0 Q d R 1 T 2 Solving for Q d in terms of temperatures T 1 and T 2 (923 K [650 C] and 1173 K [900 C]) and D 1 and D 2 ( and m 2 /s), we get Q d = R ln D 1 ln D 2 1 T 1 1 T 2 = ( ) ( ) (8.31 J/mol K) ln ln K 1173 K = 196,700 J/mol Now, solving for D 0 from Equation 5.8 (and using the 650 C value of D) D 0 = D 1 exp Q d RT , 700 J/mol = ( m /s) exp (8.31 J/mol K)(923 K) = m 2 /s

6 (b) Using these values of D 0 and Q d, D at 1148 K (875 C) is just , 700 J/mol D = ( m /s) exp (8.31 J/mol K)(1148 K) = m 2 /s Note: this problem may also be solved using the Diffusion module in the VMSE software. Open the Diffusion module, click on the D 0 and Q d from Experimental Data submodule, and then do the following: 1. In the left-hand window that appears, enter the two temperatures from the table in the book (converted from degrees Celsius to Kelvins) (viz. 923 (650ºC) and 1173 (900ºC), in the first two boxes under the column labeled T (K). Next, enter the corresponding diffusion coefficient values (viz. 5.5e-16 and 1.3e-13 ). 3. Next, at the bottom of this window, click the Plot data button. 4. A log D versus 1/T plot then appears, with a line for the temperature dependence for this diffusion system. At the top of this window are give values for D 0 and Q d ; for this specific problem these values are m 2 /s and 196 kj/mol, respectively 5. To solve the (b) part of the problem we utilize the diamond-shaped cursor that is located at the top of the line on this plot. Click-and-drag this cursor down the line to the point at which the entry under the Temperature (T): label reads 1148 (i.e., 875ºC). The value of the diffusion coefficient at this temperature is given under the label Diff Coeff (D):. For our problem, this value is m 2 /s.

7 Stress-Strain Behavior 6.3 A specimen of aluminum having a rectangular cross section 12 mm 12.9 mm is pulled in tension with 35,600 N force, producing only elastic deformation. Calculate the resulting strain. This problem calls for us to calculate the elastic strain that results for an aluminum specimen stressed in tension. The cross-sectional area is just (12 mm) (12.9 mm) = 155 mm 2 (= m 2 ); also, the elastic modulus for Al is given in Table 6.1 as 69 GPa (or N/m 2 ). Combining Equations 6.1 and 6.5 and solving for the strain yields σ F 35,600 N = = = = E A E ( m )( N/m ) 3

8 7.23 (a) From the plot of yield strength versus (grain diameter) 1/2 for a 70 Cu 30 Zn cartridge brass, Figure 7.15, determine values for the constants σ 0 and k y in Equation 7.7. (b) Now predict the yield strength of this alloy when the average grain diameter is mm. (a) Perhaps the easiest way to solve for σ 0 and k y in Equation 7.7 is to pick two values each of σ y and d -1/2 from Figure 7.15, and then solve two simultaneous equations, which may be created. For example d 1/2 (mm) 1/2 σ y (MPa) The two equations are thus σ 75 = k y σ 175 = k y of these equations yield the values of k y 1/2 = 12.5 MPa (mm) σ 0 = 25 MPa (b) When d = mm, d 1/2 = mm 1/2, and, using Equation 7.7, σ y = σ + 0 kd y 1/2 ( ) 1/2 1/2 = (25 MPa) MPa (mm) mm = 305 MPa

9 8.8 A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa m. If, during service use, the plate is exposed to a tensile stress of 200 MPa, determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.0 for Y. For this problem, we are given values of K Ic ( 55 MPa m), σ (200 MPa), and Y (1.0) for a large plate and are asked to determine the minimum length of a surface crack that will lead to fracture. All we need do is to solve for a c using Equation 8.7; therefore a c 2 1 KIc 1 55 MPa m = m 24 mm π = = = Y σ π (1.0)(200 MPa) 2

10 Cyclic Stresses (Fatigue) The S-N Curve 8.14 A fatigue test was conducted in which the mean stress was 50 MPa and the stress amplitude was 225 MPa. (a) Compute the maximum and minimum stress levels. (b) Compute the stress ratio. (c) Compute the magnitude of the stress range. (a) Given the values of σ m (50 MPa) and σ a (225 MPa) we are asked to compute σ max and σ min. From Equation 8.14 Or, σ m σ + σ 2 max min = = 50 MPa σ max + σ min = 100 MPa Furthermore, utilization of Equation 8.16 yields σ a = σ max σ min 2 = 225 MPa Or, σ max σ min = 450 MPa Simultaneously solving these two expressions leads to σ = 275 MPa max σ = 175 MPa min (b) Using Equation 8.17 the stress ratio R is determined as follows: R σ σ min = = = max 175 MPa 275 MPa 0.64

11 (c) The magnitude of the stress range σ r is determined using Equation 8.15 as σ r = σmax σmin = 275 MPa ( 175 MPa) = 450 MPa

12 9.32 For a copper silver alloy of composition 30 wt% Ag 70 wt% Cu and at 775 C (1048 K) do the following: (a) Determine the mass fractions of α and β phases. (b) Determine the mass fractions of primary α and eutectic microconstituents. (c) Determine the mass fraction of eutectic α. (a) This portion of the problem asks that we determine the mass fractions of α and β phases for an 30 wt% Ag 70 wt% Cu alloy (at 775 C (1048 K)). In order to do this it is necessary to employ the lever rule using a tie line that extends entirely across the α + β phase field. From Figure 9.7 and at 775 C (1048 K), C α = 8.0 wt% Ag, C β = 91.2 wt% Ag, and C eutectic = 71.9 wt% Sn. Therefore, the two lever rule expressions are as follows: Cβ C Wα = = = C C β α C0 Cα Wβ = = = C C β α (b) Now it is necessary to determine the mass fractions of primary α and eutectic microconstituents for this same alloy. This requires us to utilize the lever rule and a tie line that extends from the maximum solubility of Ag in the α phase at 775 C (1048 K) (i.e., 8.0 wt% Ag) to the eutectic composition (71.9 wt% Ag). Thus W α Ceutectic C = = = C C eutectic α C C = = = α We C eutectic C α (c) And, finally, we are asked to compute the mass fraction of eutectic α, W eα. This quantity is simply the difference between the mass fractions of total α and primary α as W = W W = = 0.08 eα α α

13 10.8 It is known that the kinetics of recrystallization for some alloy obey the Avrami equation and that the value of n in the exponential is 2.5. If, at some temperature, the fraction recrystallized is 0.50 after 200 min, determine the rate of recrystallization at this temperature. This problem gives us the value of y (0.40) at some time t (200 min), and also the value of n (2.5) for the recrystallization of an alloy at some temperature, and then asks that we determine the rate of recrystallization at this same temperature. It is first necessary to calculate the value of k. We first rearrange Equation as exp( kt n ) = 1 y and then take natural logarithms of both sides: kt n = ln (1 y) Now solving for k gives k = ln (1 y) t n which, using the values cited above for y, n, and t yields ln (1 0.50) k = = (200 min) 6 At this point we want to compute t 0.5, the value of t for y = 0.5, which means that it is necessary to establish a form of Equation in which t is the dependent variable. From one of the above equations t n = ln (1 y) k And solving this expression for t leads to t = ln (1 y) k 1/ n

14 For t 0.5, this equation takes the form t 0.5 ln (1 0.5) = k 1/ n and incorporation of the value of k determined above, as well as the value of n cited in the problem statement (2.5), then t 0.5 is equal to t /2.5 ln (1 0.5) = = min Therefore, from Equation 10.18, the rate is just 1 1 rate = = = (min) t min

15 10.18 Using the isothermal transformation diagram for an iron carbon alloy of eutectoid composition (Figure 10.22), specify the nature of the final microstructure (in terms of microconstituents present and approximate percentages of each) of a small specimen that has been subjected to the following time temperature treatments. In each case assume that the specimen begins at 760 C (1033 K) and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. (a) Cool rapidly to 700 C (973 K), hold for 10 4 s, then quench to room temperature. Below is Figure upon which is superimposed the above heat treatment. After cooling and holding at 700 C for 10 4 s, approximately 50% of the specimen has transformed to coarse pearlite. Upon cooling to room temperature, the remaining 50% transforms to martensite. Hence, the final microstructure consists of about 50% coarse pearlite and 50% martensite. (b) Reheat the specimen in part (a) to 700 C (973 K) for 20 h. Heating to 700 C for 20 h the specimen in part (a) will transform the coarse pearlite and martensite to spheroidite.

16 (c) Rapidly cool to 600 C (873 K), hold for 4 s, rapidly cool to 448 C (721 K), hold for 10 s, then quench to room temperature. Below is Figure upon which is superimposed the above heat treatment. After cooling to and holding at 600 C for 4 s, approximately 50% of the specimen has transformed to pearlite (medium). During the rapid cooling to 448 C no transformations occur. At 448 C we start timing again at zero time; while holding at 448 C for 10 s, approximately 50 percent of the remaining unreacted 50% (or 25% of the original specimen) will transform to bainite. And upon cooling to room temperature, the remaining 25% of the original specimen transforms to martensite. Hence, the final microstructure consists of about 50% pearlite (medium), 25% bainite, and 25% martensite. (d) Cool rapidly to 398 C (671 K), hold for 2 s, then quench to room temperature.

17 Below is Figure upon which is superimposed the above heat treatment. After cooling to and holding at 400 C for 2 s, no of the transformation begin lines have been crossed, and therefore, the specimen is 100% austenite. Upon cooling rapidly to room temperature, all of the specimen transforms to martensite, such that the final microstructure is 100% martensite. (e) Cool rapidly to 398 C (671 K), hold for 20 s, then quench to room temperature. Below is Figure upon which is superimposed the above heat treatment.

18 After cooling and holding at 400 C for 20 s, approximately 40% of the specimen has transformed to bainite. Upon cooling to room temperature, the remaining 60% transforms to martensite. Hence, the final microstructure consists of about 40% bainite and 60% martensite. (f) Cool rapidly to 398 C (671 K), hold for 200 s, then quench to room temperature. Below is Figure upon which is superimposed the above heat treatment.

19 After cooling and holding at 400 C for 200 s, the entire specimen has transformed to bainite. Therefore, during the cooling to room temperature no additional transformations will occur. Hence, the final microstructure consists of 100% bainite. (g) Rapidly cool to 575 C (848 K), hold for 20 s, rapidly cool to 350 C (623 K), hold for 100 s, then quench to room temperature. Below is Figure upon which is superimposed the above heat treatment.

20 After cooling and holding at 575 C for 20 s, the entire specimen has transformed to fine pearlite. Therefore, during the second heat treatment at 350 C no additional transformations will occur. Hence, the final microstructure consists of 100% fine pearlite. (h) Rapidly cool to 250 C (523 K), hold for 100 s, then quench to room temperature in water. Reheat to 315 C (588 K) for 1 h and slowly cool to room temperature. Below is Figure upon which is superimposed the above heat treatment.

21 After cooling and holding at 250 C for 100 s, no transformations will have occurred at this point, the entire specimen is still austenite. Upon rapidly cooling to room temperature in water, the specimen will completely transform to martensite. The second heat treatment (at 315 C for 1 h) not shown on the above plot will transform the material to tempered martensite. Hence, the final microstructure is 100% tempered martensite.

22 CHAPTER 17 CORROSION AND DEGRADATION OF MATERIALS PROBLEM SOLUTIONS Electrochemical Considerations 17.1 (a) Briefly explain the difference between oxidation and reduction electrochemical reactions. (b) Which reaction occurs at the anode and which at the cathode? (a) Oxidation is the process by which an atom gives up an electron (or electrons) to become a cation. Reduction is the process by which an atom acquires an extra electron (or electrons) and becomes an anion. (b) Oxidation occurs at the anode; reduction at the cathode.

In order to solve this problem it is first necessary to use Equation 5.5: x 2 Dt. = 1 erf. = 1.30, and x = 2 mm = 2 10-3 m. Thus,

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