SOLUTIONS FOR CHAPTER 81. FIND: When a phase transformation occurs such as a liquid phase transforming to asolid below its melting temperature, what are the two steps involved in the process?Briefly describe each.SOLUTION: During a phase transformation such as a liquid transforming to solid, there are two steps involved in the process. They are:1. nucleation of the new phase, and2. growth of the phase.Nucleation relates to the formation of the new phase and the development of theinterface seperating the two phases. Nucleation can either occur randomly throughout the structure - termed homogeneous nucleation or at specific sites such as interfaces - termed heterogeneous nucleation.Growth - Once the phases has nucleated, it begins to grow. The growth process iscontrolled by diffusion and undercooling. As in the nucleation step, there may becompeting processes that lead to a maximum growth rate at an intermediate temperature.2. FIND: We presented a derivation in Section 8.2.3 showing that the barrier for nucleation,∆G*, decreases with increasing undercooling following the proportionalityBy starting with an expression for the free energy of a distribution of spherical particles of radius r, derive equation 8.2-9a. Explain each step in the derivation. Explain anyassumptions that are made.SOLUTION: To determine the barrier to the nucleation process, ∆G* we begin bynoting that the free energy as a function of particle size for homogeneous nucleation has two terms, one that increases with r2, the interfacial energy per unit volume term, and one that decreases with r3. A maximum occurs in ∆G(r) at some r*. These graphicalrelationships are sketched below.The dependence of the various free energy terms associated with nucleation as a function of temperature: (a) the relationship between cluster radius and surface energy of agrowing spherical solid phase in liquid, (b) the relationship between the cluster radius and(c) the sum of (a) and (b). Instal l Equa tion E ditor a nd do uble -click h ere to view equat ion.The change in free energy can be written as:In this equation we assume that the nuclei can be considered as a random distribution of spheres. To locate the maximum of a function we equate the first derivative of thefunction with respect to the parameter which is the variable to zero. Here we assume thatr is the only variable. The is SL is independent of size and orentation.Using equation 8.2-4 for Instal l Equa tion E ditor a nd do uble -click h ere to view equat ion.we have:In writingwe have assumed that the heat capacity difference between the liquid and solid phases is zero. (Note: Although this may be a reasonable assumption at small undercoolings, i.e.small ∆T’s, at the large under coolings that are typical for homogeneous nucleation thatapproximation may not be valid and a more complex term is required. But for a firstorder approximation this assumption is reasonable.)In order to determine the value of ∆G(r) at r*, we introduce the expressionintoRearranging,If all the terms in parentheses are constant then,3. FIND: Explain the simultaneous influence that undercooling has on the barrier tonucleation and the atomic rearrangements necessary to initiate the transformation. Show how these competing effects lead to classical C-curve behavior in the nucleation ofdiffusional transformations.SOLUTION: With larger undercoolings, both r* and ∆G* decrease, suggesting thatsimply lowering the temperature of the system allows nucleation to occur ever morereadily. However, there are practical kinetic limits to this effect. For example, withdecreased temperature there is a corresponding reduction in atomic mobility. The random fluctuations in the local arrangements of atoms is the process that provides the clusters.Since the formation of the clusters depends on atomic mobility, a reduction in thetemperature reduces the rate of clustering. Thus, as shown in Figure (a) below, theoverall nucleation rate exhibits a maximum at an intermediate temperature. Themaximum in the nucleation rate leads to a minimum in the time required to nucleate aphase, as shown in Figure b. Because of its shape, this curve is known as a C-curve.(a) The influence of temperature on the mobility term and the nucleation barrier term.The opposing processes result in a maximum in the nucleation rate at anintermediate temperature.(b) Since the time for nucleation is inversely related to the nucleation rate, the timecurve exhibits a minimum at an intermediate temperature. Because of its shape,this curve is often referred to as a C-curve.4. FIND: Explain how the value of interfacial energy between the parent phase and thetransforming phase affects the critical radius and the barrier to nucleation.SOLUTION: Equation 8.25 gives the change in free energy as a function of r when aliquid transforms to a solid, for example. In the development of that equation it wasassumed that the transforming phase was spherical and the interfacial energy, γSL, wasisotropic. That equation consists of two terms on the right hand side, i.e.∆G(r) = (4πr2) γSL + 4/3 πr3 (∆G v)Since γSL > 0 and ∆G V < 0 for ∆T > 0, the first term increases with radius, and the second decreases. Figure 8.2-3 illustrates that there is a maximum that occurs at some r wedesignated as r* and a corresponding ∆G, we designated as ∆G* wherer* = (-2γSL)/∆G vandBoth the critical radius, r*, and the barrier to the nucleation process contain γSL in thenumerator. Thinking in terms of the barrier to nucleation, ∆G*, there is a cubicdependence on interfacial energy. The larger the S-L interfacial energy, the larger thebarrier to nucleation and hence the more difficult. The use of nucleating agents is basedon the principle of introducing particles with lower interfacial energies to stimulatenucleation.5. FIND: Compare homogeneous nucleation and heterogeneous nucleation.SOLUTION: The process of homogeneous nucleation occurs at random locations in the parent phase. The distribution of the transforming phase occurs without regard to specific sites, such as mold walls in solidification. Heterogeneous nucleation occurs at specificsites. In the case of solidification they can be at mold walls, unintentional additions such as ceramic inclusions from crucibles or it may occur at nucleating agents which areintentionally added to control the solidification microstructure.6. FIND: What is the difference between the following interfaces?a. coherentb. partially coherent, andc. incoherentSOLUTION: A coherent interface is one in which there is a one-to-one correspondence of atomic planes across the interface. This type of interface occurs when the latticeparameters in the two phases are the same or very close. When the lattice parameters are different in the two phases, the increase in strain energy that will occur as the particlegrows necessitates the periodic insertion of dislocations at the interface to accommodatethe misfit. This type of interface is referred to as partially coherent. An incoherentinterface occurs between two phases of different crystal structures and sufficientlydifferent atomic spacings that can not be accommodated by dislocations.7. FIND: How does interfacial energy vary with coherency?SOLUTION: Interfacial energy is sensitive to the nature of the interface separating thetwo phases. The interfacial energy increases going from coherent to partially coherent to incoherent, i.e. γincoh. > γpart.coh. . γcoh..8. FIND: Based upon your answer to question 7, explain how the probability forheterogeneous nucleation changes as the type of interface changes from coherent to partially coherent to incoherent.SOLUTION: Since the barrier to nucleation, ∆G*, is related to γα/β in the following way:∆G* ∝γα/β3increasing γα/β would increase the barrier to homogeneous nucleation. Consequently, the probability for heterogeneous nucleation would increase as interfacial energy increases. 9. FIND: Figure 5.3-5 contains a schematic illustration of a twin boundary in a crystal.From the point of view of coherency, what is the nature of the type of twin boundaryillustrated in the figure? Comment on the relative energy of a twin boundary compared with a random grain boundary in a polycrystalline material.GIVEN:Figure 5.3-5 illustrates a schematic of a twin in a matrix showing the two twin boundaries separating the matrix from the twin and Figure 8.2-10 illustrates an incoherent interface separating the matrix from a precipitate.Schematic of a twinMatrixPrecipitateSchematic of an incoherent boundarySOLUTION: Atoms that are on the twin plane are part of the stacking sequence in the matrix above the twin plane as well as the stacking of atoms below the twin plane. Sincea coherent interface is an interface that occurs when there is a one-to-one correspondenceacross the interface, then the twin illustrated in this figure would be classified as acoherent twin boundary. Array The incoherent boundary illustrated above occurs in asystem when there is not amatch across the boundaryseparating two phases. Since a general grain boundaryrepresents a situation where theorientation of two grains across a boundary are not the same, we would therefore expect that there would not be a match ofatoms across the boundary.Thus, a coherent twin plane would have lower interfacial energy than the interfacialenergy associated with grain boundary separating two randomly oriented grains.10. FIND: In certain nickel-base superalloys, a second phase can precipitate coherently fromthe matrix during aging because the lattice parameters of the two phases are very closeand both phases are cubic. For a coherent precipitate in this system, what is the mostlikely relationship between the crystallographic axes in the matrix phase and that of theprecipitate? Explain, using sketches.GIVEN: The matrix and precipitate are both cubic with similar lattice parameters.SOLUTION: The best match between two similar cubic structure with similar latticeparameters occurs when the two cubes are aligned with the two sets of orthogonal axesparallel to one another. Thus, the x-axis in the matrix, x m, would be parallel to the x-axis in the precipitate, x p, and the y-axis in the matrix, y m, parallel to the y-axis in theprecipitate, y p. This orientation relationship is sometimes referred to as cube//cubeorientation. That is, the two cubes defining the two crystal structures are oriented x-axisto x-axis and y-axis to y-axis.Furthermore, because the lattice parameters of the two phases are very close, the distances between the atoms in the matrix and theprecipitate match very closely.11. FIND: The transition precipitate γ'(Al2Ag) in the aluminum-silver system is a hexagonalclose-packed structure. The a axis of the HCP phase is 0.2858 nm, and the c axis is0.4607 nm. What crystallographic plane and direction in the aluminum matrix definesthe coherent interface between matrix and precipitate?GIVEN: Crystal structure of aluminum is fcc and that for γ' is hcp. Atomic radius foraluminum is 0.143 nm from Appendix C and the a and c lattice parameters for γ' are0.2858 nm and 0.4607, respectively.SOLUTION: Both structures are close-packed structures. The two structures have the same packing factor and coordination number. Both have sets of planes with the highest possible planar density, the close-packed planes. In the fcc structure these are the {111} and in the hcp they are the {0001}. Both structures have directions with the highestpossible linear density. In the fcc the highest linear density is in the 〈110〉 and in the hcp structure 〈1000〉 . The best potential match between the two structures is across the two highest packing planes. Thus the {0001} of the precipitate and the {111} of thealuminum matrix would form a boundary.Recall that the packing sequence for an fcc structure is ......ABCABCABC....., and thatfor the hcp structure is .....ABABABABAB...... . A schematic of the two structures with a common plane would be:To complete the analysis, the atomic spacings in a 〈111〉 direction of the fcc aluminummatrix must be calculated to determine if a match can be made with these planes andthose of the dense packed planes in the 〈0001〉 directions in the hcp structure of γ'.From Appendix C the atomic radius for aluminum is 0.143 nm, since the atoms touchalong 〈110〉 directions, see accompanying figure, in a fcc structure the lattice parameter, a o , for aluminum is:a o = 0.4045 nm.Schematic showing the dense packed arrangement of atoms in an fcc material like aluminum.For fcc stacking ( .....ABCABCABC.....) the distance between equivalent positions A positions can be determined using the above sketch:= ( 2a o 2 + a o2 ) ½ = a o √3 = 0.7006 nmThus the spacing between the adjacent dense packed layers in the fcc structure is= 0.7006/3 = 0.2335 nm.A schematic showing a comparison between the fcc aluminum and the hcp γ' structures are shown below.Thus the atomic spacing in the dense packed planes are similar, and the spacing between the dense packed planes for the two structures are similar.12. FIND: The expression: X = 1 - exp [ - (kt)n ] , equation 8.2-1, in the text, is a powerfulempirical function that is useful in describing the kinetics of diffusional transformations.In the equation, X is the fraction transformed, k is a rate constant having units ofreciprocal time, t is time and n is a unitless constant. Sketch the behavior of this function over a range of times that demonstrate why this expression is useful for describingmicrostructural changes like recrystallization, or the decomposition of austenite to form pearlite.SOLUTION: At the start of the transformation, t = 0, we expect no transformed phase, hence, X = 0. At very long transformation times, t →∞, we expect thetransformation to be complete, i.e. X= 1. Examining equation 9.2-1 at these twoextremes:At t = 1X = 1 - exp - (k ⋅ 0)n= 1 - 1 = 0, andAt t →∞X = 1 - exp - (k ⋅∞)n= 1 - 0 = 1.Over a range of values the function would behave like the sketch shown below:13.The value of n and k for the decomposition of a particular steel has been investigated by following the fraction transformed versus time. From the experimental data, values for n and k were determined at two temperatures. Plot the fraction transformed as a function ofo o SOLUTION: Using the equation:where X is fraction transformed, and k and n are as given in the table, to calculate the fraction transformed as a function of time.Typical values are summarized in the table below and the behavior of the function is plotted and shown on the next page.-1-114. FIND: Explain how the data from problem 10 could be used to determine the location ofthe start and finish times for the transformation.SOLUTION: Since it is often difficult to decide exactly when a reaction begins and iscompleted, the "start" and "finish" points on an isotherm transformation diagram areusually chosen when, for example, 1% of the parent phase has transformed to indicate thestart of transformation, and 99% has transformed for the finish.The start of the transformation:At 400o C0.01 = 1 - exp [ - (0.028)2 t s2]0.01 - 1 = - exp [ - (0.028)2 t s2]0.99 = exp [ - (0.028)2 t s2]- 0.01 = - (0.028)2 t s20.01 = 0.000784 t s212.75 = t22 ;t = 3.6 sec.At 360o C0.01 = 1 - exp [ - (0.085)2 t s2]0.01 = 0.007 2 t s2t s = 1.4 Sec.For the finish of the transformationAt 400o C0.99 = 1 - exp [ - (0.028)2 t f2]4.605 = 0.0078 t f2t f = 76.6 Sec.At 360o C0.99 = 1 - exp [ - 0.085)2 t f2]4.605 = 0.0072 t f2t f = 25.4 Sec.15. FIND: When fraction transformed as a function of the logarithm of time, plotted at twodifferent temperatures, results in two curves that are parallel (same n but different k), the mechanisms for the transformation are the same. Sometimes we refer to the process asbeing isokinetic. For the kinetic data shown below, plot fraction transformed versus time for the two temperatures. Does the process appear to be isokinetic?Isothermal Transformation Isothermal Transformationo oSOLUTION: The data are plotted in the form of fraction transformed versus time and shown belowSince the curves are parallel the processes appear to be isokinetic.16. FIND: The value of k in equation 8.2-12, for a set of kinetic data, can be determined bynoting:X = 1 - exp - (kt)n, andwhen kt = 1X = 1 - exp (-1) - 1 - 1/e orX = 0.632.What are the values for the rate constants for the data in problem 12 ?SOLUTION: To find the value of k, we need to find the time (t*) at which kt =1. The data is nearly linear at X = .632, therefore we can find this time by@ 415o Ct* = 10.5 sec.kt* = 1 k (10.5) = 1 k = 0.095@ 375o Ct* = 23.9 kt* = 1 k = 0.04217. FIND: When the mechanisms controlling a particular transformation are independent oftemperature, we can define an empirical activation energy, Q, for the process by noting:k = A exp (-Q/RT).Determine the empirical activation energy for the data given in problem 12 .SOLUTION: Since we have two value of k for two different temperatures we can solvetwo equations for Q.from problem 13:k 1 = .042 k 2 = .095 T 1 = 648 K T 2 = 688 KR = 8.314 J/(mole K)Q = 75,600 J/mole.18. FIND: The value of n from a set of data can be found by noting:X = 1 - exp - (kt)n 1-X = exp - (kt)n ln (1 - x) = - (kt)n-ln (1 - x) = ln (1/(1-x)), then ln (1/(1-x)) = (kt)nln ln (1/(1-x)) = n lnt + n lnk.Determine n at the two temperatures from the data in problem 12.SOLUTION: We have data for X vs t and we have already determined k at each temperature (problem 13). Computeand ln(t) for each data point. The slope of a plot of Instal l Equa tion E ditor a nd do uble -click h ere to view equat ion. vs ln(t) is equal ton.@ 415o C n = 2.08@ 375o Cn = 2.0819.FIND: In a diffusion-controlled transformation, the time t at different temperatures to yield the same fraction transformed X is as follows:uble -SOLUTION: By noting that the time for a given fraction transformed is inversely related to the rate of the reaction if the processes have the same activation energy. Then using the Arrhenius:Rate = constant exp (-Q/RT)where Q is the required activation energy, R the gas constant and T the absolute temperature, the activation energy is determined graphically by plotting, ln 1/t versus 1/T.From the graph: slope = - 3629.9Then, Q/R = 3629.9R = 8.314 Joules/mole⋅KQ = 30,180 Joules/mole⋅K20. FIND: Using the data from Problem 19, calculate the time required to yield the samefraction transformed at 125︒C.GIVEN: Data from problem 19.SOLUTION: Least squares fit of the experimental data give an equation of the form y (ln(reciprocal of the time)) = m⋅x (1/T) + bWith the values the expression becomes:y = -3630 ⋅ ( 1/ T ) + 9.8572To calculate the time at 125︒Cy = -3630 ⋅ ( 1/ (125 + 273)) + 9.8572y = 0.7366Taking the inverse ln gives:1/t = 2.089,or, t = .4787 hours21. FIND: The precipitation of carbides in certain steels can increase their strength. Shownbelow are data relating time to reach peak strength and the isothermal hold temperature.From the data, determine the activation energy for the precipitation process. Compareyour results with the activation energy for diffusion of carbon in bcc iron. Explain why the similarities between the two activation energies are not too surprising.T (︒C)Time to Peak Hardness (min.)353 0.9375 3.0400 4.0425 10.0450 20.0Applications to Engineering Materials (8.3)22. FIND: Sketch an isothermal transformation (IT) diagram for a plain carbon eutectoid steel. Label theSOLUTION: An IT diagram for a plain carbon eutectoid steel should contain thefollowing information:1. The temperature of the eutectoid - 727o C.2. The start and finish lines for the diffusional reactions:γ→ pearlite and γ→ bainite.3. The start and finish lines for the athermal martensitetransformation.Shown below is the IT diagram.23. FIND: Sketch a section of the Fe - Fe3C diagram over the composition range 0 to 2 wt%carbon and over the temperature range from 900o C to room temperature. Label the phase fields and compare the information that can be extracted from the phase diagram and the IT diagram.SOLUTION: An IT diagram such as that for a steel provides kinetic information that is, how rapidly austenite decomposes into a particular microstructure. It also provides theinformation regarding the morphology of the phases present. For example, dependingupon the quench temperature either coarse pearlite or fine pearlite may form. These are the same microstructures except for the scale. The lower the transformation temperature, the finer the lamellar spacing. Alternatively, rather than pearlite forming the austenitemight decompose into an alternative two phase structure of finely dispersed carbides in ferrite.Also noted on an IT diagram is the presence of metastable phases such as martensite. On an IT diagram, the temperature range (M s - M f) of the metastable phase is presented.In contrast, the equilibrium diagram provides no information regarding kinetics,morphology or the the presence of metastable phases. Since the diagram is constructed on the requirement of equilibrium, only equilibrate phases will be present.24. FIND: Explain experimentally how you would determine an IT diagram for a particularsteel.SOLUTION: Isothermal transformation (IT) diagrams are constructed by following the isothermal changes in microstructure at different temperatures. Since it is difficult todecide exactly when a reaction starts and is completed the start and finish lines on an IT diagram are arbitrarily defined. The microstructural changes occurring can be followed by monitoring microstructural changes directly using metallographic techniques ortracking a property such as electrical resistivity which is affected by the amount of solute in solution or hardness, which is controlled by the nature of the microstructure. As many microstructure sensitive properties as possible are used to provide sufficient collaborative information for a desired degree of accuracy.25. FIND: Based upon the methods you outlined in question 19, explain the limitations ofusing such a diagram. In particular, explain why the application of an IT diagram isrestricted and may not be applied directly to the production environment.SOLUTION: Since an IT diagram is constructed from numerous samples transformed under isothermal conditions, they can only be used that way. In general, the data for ITdiagrams are collected using thin specimens in order to assure that the requirement ofinstantaneous quench are realized. When larger pieces are considered, especially when considering a production environment where a variety of cross-sections are cooled,instantaneous temperature changes can not be realized. Depending upon the mass of the part, slow cooling may be achieved at the center, whereas rapid cooling may occur near the surface. If the cooling rate is slow, then some microstructural changes will occur as the sample cools down to the desired temperature. Consequently, the continuous cooling (CC) curves were established. These curves recognize that the effects of time andtemperature must be integrated.26. FIND: In our discussions in Chapter 7, we introduce the concept of a phase, and in thischapter we have been concerned with the microstructure of an alloy and ultimately how a particular microstructure affects properties. Explain the difference between the concept of a phase and the phases that are present, and the microstructure of the material.SOLUTION: A phase is a homogeneous portion of matter. A particular microstructure such as 100% martensite can be a single phase microstructure. The microstructure iscomposed of martensite and only martensite. However, a microstructure can bemultiphase. For example, pearlite is a two-phase microstructure which consists oftwophases that form as alternate lamellae of ferrite and cementite. So the concept of amicrostructure includes information about the phase or phases present and theirdistribution.27. FIND: Using the concepts associated with phases and microstructure, explain what ismeant by the terms:a. austeniteb. ferritec. pearlited. bainitee. martensitef. cementite, andg. spheroiditeSOLUTION:(a) austenite - The face-centered cubic (FCC) form of iron.(b) ferrite - The body-centered cubic (BCC) form of iron.(c) pearlite - A two-phase microstructure of alternate ferrite and cementite lamellaeoccurring in some steels. Pearlite forms by the decomposition of austenite.(d) bainite - A two-phase microstructure of ferrite and cementite. Bainite forms whenan austenitic steel is quenched to a temperature below the pearlite region, butabove the martensite start, M s, temperature.(e) martensite - Metastable body-centered tetragonal (BCT) iron phase that issaturated in carbon. Iron carbon martensite forms when austenite is rapidlyquenched to low temperatures. (Note: A wide variety of martensite-liketransformations occur in other systems. Consequently, the term martensite hastaken on a more general concept.)(f) cementite - An iron carbide phase, Fe3C.(g) spheroidite - A two-phase microstructure consisting of spheroidized carbide inferrite, which is formed by heat-treating pearlite, bainite, or martensite at atemperature below the eutectoid temperature.It should be noted that in this list of terms there is a distinction between single-phasematerials - austenite, ferrite, martensite and cementite and microstructures which can be combinations of phases such as pearlite and spheroidite.Problems 28 - 31 can be solved using Figures 8.3-6 and 8.3-11.28. FIND: Thin specimens of a plain carbon steel having eutectoid composition are held at800o C and have been at that temperature long enough to have achieved a complete andhomogeneous austenitic structure. Describe the phases present and the microstructuresthat would occur using the quench paths given below.a. Instantaneous quench to 650o C, hold at that temperature for 200 seconds, andquench to room temperature.b. Instantaneous quench to 300o C, hold for 1000 seconds, and quench to roomtemperature.c. Instantaneous quench to room temperature.d. Instantaneous quench to 500o C, hold for 3 seconds and quench to roomtemperature.SOLUTION: When a eutectoid steel containing approximately 0.8 wt% carbon is taken to 800o C and held, whatever microstructure was present prior to the 800o C exposure isreplaced by single phase, austenite containing 0.8% C.(a) When austenite containing 0.8 wt% carbon is instantaneously quenched to 650o Cand held for 200 seconds it forms the two-phase pearlite microstructure. Since allthe austenite has transformed to pearlite at this temperature and time, no othermicrostructures can form on the quench to room temperature. Consequently, themicrostructure consists of alternate ferrite and cementite lamellae - termed pearlite.The microstructure contains 100% pearlite.(b) When austenite containing 0.8% C is quenched to 300o C and held for 1000seconds approximately 75% of the austenite has transformed to bainite. Theremaining austenite transforms to martensite when the material is quenched toroom temperature. Consequently, the microstructure contains 75% bainite and25% martensite.(c) When the 0.8% C alloy is quenched from 800o C to room temperatureinstantaneously, all the austenite transforms to martensite. Consequently, themicrostructure is 100% martensitic.(d) When the 0.8% C alloy is instantaneously quenched from 800o C to 500o C andheld for 3 seconds, approximately 50% of the austenite transforms to amicrostructure that consists of fine pearlite. The remaining austenite transformsto martensite when the material is quenched to room temperature. Consequently,the microstructure is a mixture of 50% fine pearlite and 50% martensite.29. FIND: Using the same initial conditions outlined in problem 23, describe the phasespresent and the microstructures that would occur using the quench paths given below.a. Instantaneous quench to 650o C, hold for 15 seconds, and quench to roomtemperature.b. Instantaneous quench to 500o C hold for 60 seconds and quench to roomtemperature.c. Instantaneous quench to 170o C hold for 100 seconds and quench to roomtemperature.d. Instantaneous quench to 170o C and hold at that temperature.SOLUTION:。
