Module 29. Precipitation from solid solution I. Lecture 29. Precipitation from solid solution I

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1 Module 29 Precipitation from solid solution I Lecture 29 Precipitation from solid solution I 1

2 Keywords : Properties of two phase alloys, super saturated solid solutions, historical perspective, solution treatment, homogeneous & heterogeneous nucleation, age hardening, thermodynamics & kinetics of ageing, types of precipitate, coherent & incoherent precipitate, sequence of precipitation Introduction One of the main objectives of alloying is to improve the strength of metal. There are alloys that have unlimited solubility. Slide 1 shows the effect of composition on the strength of such an alloy. This shows the phase diagram of such an alloy made of two metals & B. The strength of pure is and that of pure B is B. The strength of metal has a direct relation with its melting point. Note that the melting point of is higher than that of B. This is why > B. lloy addition results in strengthening. This is associated with the interaction between the stress field surrounding solute atoms and that of the dislocations. The flow stress of alloys is directly proportional to the concentration of solute atoms. Thus the addition of to B increases the strength of rich solid solution. Following the same logic the strength of B rich solid solution increases with the addition of to B. n alloy having an intermediate composition has the highest strength. Solid solution T L T B UTS Slide 1 B lloy addition increases strength of both. Intermediate composition has the maximum strength 2 However unlimited solubility is rare. Bulk of the binary alloy system has terminal solid solutions. Solid solution rich in is represented as and the one rich in B is known as. Since there are

3 more than two phases involved there must be at least one isothermal reaction involving three phases. Binary eutectic and peritectic systems are the most common examples of alloys having terminal solid solutions. In such a system apart from having two single phase domains at the two ends of the phase diagrams there is a region where the alloy is a mixture of two phases. The properties of alloys within this zone are given by the rule of mixture. This is illustrated in slides 2. Two phase alloys T L T B L UTS UTS B B Slide 2 %B B % B B Two phase alloys: strength is given by rule of mixture The limits of solubility of the terminal solid solutions in the two phase diagrams given in slide 2 are independent of temperatures below the eutectic or the peritectic reaction isotherms. Therefore if an alloy in this regime is heated to a temperature close to the eutectic (or peritectic) and subsequently cooled to room temperature there will not be any noticeable change in its structure. However if the solvus curve is like the one shown in slide 3 an interesting situation may arise. Consider an alloy X. t room temperature it has two phases, dispersed in a matrix of. Heat it to the region where it consists of 100% and then quench it to room temperature to suppress precipitation. Transformation in solid state is slow. Therefore the above treatment may give a super saturated solid solution. Can it remain in this state indefinitely? This is what we are going to learn at in this module. 3

4 What happens in a supersaturated terminal solid solution? T + L Eutectic Slide 3 X % B Historical perspective: geing of supersaturated solid solutions gives rise to precipitation. If the precipitate fulfills certain characteristics it gives to precipitation hardening. This was discovered accidentally during the first decade of the 20 th century. Wilm while studying solid solution strengthening of l Cu alloy by adding Mg in 1906 found that the strength of a quenched alloy increased with ageing. Duralumin was the first commercial alloy produced in It has 3 4.5Cu 0.4 1Mg 0 0.7Mn and impurities such as 0.4 1Fe Si. Merica was first to suggest in 1920 that the increase in strength is due to sub microscopic precipitates that form below the solvus temperature. This was conclusively proved in 1950s with the help of Electron microscope. Supersaturated solid solution: 4

5 Terminal solid solutions b T T E L L L TB T Soaking & quenching Slide 3 a Supersaturated solid solution Solutionizing heat treatment B time c Slide 3 shows a binary eutectic phase diagram. The alloys of interest are in the region beyond the eutectic reaction isotherm. These are marked by dotted lines. This illustrates the steps involved in the dissolution of precipitates and subsequent cooling to get a supersaturated solid solution. The process is known as solutionizing heat treatment. t room temperature the alloy consists of dispersed in a matrix of. This is represented by the microstructure marked as a in slide 3. The sketch at the extreme right of slide 3 shows the heating / cooling cycle in the form of a graph temperature (T) versus time (t). The alloy is heated to the region as shown in the slide 3. It is kept at this temperature for a sufficient length of time so that the precipitates dissolve. The process is known as soaking. Finer precipitates dissolve faster than the coarser precipitates. The figure marked b is the microstructure of the alloy at this stage. llow sufficient time so that it becomes homogeneous. It is then quenched to room temperature. Precipitation from solid solutions occurs through diffusion. Solute atoms need time to diffuse over a sizable distance. Fast cooling does not allow this time. Therefore excess solute remains within the matrix. The sketch marked c in slide 3 is the typical microstructure of the alloy at this stage. The super saturated solution is often represented as. Similar heat treatment can also be given to an alloy where is the matrix and is the precipitate. 5 Stability of super saturated solid solution:

6 Stability of supersaturated SS Phase diagram Solvus temp ging temp T L L B SS Equilibrium Slide 4 Free energy composition diagram G x X G G x B The most stable structure of the alloy X (see slide 4) consists of precipitates of in a matrix of as shown in slide 4. If it is heated above the solvus but below the solidus the precipitates dissolve. Keep it at this temperature for some time so that it becomes homogeneous. Quench it to room temperature. The cooling rate is fast. Therefore precipitation is suppressed. What we have now is a supersaturated solid solution of B in. This is represented as. The structure is unstable. Its stability at a temperature is given by the difference between the free energy of and that of the mixture of and. This is illustrated with the help of free energy composition diagram at the ageing temperature in slide 4. The free energy plots for & are much lower than that of the liquid. The common tangent meets the plots for & at x and x. They represent the composition of the two phases under the state of equilibrium. The degree of super saturation is given by (X x ). The free energy of the stable state is given by G. The vertical line at %B = X intersects the free energy plot of at G. This denotes the free energy of the super saturated solid solution. The difference is a measure of its thermodynamic instability. This is given by: (1) 6 This can be expressed as a function either the excess solute (X = X x ) or the degree of super cooling (T). This is the main driving force for the transformation of the super saturated solid solution ( ) into a mixture of &. This is further illustrated by a set of sketches in slide 5.

7 Driving force for precipitation T (a) G X X G + L T L % Solute T (b) ctivation hill Solution treatment geing time Hv Excess solute Super-cooling T Slide 5 (c) SS Equil. (d) time The sketch (a) in slide 5 gives the phase diagram. The filled circle within the region denotes the state of the alloy after homogenization. The sketch (b) gives the heat treatment cycle. It consists of two stages. During solution treatment the alloy is kept at a temperature above the solvus but below the liquidus so that the precipitates dissolve and the alloy becomes homogeneous (. This is subsequently quenched. The alloy now is in a supersaturated state. It is subsequently heated to the ageing temperature. The sketch (c) represents free energy as a function of state. Under super saturated state it has higher free energy than that in its equilibrium state. Yet the transformation is not spontaneous. This is because the precipitation is associated with the creation of a new interface. This needs additional thermal activation. This is schematically shown in the sketch (c) of slide 5. The kinetics of ageing is a function of both temperature and time. This is accompanied by a change in the structure and properties of the alloy. This is shown by a set of plots of hardness (Hv) versus time in sketch (d) of slide 5. Note the location of the peak and the time to reach the peak. t a higher ageing temperature the peak hardness is lower but it is attained quickly (faster). If aged beyond the peak the precipitates grow and hardness keeps decreasing. It gives rise to an over aged structure. The precipitates are now coarse enough to be seen under optical microscope. 7 Homogeneous nucleation: Nucleation is the first stage of precipitation. The formation of a new phase within a matrix is associated with the creation of a new surface. ll surfaces have stored energy. Besides this the nucleation of precipitate is also accompanied by a change in volume. The density of the

8 precipitate is different from that of the matrix. Therefore the nucleus would be surrounded by an elastic strain field. This too has a stored energy. The nucleation of precipitate would take place only if the net free energy change is negative. If the total change in free energy = ; = change in free energy per unit volume due to the presence of excess solute in; E S = elastic strain energy per unit volume of the precipitate; = surface energy per unit are of the precipitate; and r = the radius of the precipitate; the relation between these is given by the following expression: 4 (2) It has two parts one is proportional to the cube of the radius and the other is proportional to the square of the radius. The net free energy of transformation therefore is likely to have a maximum at a specific value of r. This can be estimated by equating the first differential to zero. This gives the critical size of the nucleus (r*) and the maximum value of the total energy is the activation hill (G*) which must be exceeded for the nucleus to be stable. These are given by the following expressions. (3) (4) The effects of the volume free energy and strain energy on the critical size of the nucleus and the activation hill are shown in fig 1. Note that G v is negative and E s is positive but the magnitude of G v is much larger than that of E s. G* E s = constant G* G v = constant G* G* G T G T (a) 0 r* r* r Fig 1 (b) 0 r* r* r 8 The degree of super cooling (or super saturation) controls the magnitude of G v which is the main driving force for the nucleation of precipitates. G v increases as the degree of super cooling increases. If the strain energy remains constant the size of the critical nucleus decreases with increasing degree of super cooling. The height of the activation hill also decreases. This is

9 shown in the form of a set of plots in fig 1(a). If G v remains constant the size of the critical nucleus increases as E s increases. The height of the activation too increases when E s increases. This is shown in the form of a set of plots in fig 1(b). The homogeneous nucleation will therefore be preferred if the height of the activation hill is lower and the size of the critical nucleus is low. From the expressions it is evident that low surface energy, higher degree of super saturation, and low strain energy would promote homogeneous nucleation. Usually the grains of metals are weaker that the grain boundaries. It is necessary to promote precipitation within the grains to make them stronger. Grain boundaries are always the favored sites for precipitation as a part of surface energy comes from the preexisting boundaries. It means that homogeneous nucleation must be promoted for precipitation hardening. Heterogeneous nucleation: The presence of interfaces such as grain boundaries in an alloy acts as potential sites for nucleation. This is often the more preferred mode of precipitation. This is because a part of the energy needed to create new interfaces so that a stable embryo of the precipitate may nucleate is provided by the existing interfaces. This is known as heterogeneous nucleation. (a) G* r (b) Heterogeneous nucleation 2 cos G V G E 2 V cos v S 3 2 r 3 23cos cos r sin r r r * * het hom * * het hom G G S S 2 ( GB ES) 3 0.5* 2 cos 3cos Slide 6 9 Slide 6 explains with the help of a set of sketches the derivation of the expressions for the critical radius and the activation barrier (or hill) for heterogeneous nucleation. The steps involved are exactly same as that of homogeneous nucleation. The net free energy change is

10 the sum of the energy associated with the volume of the precipitate and that associated with the creation of new interfaces. The sketch (a) in slide 6 shows how the interface gets modified when a lens shaped precipitate () forms at the grain boundary. The radius of curvature of the two faces of the precipitate is r. Note the directions of the forces due to the surface tensions (or the grain boundary energy) acting at the point of intersection of and boundaries. The respective surface energies are represented as and. The angle between the directions of and the initial grain boundary is. This is given by the respective surface energies &. Note the expression given in the slide. When a precipitate forms, a part of the original boundary is replaced by two boundaries. Let the respective areas be represented as and. The magnitudes of these in terms of r and are given in the slide 6. These are derived by assuming the surfaces to be a part of a sphere of radius r. The volume of a lens shaped precipitate can also be obtained by integration. The expression is given in the slide. On substitution of these the express for the change in energy is given by: 4 (5) S() is known as the shape factor (or function). It depends on the shape of the precipitate. The shape factor for a lens shaped precipitate is given in slide 6. Equate 0 to find the critical radius of curvature of the precipitate. Substitute the same in equation 5 to find the magnitude of the activation barrier which must be overcome so that the nucleus is stable. The expressions are given in slide 6. Note that the magnitude of the radius of curvature of the precipitate for heterogeneous nucleation is exactly same as the critical radius for homogeneous nucleation. However the activation barrier for heterogeneous nucleation is less than that of homogeneous nucleation. Note that when approaches zero S() = 0. This represents a situation where there is no activation barrier. In such a case the precipitate is likely to form as a thin film around the interface. The implication of heterogeneous nucleation can easily be under stood with the help of slide 7. 10

11 Heterogeneous nucleation (a) 1 60 G G * * het het S * * hom Vhom 2 cos V GC (b) 1 GB GE 0 cos 1 Slide 7 Grain boundary Grain edge Grain corner (c) (d) (e) Note that the angle can have values from 0 to 90. Therefore cos can have values between 0 and 1. S() represents the ratio of the volume of a lens shaped heterogeneous nucleus and that of a spherical homogeneous nucleus. When approaches 90, S() approaches 1. This corresponds to homogeneous nucleation. The sketch (a) in slide 7 shows a plot of / aa against. When = 0 the ratio is the lowest. It has a value of 0.5. It increases with. The sketch (b) shows the variation of the ratio of the activation barrier of heterogeneous nucleation to that of homogeneous nucleation as a function of cos. This shows that high angle boundaries are the most preferred sites for nucleation. This is because the activation barrier tends to zero. The sketch (b) shows a set of three plots for three different types of sites given in sketches (c), (d) & (e). The solution for precipitation at grain boundary has been worked out in this module. The same concept can be extended to precipitation at a grain edge or a grain corner. In a microstructure we do see points where 3 grains meet. These are the grain corners. It is a more preferred site than a grain boundary because the area of the high energy boundary that is replaced by relatively new low energy interfaces is much more. The same logic can be extended to grain corners. These are points where 4 grains meet in 3D. It is a more preferred site than the grain edges. 11 Types of precipitates The alloy and the precipitates are crystalline. When a precipitate forms in a matrix it may or may not have a definite orientation relationship. On the basis of the relation between the way crystal planes are arranged in the precipitate and the matrix they can be classified as coherent,

12 semi coherent and incoherent precipitates. Slide 8 describes the main features of the three types of precipitates. Coherent (a) Types of precipitate D/2 Partially coherent (b) a a a a D Incoherent small D indicates total miss - match. Such boundaries are called incoherent boundary a (c) a Slide 8 The sketch (a) in slide 8 shows the way the crystal planes are arranged in the precipitate () and the matrix (). Note that the inter planar spacing along a particular direction in the precipitate matches exactly with the inter planar spacing of the matrix along another direction. There is a specific crystallographic orientation relationship between the two. They form on specific crystallographic planes of the matrix called habit planes. There is no lattice strain because of perfect matching. Such precipitates are called coherent precipitates. The interface between the two has little surface energy. It may be taken as zero. 12 The sketch (b) in slide 8 illustrates a case where the lattice spacing of the precipitate (a ) is a little less than that of the matrix (a ). Note that the length of the precipitate is D. Within this it has one extra plane than that of the matrix. The extra plane looks exactly similar to that in an edge dislocation. It is in fact an interface dislocation. The lattice around it is a little strained. It extends over both the matrix and the precipitate. Here is an example of a precipitate where the lattice mismatch is adjusted with the introduction of an edge dislocation. It is known as semicoherent (or partially coherent) precipitate. The degree of mismatch is defined as the ratio of the difference in lattice spacing over the spacing of one of the lattice. This is represented as. The expression for is given in slide 8. Look at the sketch (b). Note that within a length of D there is one dislocation. precipitate would consist of several such blocks. The distance

13 between the interface dislocations should be equal to a /. If D approaches zero the nature of the boundary changes from semi coherent to incoherent boundary. The sketch (c) in slide 8 gives a schematic diagram of the way the atomic planes within the precipitate and the matrix are arranged at the interface of a normal precipitate. There is no apparent relation between the two. It occurs when the difference between the lattice spacing is large. The boundary is very much like a high angle boundary. It may be visualized as an irregular array of a large number of dislocations. Sequence of precipitation in an age hardenable alloy: The precipitates that form during the initial stages of the ageing treatment are coherent. Its composition is entirely different from that of the final precipitate (). It is known as a metastable precipitate. Its free energy is higher than that of the stable precipitate. The nucleation of such precipitates occurs because it has lower surface energy and a lower activation barrier. However depending on the extent of the lattice miss match such precipitates have a certain amount of strain energy. This imposes a restriction on the size of the precipitates. These are extremely fine in the initial stages. They cannot be detected by optical microscope. The presence of such precipitates was first detected by the change in the shapes of the X Ray diffraction spots by Guiner & Preston. Therefore the precipitates that form during the initial stages are known as GP zones. The sequence of precipitation from a super saturated solid solution may thus be represented as follows: (6) Note that during the process of ageing the composition of the alloy keeps changing from whereas that of the precipitate changes from. The intermediate precipitates ( & ) are coherent but the final and the most stable precipitate () is incoherent. During ageing hardness increases with the formation of coherent / semi coherent precipitates. These are extremely small. They cannot be seen under optical microscopes. Beyond the peak in the hardness versus time plot, the precipitates grow. This is accompanied by loss of strength. The over aged precipitates can be seen under optical microscopes. On the basis of the above it may be concluded that the following conditions must be fulfilled for an alloy to exhibit age hardening characteristics. 13 1) It should be possible to transform the alloy into a super saturated state having more than the equilibrium amount of solute and it should be homogeneous. 2) On ageing it should form coherent precipitates having low surface energy () & low coherency strain (E s ). This is necessary to promote that homogeneous precipitation so that there is a uniform distribution of precipitates within the grain.

14 Summary: In this module we looked at the stability of super saturated solid solutions. Terminal solid solutions where the solubility decreases with temperature can be converted to such a state by homogenization in the solid state and subsequent quenching. This is known as solutionizing heat treatment. On subsequent ageing it transforms to a more stable state by precipitation of excess solute in the form of precipitates. The strength (or the hardness) of the alloy increases if the precipitates are extremely fine and coherent. During the process the composition of precipitate and their characteristics keep changing. These are too fine to be seen under optical microscope. Nevertheless its effect is evident from the initial increase in hardness. It increases with increasing volume fraction of fine coherent precipitates. Once these are replaced by stable precipitates the hardness starts dropping. This is associated with the coarsening of precipitates. When it happens the alloy is said to be over aged. ny alloy that fulfills this condition exhibits age hardening behavior. Exercise: 1. What types of alloys would respond to precipitation hardening? 2. Why luminium alloy rivets are strored in refigerator? 3. For a lens shaped nucleus formed at interface estimate the angle if = 500 and = 600mJ/m 2 and hence find out the shape factor of the precipitate. 4. Critical radii for both homogeneous and heterogeneous nucleation are the same yet the latter is more likely to occur. Give reason. 5. batch of age hadrenable alloy has been overagred by mistake. Is there any way to slavage these? nswers: 1. lloys which at room temperture has at least two phases and on heating beyond a temperature attains a single phase structure (solid) can respond to precipitation hardening provided it satisfies the following conditions i) on queching precipitation could be suppressed ii) on heating or aging precipitates that nucleates are cohent with the matrix lumunium alloy rivets are first solution treated to get supersaturated solid solution so that they are soft and could be easily deformed durring rivetting. If these are not stored at subzero temperature after solution treatment they would age and become strong and relatively brittle. Therefore it would become unusable.

15 3. It is seen from the following figure 2 500, / lthough the critical radii for both homogeneous & heterogeneous nucleii are the same their volumes are significantly different. This is clear from the following sketch. Number of atoms required to form a stable nucleus is much less than that for homogeneous nucleus. r* r* Dotted circle represents homogeneous nucleus. Horizontal dotted line is grain boundary & the hatched region is a lens shaped heterogeneously formed nucleus. 5. Yes. The overaged alloy can be solution treated and quenched before using them as rivets. 15