0 50 1000

0.5

1

Time

Fra

ctio

n (R

e)cr

ysta

llize

d 0.999

0

1 exp tn

1000 t

New grain

New grains nucleateand grow into regions of high dislocation density.

High dislocation density

Experimental data usually fits a sigmoid curve

0 50 1000.01

0

0.01

0.02

0.03

Time

Rat

e of

(R

e)cr

ysta

lliza

tion

(fra

ctio

n/s)

0.022

1.038 1014

t1 exp t

n d

d

1000 t

Rate of recrystallization in the

early stages is limited by nucleation and grain boundary area

Rate of recrystallization in the middle stage is a maximum because of the presence of many nuclei, high grain boundary area and limited impingement

Rate of recrystallization in the late stage is limited by impingement and the lack of contact between grain boundaries and high dislocation density areas

Heat Treatment

It is often possible to engineer the microstructure and properties of materials through the use ofthermal treatments. The thermal treatment referred to as annealing is used to control the recoveryand grain growth of mechanically deformed materials. The thermal treatment referred to ashomogenization is used to distribute or 'smooth out' composition variations associated withsegregation during solidification. The thermal treatment referred to as precipitation hardening isused to cause formation of multiple phases to impede dislocation motion.

Annealinga) Recrystallization.Cold worked or glassy materials can be heat treated to control the degree of atomic disorder andthe grain size of the material. For a ductile material, the degree of disorder in the form of cold workof metals and the grain size has a large effect on the mechanical behavior of the material. Forexample, the flow stress of the material can be related to the grain size through the Hall-Petch

equation ik

d

1

2

where i is the flow stress for a single crystal of the specific material, d

is grain size and k is a material dependent property. When a disordered material is heated, theatoms/molecules are provided with enough thermal energy to rearrange to a lower energyconfiguration. This low energy configuration is often a regular crystalline arrangement. Thecrystalline regions nucleate at different points in the material and generally these differentcrystalline regions will have different orientations. When the regions grow together, a grainboundary is formed. This process is known as recrystallization for annealing of cold workedmaterials or crystallization for glassy materials. The nucleation and growth process for thecrystalline regions at the expense of the disordered material is a thermally activated process. Inmost cases, a plot of volume fraction recrystallized versus time has a sigmoidal ('S') shape. Thekinetics of thermally activated, nucleation and growth processes can be approximated using theJohnson-Mehl-Avrami (JMA) equation. The simplest form of the JMA equation for the volumefraction recrystallized, X(t), is

[1] X t( ) 1 exp T G tn

where is a function of temperature, T, and free energy change, G for the (re)crystallization

process, t is the time and n is the exponent for the process. An example of the form of timedependent (x-axis) recrystallization fraction (y-axis) predicted by the JMA equation is given with

.000007 n 3 t 0 100

Volume fraction transformed fits the curve

The activation energy for the process is implicitly contained in and may be writtenexplicitly as

[2] a expQ

R T

where Q is the activation energy, which can be a function of the free energy ofrecrystallization. The JMA equation is useful if the exponent, n, and the activation energyfor can be determined. This allows one to examine how the temperature of the heattreatment controls the rate of the process without actually having to do all theexperiments. It is also possible to use the values of Q and n to evaluate the mechanism ofthe crystallization process. To determine Q and n from equations [1] and [2], the data fromexperiments at three or four temperatures needs to be massaged. Note that theexperimental data is the lhs of equation [1] which we attempt to fit to the rhs of equation[1]. We can rearrange equation [1] by substituting equation [2], subtracting 1 from eachside, multiply each side by -1, take the natural log of each side, multiply by -1 on eachside, and then take the natural log of each side again. Now the data will have the form of aplane on a 3-D plot of the massaged data (z-axis) vs the natural log of time (x-axis) andthe inverse of temperature (1/T). This has the form of

[3] ln ln data t1

T

1

n ln t ln a( )Q

R T

Thus, n is the slope of the massaged data on the rhs of equation [3] versus ln(t) and theslope data versus 1/T is -Q/R. The intercept projected back to ln t =0 and 1/T=0 is ln(a).An example of such a plot with n=3, a=0.01, Q/R=20 K.

Ќ is assumed to described a thermally activated process (diffusion, nucleation)

Let X(t) specify the experimental data of volume fraction transformed as a function of time and Temperature

The activation energy for the process is implicitly contained in and may be writtenexplicitly as

[2] a expQ

R T

where Q is the activation energy, which can be a function of the free energy ofrecrystallization. The JMA equation is useful if the exponent, n, and the activation energyfor can be determined. This allows one to examine how the temperature of the heattreatment controls the rate of the process without actually having to do all theexperiments. It is also possible to use the values of Q and n to evaluate the mechanism ofthe crystallization process. To determine Q and n from equations [1] and [2], the data fromexperiments at three or four temperatures needs to be massaged. Note that theexperimental data is the lhs of equation [1] which we attempt to fit to the rhs of equation[1]. We can rearrange equation [1] by substituting equation [2], subtracting 1 from eachside, multiply each side by -1, take the natural log of each side, multiply by -1 on eachside, and then take the natural log of each side again. Now the data will have the form of aplane on a 3-D plot of the massaged data (z-axis) vs the natural log of time (x-axis) andthe inverse of temperature (1/T). This has the form of

[3] ln ln data t1

T

1

n ln t ln a( )Q

R T

Thus, n is the slope of the massaged data on the rhs of equation [3] versus ln(t) and theslope data versus 1/T is -Q/R. The intercept projected back to ln t =0 and 1/T=0 is ln(a).An example of such a plot with n=3, a=0.01, Q/R=20 K.

The activation energy for the process is implicitly contained in and may be writtenexplicitly as

[2] a expQ

R T

where Q is the activation energy, which can be a function of the free energy ofrecrystallization. The JMA equation is useful if the exponent, n, and the activation energyfor can be determined. This allows one to examine how the temperature of the heattreatment controls the rate of the process without actually having to do all theexperiments. It is also possible to use the values of Q and n to evaluate the mechanism ofthe crystallization process. To determine Q and n from equations [1] and [2], the data fromexperiments at three or four temperatures needs to be massaged. Note that theexperimental data is the lhs of equation [1] which we attempt to fit to the rhs of equation[1]. We can rearrange equation [1] by substituting equation [2], subtracting 1 from eachside, multiply each side by -1, take the natural log of each side, multiply by -1 on eachside, and then take the natural log of each side again. Now the data will have the form of aplane on a 3-D plot of the massaged data (z-axis) vs the natural log of time (x-axis) andthe inverse of temperature (1/T). This has the form of

[3] ln ln data t1

T

1

n ln t ln a( )Q

R T

Thus, n is the slope of the massaged data on the rhs of equation [3] versus ln(t) and theslope data versus 1/T is -Q/R. The intercept projected back to ln t =0 and 1/T=0 is ln(a).An example of such a plot with n=3, a=0.01, Q/R=20 K.

ln(t)

1/T

How can we get information about the physical process from n and Q?

JohnsonMehlAvrami EquationSo how can we get information about the process from the value of n and Q? As an example,consider spherical growth of grains with boundary velocity, G, nucleating at a constant rate perunit volume, N. If we pretend that growing grains do not impinge (run into) one another and thatthe total number of nuclei formed per unit time does not depend on the the fractionrecrystallized, then we can write an equation for this imaginary (virtual) volume fractionrecrystallized, Xv t( ) as

[4a] X v t( )

0

t

N4

3 G

3 t 3

d

0

ts

N4

3 G ts 3

d1

3ts

4N G

where marks the time at which a nucleation event occurs. To be more explicit about thedifficult concept of a nucleation time, , as opposed to a clock time, t, consider what the radiusof a grain would be after a 20 second (t) heat treatment if it nucleated 10 seconds () after the

heat treatment began. If the value of G is 106 m

s this radius would be given by

G t 106 m

s 20 s 10 s( )

105

m

The volume of this grain can then be determined from the radius. What we have to do to get thetotal volume fraction transformed is to add up the volumes of all the grains nucleated at all thevarious times, leading to the integral in eq. [4]. So how does the imaginary volume fraction helpus? There is a theory based on Poisson statistics which proposes that the ratio of the rate of

the real volume fraction recrystallization, tX t( )d

d, with the rate of the virtual volume fraction

transformed, tXv t( )d

d, has the specific form

[4b] tX t( )d

d

tXv t( )d

d

1 X t( ) or

X 0( )

X t( )

X1

1 X t( )

dXv 0( )

Xv t( )

Xv1

d

leading to the general solution X t( ) 1 exp Xv t( )

Noting the result in equation [4b] and substituting equation [4a] for the specific case ofspherical growth and taking the derivative leads to the differential equation

tX t( )d

d tXv t( )d

d exp Xv t( ) 1 X t( )( )

t0

t

N4

3 G ts 3

dd

d

tX t( )d

d1 X t( )( )

4

3t3 N G

The solution is

[5a] X t( ) 1 exp1

3N G

3 t4

which is equivalent to the general result [5b] X t( ) 1 exp Xv t( )

Comparing equations [1] and [2] with equation [5] allows us to relate the JMA equationparameters with physically relavant parameters. For example, if we have growth of sphericalgrains and a constant nucleation rate, the process exponent, n, would be equal to 4. Since Nand G are thermally activated, a simple form is

G b expQG

R T

and N c expQN

R T

then the activation energy, Q, in the JMA equation is Q QN n 1( ) QG

It can also be seen that for this case,

a expQ

R T

n 1( ) bn 1 c exp

QN n 1( ) QG

R T

It is also possible to consider time dependent behavior of parameters within the JMA equation.For example, QN, can be represented as a function of time if heterogeneous nucleation sites

have a range of activation energies, with the lowest activation energy sites being used up first.Another common case of a time dependent parameter is where the temperature changes duringthe recrystallization process. In these cases, is a function of time, t( ). The general

approach to deal with this is to integrate the virtual recrystallization rate over the nucleationtime, or

X t( ) 1 exp

0

t

Xv d

d

d

Define the virtual volume fraction transformed-No impingement. For spherical grains this is

G is the interface velocity (growth rate) of the grain boundary (m/s)N is the nucleation rate per unit volume (1/(m^3 s))t is the clock timeτ is the nucleation time

time

R

1:00 am 1:01 am 1:02 am 1:10 am

Nucleation time for the upper particle is 1:01 am and lower particle is 1:02 amClock time for both particles is 1:10 am

R G t

R=G(540 s)

R=G(480 s)

4

3 G t 3Volume of individual grain is

How can we get information about the physical process from n and Q?

How can we get information about the physical process from n and Q?

So how can we get information about the process from the value of n and Q? As an example,consider spherical growth of grains with boundary velocity, G, nucleating at a constant rate perunit volume, N. If we pretend that growing grains do not impinge (run into) one another and thatthe total number of nuclei formed per unit time does not depend on the the fractionrecrystallized, then we can write an equation for this imaginary (virtual) volume fractionrecrystallized, Xv t( ) as

[4a] Xv t( )

0

t

N4

3 G

3 t 3

d

0

ts

N4

3 G ts 3

d1

3ts

4N G

where marks the time at which a nucleation event occurs. To be more explicit about thedifficult concept of a nucleation time, , as opposed to a clock time, t, consider what the radiusof a grain would be after a 20 second (t) heat treatment if it nucleated 10 seconds () after the

heat treatment began. If the value of G is 106 m

s this radius would be given by

G t 106 m

s 20 s 10 s( )

105

m

The volume of this grain can then be determined from the radius. What we have to do to get thetotal volume fraction transformed is to add up the volumes of all the grains nucleated at all thevarious times, leading to the integral in eq. [4]. So how does the imaginary volume fraction helpus? There is a theory based on Poisson statistics which proposes that the ratio of the rate of

the real volume fraction recrystallization, tX t( )d

d, with the rate of the virtual volume fraction

transformed, tXv t( )d

d, has the specific form

[4b] tX t( )d

d

tX v t( )d

d

1 X t( ) or

X 0( )

X t( )

X1

1 X t( )

dXv 0( )

Xv t( )

Xv1

d

leading to the general solution X t( ) 1 exp Xv t( )

Noting the result in equation [4b] and substituting equation [4a] for the specific case ofspherical growth and taking the derivative leads to the differential equation

tX t( )d

d tXv t( )d

d exp Xv t( ) 1 X t( )( )

t0

t

N4

3 G ts 3

dd

d

tX t( )d

d1 X t( )( )

4

3t3 N G

The solution is

[5a] X t( ) 1 exp1

3N G

3 t4

which is equivalent to the general result [5b] X t( ) 1 exp Xv t( )

Comparing equations [1] and [2] with equation [5] allows us to relate the JMA equationparameters with physically relavant parameters. For example, if we have growth of sphericalgrains and a constant nucleation rate, the process exponent, n, would be equal to 4. Since Nand G are thermally activated, a simple form is

G b expQG

R T

and N c expQN

R T

then the activation energy, Q, in the JMA equation is Q QN n 1( ) QG

It can also be seen that for this case,

a expQ

R T

n 1( ) bn 1 c exp

QN n 1( ) QG

R T

It is also possible to consider time dependent behavior of parameters within the JMA equation.For example, QN, can be represented as a function of time if heterogeneous nucleation sites

have a range of activation energies, with the lowest activation energy sites being used up first.Another common case of a time dependent parameter is where the temperature changes duringthe recrystallization process. In these cases, is a function of time, t( ). The general

approach to deal with this is to integrate the virtual recrystallization rate over the nucleationtime, or

X t( ) 1 exp

0

t

Xv d

d

d

Heat Treatment

It is often possible to engineer the microstructure and properties of materials through the use ofthermal treatments. The thermal treatment referred to as annealing is used to control the recoveryand grain growth of mechanically deformed materials. The thermal treatment referred to ashomogenization is used to distribute or 'smooth out' composition variations associated withsegregation during solidification. The thermal treatment referred to as precipitation hardening isused to cause formation of multiple phases to impede dislocation motion.

Annealinga) Recrystallization.Cold worked or glassy materials can be heat treated to control the degree of atomic disorder andthe grain size of the material. For a ductile material, the degree of disorder in the form of cold workof metals and the grain size has a large effect on the mechanical behavior of the material. Forexample, the flow stress of the material can be related to the grain size through the Hall-Petch

equation ik

d

1

2

where i is the flow stress for a single crystal of the specific material, d

is grain size and k is a material dependent property. When a disordered material is heated, theatoms/molecules are provided with enough thermal energy to rearrange to a lower energyconfiguration. This low energy configuration is often a regular crystalline arrangement. Thecrystalline regions nucleate at different points in the material and generally these differentcrystalline regions will have different orientations. When the regions grow together, a grainboundary is formed. This process is known as recrystallization for annealing of cold workedmaterials or crystallization for glassy materials. The nucleation and growth process for thecrystalline regions at the expense of the disordered material is a thermally activated process. Inmost cases, a plot of volume fraction recrystallized versus time has a sigmoidal ('S') shape. Thekinetics of thermally activated, nucleation and growth processes can be approximated using theJohnson-Mehl-Avrami (JMA) equation. The simplest form of the JMA equation for the volumefraction recrystallized, X(t), is

[1] X t( ) 1 exp T G tn

where is a function of temperature, T, and free energy change, G for the (re)crystallization

process, t is the time and n is the exponent for the process. An example of the form of timedependent (x-axis) recrystallization fraction (y-axis) predicted by the JMA equation is given with

.000007 n 3 t 0 100

Johnson, Mehl, Avrami

So how can we get information about the process from the value of n and Q? As an example,consider spherical growth of grains with boundary velocity, G, nucleating at a constant rate perunit volume, N. If we pretend that growing grains do not impinge (run into) one another and thatthe total number of nuclei formed per unit time does not depend on the the fractionrecrystallized, then we can write an equation for this imaginary (virtual) volume fractionrecrystallized, Xv t( ) as

[4a] Xv t( )

0

t

N4

3 G

3 t 3

d

0

ts

N4

3 G ts 3

d1

3ts

4N G

where marks the time at which a nucleation event occurs. To be more explicit about thedifficult concept of a nucleation time, , as opposed to a clock time, t, consider what the radiusof a grain would be after a 20 second (t) heat treatment if it nucleated 10 seconds () after the

heat treatment began. If the value of G is 106 m

s this radius would be given by

G t 106 m

s 20 s 10 s( )

105

m

The volume of this grain can then be determined from the radius. What we have to do to get thetotal volume fraction transformed is to add up the volumes of all the grains nucleated at all thevarious times, leading to the integral in eq. [4]. So how does the imaginary volume fraction helpus? There is a theory based on Poisson statistics which proposes that the ratio of the rate of

the real volume fraction recrystallization, tX t( )d

d, with the rate of the virtual volume fraction

transformed, tXv t( )d

d, has the specific form

[4b] tX t( )d

d

tXv t( )d

d

1 X t( ) or

X 0( )

X t( )

X1

1 X t( )

dXv 0( )

Xv t( )

Xv1

d

leading to the general solution X t( ) 1 exp X v t( )

Noting the result in equation [4b] and substituting equation [4a] for the specific case ofspherical growth and taking the derivative leads to the differential equation

tX t( )d

d tXv t( )d

d exp Xv t( ) 1 X t( )( )

t0

t

N4

3 G ts 3

dd

d

tX t( )d

d1 X t( )( )

4

3t3 N G

The solution is

[5a] X t( ) 1 exp1

3N G

3 t4

which is equivalent to the general result [5b] X t( ) 1 exp Xv t( )

Comparing equations [1] and [2] with equation [5] allows us to relate the JMA equationparameters with physically relavant parameters. For example, if we have growth of sphericalgrains and a constant nucleation rate, the process exponent, n, would be equal to 4. Since Nand G are thermally activated, a simple form is

G b expQG

R T

and N c expQN

R T

then the activation energy, Q, in the JMA equation is Q QN n 1( ) QG

It can also be seen that for this case,

a expQ

R T

n 1( ) bn 1 c exp

QN n 1( ) QG

R T

It is also possible to consider time dependent behavior of parameters within the JMA equation.For example, QN, can be represented as a function of time if heterogeneous nucleation sites

have a range of activation energies, with the lowest activation energy sites being used up first.Another common case of a time dependent parameter is where the temperature changes duringthe recrystallization process. In these cases, is a function of time, t( ). The general

approach to deal with this is to integrate the virtual recrystallization rate over the nucleationtime, or

X t( ) 1 exp

0

t

Xv d

d

d

So how can we get information about the process from the value of n and Q? As an example,consider spherical growth of grains with boundary velocity, G, nucleating at a constant rate perunit volume, N. If we pretend that growing grains do not impinge (run into) one another and thatthe total number of nuclei formed per unit time does not depend on the the fractionrecrystallized, then we can write an equation for this imaginary (virtual) volume fractionrecrystallized, Xv t( ) as

[4a] X v t( )

0

t

N4

3 G

3 t 3

d

0

ts

N4

3 G ts 3

d1

3ts

4N G

where marks the time at which a nucleation event occurs. To be more explicit about thedifficult concept of a nucleation time, , as opposed to a clock time, t, consider what the radiusof a grain would be after a 20 second (t) heat treatment if it nucleated 10 seconds () after the

heat treatment began. If the value of G is 106 m

s this radius would be given by

G t 106 m

s 20 s 10 s( )

105

m

The volume of this grain can then be determined from the radius. What we have to do to get thetotal volume fraction transformed is to add up the volumes of all the grains nucleated at all thevarious times, leading to the integral in eq. [4]. So how does the imaginary volume fraction helpus? There is a theory based on Poisson statistics which proposes that the ratio of the rate of

the real volume fraction recrystallization, tX t( )d

d, with the rate of the virtual volume fraction

transformed, tXv t( )d

d, has the specific form

[4b] tX t( )d

d

tXv t( )d

d

1 X t( ) or

X 0( )

X t( )

X1

1 X t( )

dXv 0( )

Xv t( )

Xv1

d

leading to the general solution X t( ) 1 exp Xv t( )

Noting the result in equation [4b] and substituting equation [4a] for the specific case ofspherical growth and taking the derivative leads to the differential equation

tX t( )d

d tXv t( )d

d exp Xv t( ) 1 X t( )( )

t0

t

N4

3 G ts 3

dd

d

tX t( )d

d1 X t( )( )

4

3t3 N G

The solution is

[5a] X t( ) 1 exp1

3N G

3 t4

which is equivalent to the general result [5b] X t( ) 1 exp Xv t( )

Comparing equations [1] and [2] with equation [5] allows us to relate the JMA equationparameters with physically relavant parameters. For example, if we have growth of sphericalgrains and a constant nucleation rate, the process exponent, n, would be equal to 4. Since Nand G are thermally activated, a simple form is

G b expQG

R T

and N c expQN

R T

then the activation energy, Q, in the JMA equation is Q QN n 1( ) QG

It can also be seen that for this case,

a expQ

R T

n 1( ) bn 1 c exp

QN n 1( ) QG

R T

It is also possible to consider time dependent behavior of parameters within the JMA equation.For example, QN, can be represented as a function of time if heterogeneous nucleation sites

have a range of activation energies, with the lowest activation energy sites being used up first.Another common case of a time dependent parameter is where the temperature changes duringthe recrystallization process. In these cases, is a function of time, t( ). The general

approach to deal with this is to integrate the virtual recrystallization rate over the nucleationtime, or

X t( ) 1 exp

0

t

Xv d

d

d

So how can we get information about the process from the value of n and Q? As an example,consider spherical growth of grains with boundary velocity, G, nucleating at a constant rate perunit volume, N. If we pretend that growing grains do not impinge (run into) one another and thatthe total number of nuclei formed per unit time does not depend on the the fractionrecrystallized, then we can write an equation for this imaginary (virtual) volume fractionrecrystallized, Xv t( ) as

[4a] X v t( )

0

t

N4

3 G

3 t 3

d

0

ts

N4

3 G ts 3

d1

3ts

4N G

where marks the time at which a nucleation event occurs. To be more explicit about thedifficult concept of a nucleation time, , as opposed to a clock time, t, consider what the radiusof a grain would be after a 20 second (t) heat treatment if it nucleated 10 seconds () after the

heat treatment began. If the value of G is 106 m

s this radius would be given by

G t 106 m

s 20 s 10 s( )

105

m

The volume of this grain can then be determined from the radius. What we have to do to get thetotal volume fraction transformed is to add up the volumes of all the grains nucleated at all thevarious times, leading to the integral in eq. [4]. So how does the imaginary volume fraction helpus? There is a theory based on Poisson statistics which proposes that the ratio of the rate of

the real volume fraction recrystallization, tX t( )d

d, with the rate of the virtual volume fraction

transformed, tXv t( )d

d, has the specific form

[4b] tX t( )d

d

tXv t( )d

d

1 X t( ) or

X 0( )

X t( )

X1

1 X t( )

dXv 0( )

Xv t( )

Xv1

d

leading to the general solution X t( ) 1 exp Xv t( )

Noting the result in equation [4b] and substituting equation [4a] for the specific case ofspherical growth and taking the derivative leads to the differential equation

tX t( )d

d tXv t( )d

d exp Xv t( ) 1 X t( )( )

t0

t

N4

3 G ts 3

dd

d

tX t( )d

d1 X t( )( )

4

3t3 N G

The solution is

[5a] X t( ) 1 exp1

3N G

3 t4

which is equivalent to the general result [5b] X t( ) 1 exp Xv t( )

Comparing equations [1] and [2] with equation [5] allows us to relate the JMA equationparameters with physically relavant parameters. For example, if we have growth of sphericalgrains and a constant nucleation rate, the process exponent, n, would be equal to 4. Since Nand G are thermally activated, a simple form is

G b expQG

R T

and N c expQN

R T

then the activation energy, Q, in the JMA equation is Q QN n 1( ) QG

It can also be seen that for this case,

a expQ

R T

n 1( ) bn 1 c exp

QN n 1( ) QG

R T

It is also possible to consider time dependent behavior of parameters within the JMA equation.For example, QN, can be represented as a function of time if heterogeneous nucleation sites

have a range of activation energies, with the lowest activation energy sites being used up first.Another common case of a time dependent parameter is where the temperature changes duringthe recrystallization process. In these cases, is a function of time, t( ). The general

approach to deal with this is to integrate the virtual recrystallization rate over the nucleationtime, or

X t( ) 1 exp

0

t

Xv d

d

d

So how can we get information about the process from the value of n and Q? As an example,consider spherical growth of grains with boundary velocity, G, nucleating at a constant rate perunit volume, N. If we pretend that growing grains do not impinge (run into) one another and thatthe total number of nuclei formed per unit time does not depend on the the fractionrecrystallized, then we can write an equation for this imaginary (virtual) volume fractionrecrystallized, Xv t( ) as

[4a] Xv t( )

0

t

N4

3 G

3 t 3

d

0

ts

N4

3 G ts 3

d1

3ts

4N G

where marks the time at which a nucleation event occurs. To be more explicit about thedifficult concept of a nucleation time, , as opposed to a clock time, t, consider what the radiusof a grain would be after a 20 second (t) heat treatment if it nucleated 10 seconds () after the

heat treatment began. If the value of G is 106 m

s this radius would be given by

G t 106 m

s 20 s 10 s( )

105

m

The volume of this grain can then be determined from the radius. What we have to do to get thetotal volume fraction transformed is to add up the volumes of all the grains nucleated at all thevarious times, leading to the integral in eq. [4]. So how does the imaginary volume fraction helpus? There is a theory based on Poisson statistics which proposes that the ratio of the rate of

the real volume fraction recrystallization, tX t( )d

d, with the rate of the virtual volume fraction

transformed, tXv t( )d

d, has the specific form

[4b] tX t( )d

d

tXv t( )d

d

1 X t( ) or

X 0( )

X t( )

X1

1 X t( )

dXv 0( )

Xv t( )

Xv1

d

leading to the general solution X t( ) 1 exp Xv t( )

Noting the result in equation [4b] and substituting equation [4a] for the specific case ofspherical growth and taking the derivative leads to the differential equation

tX t( )d

d tXv t( )d

d exp Xv t( ) 1 X t( )( )

t0

t

N4

3 G ts 3

dd

d

tX t( )d

d1 X t( )( )

4

3t3 N G

The solution is

[5a] X t( ) 1 exp1

3N G

3 t( )4

which is equivalent to the general result [5b] X t( ) 1 exp Xv t( )

Comparing equations [1] and [2] with equation [5] allows us to relate the JMA equationparameters with physically relavant parameters. For example, if we have growth of sphericalgrains and a constant nucleation rate, the process exponent, n, would be equal to 4. Since Nand G are thermally activated, a simple form is

G b expQG

R T

and N c expQN

R T

then the activation energy, Q, in the JMA equation is Q QN n 1( ) QG

It can also be seen that for this case,

a expQ

R T

n 1( ) bn 1 c exp

QN n 1( ) QG

R T

It is also possible to consider time dependent behavior of parameters within the JMA equation.For example, QN, can be represented as a function of time if heterogeneous nucleation sites

have a range of activation energies, with the lowest activation energy sites being used up first.Another common case of a time dependent parameter is where the temperature changes duringthe recrystallization process. In these cases, is a function of time, t( ). The general

approach to deal with this is to integrate the virtual recrystallization rate over the nucleationtime, or

X t( ) 1 exp

0

t

Xv d

d

d

How can we get information about the physical process from n and Q?

Heat Treatment

It is often possible to engineer the microstructure and properties of materials through the use ofthermal treatments. The thermal treatment referred to as annealing is used to control the recoveryand grain growth of mechanically deformed materials. The thermal treatment referred to ashomogenization is used to distribute or 'smooth out' composition variations associated withsegregation during solidification. The thermal treatment referred to as precipitation hardening isused to cause formation of multiple phases to impede dislocation motion.

Annealinga) Recrystallization.Cold worked or glassy materials can be heat treated to control the degree of atomic disorder andthe grain size of the material. For a ductile material, the degree of disorder in the form of cold workof metals and the grain size has a large effect on the mechanical behavior of the material. Forexample, the flow stress of the material can be related to the grain size through the Hall-Petch

equation ik

d

1

2

where i is the flow stress for a single crystal of the specific material, d

is grain size and k is a material dependent property. When a disordered material is heated, theatoms/molecules are provided with enough thermal energy to rearrange to a lower energyconfiguration. This low energy configuration is often a regular crystalline arrangement. Thecrystalline regions nucleate at different points in the material and generally these differentcrystalline regions will have different orientations. When the regions grow together, a grainboundary is formed. This process is known as recrystallization for annealing of cold workedmaterials or crystallization for glassy materials. The nucleation and growth process for thecrystalline regions at the expense of the disordered material is a thermally activated process. Inmost cases, a plot of volume fraction recrystallized versus time has a sigmoidal ('S') shape. Thekinetics of thermally activated, nucleation and growth processes can be approximated using theJohnson-Mehl-Avrami (JMA) equation. The simplest form of the JMA equation for the volumefraction recrystallized, X(t), is

[1] X t( ) 1 exp T G tn

where is a function of temperature, T, and free energy change, G for the (re)crystallization

process, t is the time and n is the exponent for the process. An example of the form of timedependent (x-axis) recrystallization fraction (y-axis) predicted by the JMA equation is given with

.000007 n 3 t 0 100

The activation energy for the process is implicitly contained in and may be writtenexplicitly as

[2] a expQ

R T

where Q is the activation energy, which can be a function of the free energy ofrecrystallization. The JMA equation is useful if the exponent, n, and the activation energyfor can be determined. This allows one to examine how the temperature of the heattreatment controls the rate of the process without actually having to do all theexperiments. It is also possible to use the values of Q and n to evaluate the mechanism ofthe crystallization process. To determine Q and n from equations [1] and [2], the data fromexperiments at three or four temperatures needs to be massaged. Note that theexperimental data is the lhs of equation [1] which we attempt to fit to the rhs of equation[1]. We can rearrange equation [1] by substituting equation [2], subtracting 1 from eachside, multiply each side by -1, take the natural log of each side, multiply by -1 on eachside, and then take the natural log of each side again. Now the data will have the form of aplane on a 3-D plot of the massaged data (z-axis) vs the natural log of time (x-axis) andthe inverse of temperature (1/T). This has the form of

[3] ln ln data t1

T

1

n ln t ln a( )Q

R T

Thus, n is the slope of the massaged data on the rhs of equation [3] versus ln(t) and theslope data versus 1/T is -Q/R. The intercept projected back to ln t =0 and 1/T=0 is ln(a).An example of such a plot with n=3, a=0.01, Q/R=20 K.

So how can we get information about the process from the value of n and Q? As an example,consider spherical growth of grains with boundary velocity, G, nucleating at a constant rate perunit volume, N. If we pretend that growing grains do not impinge (run into) one another and thatthe total number of nuclei formed per unit time does not depend on the the fractionrecrystallized, then we can write an equation for this imaginary (virtual) volume fractionrecrystallized, Xv t( ) as

[4a] Xv t( )

0

t

N4

3 G

3 t 3

d

0

ts

N4

3 G ts 3

d1

3ts

4N G

where marks the time at which a nucleation event occurs. To be more explicit about thedifficult concept of a nucleation time, , as opposed to a clock time, t, consider what the radiusof a grain would be after a 20 second (t) heat treatment if it nucleated 10 seconds () after the

heat treatment began. If the value of G is 106 m

s this radius would be given by

G t 106 m

s 20 s 10 s( )

105

m

The volume of this grain can then be determined from the radius. What we have to do to get thetotal volume fraction transformed is to add up the volumes of all the grains nucleated at all thevarious times, leading to the integral in eq. [4]. So how does the imaginary volume fraction helpus? There is a theory based on Poisson statistics which proposes that the ratio of the rate of

the real volume fraction recrystallization, tX t( )d

d, with the rate of the virtual volume fraction

transformed, tXv t( )d

d, has the specific form

[4b] tX t( )d

d

tXv t( )d

d

1 X t( ) or

X 0( )

X t( )

X1

1 X t( )

dXv 0( )

Xv t( )

Xv1

d

leading to the general solution X t( ) 1 exp Xv t( )

Noting the result in equation [4b] and substituting equation [4a] for the specific case ofspherical growth and taking the derivative leads to the differential equation

tX t( )d

d tXv t( )d

d exp Xv t( ) 1 X t( )( )

t0

t

N4

3 G ts 3

dd

d

tX t( )d

d1 X t( )( )

4

3t3 N G

The solution is

[5a] X t( ) 1 exp1

3N G

3 t( )4

which is equivalent to the general result [5b] X t( ) 1 exp Xv t( )

Comparing equations [1] and [2] with equation [5] allows us to relate the JMA equationparameters with physically relavant parameters. For example, if we have growth of sphericalgrains and a constant nucleation rate, the process exponent, n, would be equal to 4. Since Nand G are thermally activated, a simple form is

G b expQG

R T

and N c expQN

R T

then the activation energy, Q, in the JMA equation is Q QN n 1( ) QG

It can also be seen that for this case,

a expQ

R T

n 1( ) bn 1 c exp

QN n 1( ) QG

R T

It is also possible to consider time dependent behavior of parameters within the JMA equation.For example, QN, can be represented as a function of time if heterogeneous nucleation sites

have a range of activation energies, with the lowest activation energy sites being used up first.Another common case of a time dependent parameter is where the temperature changes duringthe recrystallization process. In these cases, is a function of time, t( ). The general

approach to deal with this is to integrate the virtual recrystallization rate over the nucleationtime, or

X t( ) 1 exp

0

t

Xv d

d

d

Q is the activation energy for N and G

n indicates the geometry of the particle--dimension = n-1

So how can we get information about the process from the value of n and Q? As an example,consider spherical growth of grains with boundary velocity, G, nucleating at a constant rate perunit volume, N. If we pretend that growing grains do not impinge (run into) one another and thatthe total number of nuclei formed per unit time does not depend on the the fractionrecrystallized, then we can write an equation for this imaginary (virtual) volume fractionrecrystallized, Xv t( ) as

[4a] Xv t( )

0

t

N4

3 G

3 t 3

d

0

ts

N4

3 G ts 3

d1

3ts

4N G

where marks the time at which a nucleation event occurs. To be more explicit about thedifficult concept of a nucleation time, , as opposed to a clock time, t, consider what the radiusof a grain would be after a 20 second (t) heat treatment if it nucleated 10 seconds () after the

heat treatment began. If the value of G is 106 m

s this radius would be given by

G t 106 m

s 20 s 10 s( )

105

m

The volume of this grain can then be determined from the radius. What we have to do to get thetotal volume fraction transformed is to add up the volumes of all the grains nucleated at all thevarious times, leading to the integral in eq. [4]. So how does the imaginary volume fraction helpus? There is a theory based on Poisson statistics which proposes that the ratio of the rate of

the real volume fraction recrystallization, tX t( )d

d, with the rate of the virtual volume fraction

transformed, tXv t( )d

d, has the specific form

[4b] tX t( )d

d

tXv t( )d

d

1 X t( ) or

X 0( )

X t( )

X1

1 X t( )

dXv 0( )

Xv t( )

Xv1

d

leading to the general solution X t( ) 1 exp Xv t( )

Noting the result in equation [4b] and substituting equation [4a] for the specific case ofspherical growth and taking the derivative leads to the differential equation

tX t( )d

d tXv t( )d

d exp Xv t( ) 1 X t( )( )

t0

t

N4

3 G ts 3

dd

d

tX t( )d

d1 X t( )( )

4

3t3 N G

The solution is

[5a] X t( ) 1 exp1

3N G

3 t( )4

which is equivalent to the general result [5b] X t( ) 1 exp Xv t( )

Comparing equations [1] and [2] with equation [5] allows us to relate the JMA equationparameters with physically relavant parameters. For example, if we have growth of sphericalgrains and a constant nucleation rate, the process exponent, n, would be equal to 4. Since Nand G are thermally activated, a simple form is

G b expQG

R T

and N c expQN

R T

then the activation energy, Q, in the JMA equation is Q QN n 1( ) QG

It can also be seen that for this case,

a expQ

R T

n 1( ) bn 1 c exp

QN n 1( ) QG

R T

It is also possible to consider time dependent behavior of parameters within the JMA equation.For example, QN, can be represented as a function of time if heterogeneous nucleation sites

have a range of activation energies, with the lowest activation energy sites being used up first.Another common case of a time dependent parameter is where the temperature changes duringthe recrystallization process. In these cases, is a function of time, t( ). The general

approach to deal with this is to integrate the virtual recrystallization rate over the nucleationtime, or

X t( ) 1 exp

0

t

Xv d

d

d

So how can we get information about the process from the value of n and Q? As an example,consider spherical growth of grains with boundary velocity, G, nucleating at a constant rate perunit volume, N. If we pretend that growing grains do not impinge (run into) one another and thatthe total number of nuclei formed per unit time does not depend on the the fractionrecrystallized, then we can write an equation for this imaginary (virtual) volume fractionrecrystallized, Xv t( ) as

[4a] Xv t( )

0

t

N4

3 G

3 t 3

d

0

ts

N4

3 G ts 3

d1

3ts

4N G

where marks the time at which a nucleation event occurs. To be more explicit about thedifficult concept of a nucleation time, , as opposed to a clock time, t, consider what the radiusof a grain would be after a 20 second (t) heat treatment if it nucleated 10 seconds () after the

heat treatment began. If the value of G is 106 m

s this radius would be given by

G t 106 m

s 20 s 10 s( )

105

m

The volume of this grain can then be determined from the radius. What we have to do to get thetotal volume fraction transformed is to add up the volumes of all the grains nucleated at all thevarious times, leading to the integral in eq. [4]. So how does the imaginary volume fraction helpus? There is a theory based on Poisson statistics which proposes that the ratio of the rate of

the real volume fraction recrystallization, tX t( )d

d, with the rate of the virtual volume fraction

transformed, tXv t( )d

d, has the specific form

[4b] tX t( )d

d

tXv t( )d

d

1 X t( ) or

X 0( )

X t( )

X1

1 X t( )

dXv 0( )

Xv t( )

Xv1

d

leading to the general solution X t( ) 1 exp Xv t( )

Noting the result in equation [4b] and substituting equation [4a] for the specific case ofspherical growth and taking the derivative leads to the differential equation

tX t( )d

d tXv t( )d

d exp Xv t( ) 1 X t( )( )

t0

t

N4

3 G ts 3

dd

d

tX t( )d

d1 X t( )( )

4

3t3 N G

The solution is

[5a] X t( ) 1 exp1

3N G

3 t( )4

which is equivalent to the general result [5b] X t( ) 1 exp Xv t( )

Comparing equations [1] and [2] with equation [5] allows us to relate the JMA equationparameters with physically relavant parameters. For example, if we have growth of sphericalgrains and a constant nucleation rate, the process exponent, n, would be equal to 4. Since Nand G are thermally activated, a simple form is

G b expQ G

R T

and N c expQN

R T

then the activation energy, Q, in the JMA equation is Q QN n 1( ) QG

It can also be seen that for this case,

a expQ

R T

n 1( ) bn 1 c exp

QN n 1( ) QG

R T

It is also possible to consider time dependent behavior of parameters within the JMA equation.For example, QN, can be represented as a function of time if heterogeneous nucleation sites

have a range of activation energies, with the lowest activation energy sites being used up first.Another common case of a time dependent parameter is where the temperature changes duringthe recrystallization process. In these cases, is a function of time, t( ). The general

approach to deal with this is to integrate the virtual recrystallization rate over the nucleationtime, or

X t( ) 1 exp

0

t

Xv d

d

d

So how can we get information about the process from the value of n and Q? As an example,consider spherical growth of grains with boundary velocity, G, nucleating at a constant rate perunit volume, N. If we pretend that growing grains do not impinge (run into) one another and thatthe total number of nuclei formed per unit time does not depend on the the fractionrecrystallized, then we can write an equation for this imaginary (virtual) volume fractionrecrystallized, Xv t( ) as

[4a] Xv t( )

0

t

N4

3 G

3 t 3

d

0

ts

N4

3 G ts 3

d1

3ts

4N G

where marks the time at which a nucleation event occurs. To be more explicit about thedifficult concept of a nucleation time, , as opposed to a clock time, t, consider what the radiusof a grain would be after a 20 second (t) heat treatment if it nucleated 10 seconds () after the

heat treatment began. If the value of G is 106 m

s this radius would be given by

G t 106 m

s 20 s 10 s( )

105

m

The volume of this grain can then be determined from the radius. What we have to do to get thetotal volume fraction transformed is to add up the volumes of all the grains nucleated at all thevarious times, leading to the integral in eq. [4]. So how does the imaginary volume fraction helpus? There is a theory based on Poisson statistics which proposes that the ratio of the rate of

the real volume fraction recrystallization, tX t( )d

d, with the rate of the virtual volume fraction

transformed, tXv t( )d

d, has the specific form

[4b] tX t( )d

d

tXv t( )d

d

1 X t( ) or

X 0( )

X t( )

X1

1 X t( )

dXv 0( )

Xv t( )

Xv1

d

leading to the general solution X t( ) 1 exp Xv t( )

Noting the result in equation [4b] and substituting equation [4a] for the specific case ofspherical growth and taking the derivative leads to the differential equation

tX t( )d

d tXv t( )d

d exp Xv t( ) 1 X t( )( )

t0

t

N4

3 G ts 3

dd

d

tX t( )d

d1 X t( )( )

4

3t3 N G

The solution is

[5a] X t( ) 1 exp1

3N G

3 t( )4

which is equivalent to the general result [5b] X t( ) 1 exp Xv t( )

Comparing equations [1] and [2] with equation [5] allows us to relate the JMA equationparameters with physically relavant parameters. For example, if we have growth of sphericalgrains and a constant nucleation rate, the process exponent, n, would be equal to 4. Since Nand G are thermally activated, a simple form is

G b expQG

R T

and N c expQ N

R T

then the activation energy, Q, in the JMA equation is Q QN n 1( ) QG

It can also be seen that for this case,

a expQ

R T

n 1( ) bn 1 c exp

QN n 1( ) QG

R T

It is also possible to consider time dependent behavior of parameters within the JMA equation.For example, QN, can be represented as a function of time if heterogeneous nucleation sites

have a range of activation energies, with the lowest activation energy sites being used up first.Another common case of a time dependent parameter is where the temperature changes duringthe recrystallization process. In these cases, is a function of time, t( ). The general

approach to deal with this is to integrate the virtual recrystallization rate over the nucleationtime, or

X t( ) 1 exp

0

t

Xv d

d

d

Q Q N 3( ) Q G

Growth rate controlled by diffusionActivation energy

Nucleation rate/volume controlled by nucleation activation energy

X t( ) 1 exp

0

t

N4

3 G

3 t 3

d

Sphere n=4

Plate n=3

Rod n=2

0

t

N4

3 G

3 t 3

d

0

t

N h 2 G2 t 2

d

0

t

N 2 r2 G t

d

So how can we get information about the process from the value of n and Q? As an example,consider spherical growth of grains with boundary velocity, G, nucleating at a constant rate perunit volume, N. If we pretend that growing grains do not impinge (run into) one another and thatthe total number of nuclei formed per unit time does not depend on the the fractionrecrystallized, then we can write an equation for this imaginary (virtual) volume fractionrecrystallized, Xv t( ) as

[4a] Xv t( )

0

t

N4

3 G

3 t 3

d

0

ts

N4

3 G ts 3

d1

3ts

4N G

where marks the time at which a nucleation event occurs. To be more explicit about thedifficult concept of a nucleation time, , as opposed to a clock time, t, consider what the radiusof a grain would be after a 20 second (t) heat treatment if it nucleated 10 seconds () after the

heat treatment began. If the value of G is 106 m

s this radius would be given by

G t 106 m

s 20 s 10 s( )

105

m

The volume of this grain can then be determined from the radius. What we have to do to get thetotal volume fraction transformed is to add up the volumes of all the grains nucleated at all thevarious times, leading to the integral in eq. [4]. So how does the imaginary volume fraction helpus? There is a theory based on Poisson statistics which proposes that the ratio of the rate of

the real volume fraction recrystallization, tX t( )d

d, with the rate of the virtual volume fraction

transformed, tXv t( )d

d, has the specific form

[4b] tX t( )d

d

tXv t( )d

d

1 X t( ) or

X 0( )

X t( )

X1

1 X t( )

dXv 0( )

Xv t( )

Xv1

d

leading to the general solution X t( ) 1 exp Xv t( )

Noting the result in equation [4b] and substituting equation [4a] for the specific case ofspherical growth and taking the derivative leads to the differential equation

tX t( )d

d tXv t( )d

d exp Xv t( ) 1 X t( )( )

t0

t

N4

3 G ts 3

dd

d

tX t( )d

d1 X t( )( )

4

3t3 N G

The solution is

[5a] X t( ) 1 exp1

3N G

3 t( )4

which is equivalent to the general result [5b] X t( ) 1 exp Xv t( )

Comparing equations [1] and [2] with equation [5] allows us to relate the JMA equationparameters with physically relavant parameters. For example, if we have growth of sphericalgrains and a constant nucleation rate, the process exponent, n, would be equal to 4. Since Nand G are thermally activated, a simple form is

G b expQG

R T

and N c expQN

R T

then the activation energy, Q, in the JMA equation is Q QN n 1( ) QG

It can also be seen that for this case,

a expQ

R T

n 1( ) bn 1 c exp

QN n 1( ) QG

R T

It is also possible to consider time dependent behavior of parameters within the JMA equation.For example, QN, can be represented as a function of time if heterogeneous nucleation sites

have a range of activation energies, with the lowest activation energy sites being used up first.Another common case of a time dependent parameter is where the temperature changes duringthe recrystallization process. In these cases, is a function of time, t( ). The general

approach to deal with this is to integrate the virtual recrystallization rate over the nucleationtime, or

X t( ) 1 exp

0

t

Xv d

d

d

X t( ) 1 exp

0

t

N4

3 G

3 t 3

d

r

h

n corresponds to shape

So how does this data on n and Q apply to a ‘real’ problem?

X t( ) 1 exp

0

t

a expQ

RT

n

d

d

d

T variation with time (position)

Hot drawing of wire

Modify the JMA equation so that we add up the amount of recrystallization occurring at each place (at each temperature)on the

wire.