Diffmionless transformations
440
6
.
6
are two different but equivalent ways of producing the lattice invariant shear. Show exactly what is meant by this. What is the experimental proof of both types of shear? Draw a diagram to illustrate Bain's homogeneous deformation model for the fee -> bec diffusionless transformation. Assuming a7 = 3 56 A and flu = 2.86 A, and that cia for martensite is 1.15 calculate the maximum movement experienced by atoms during the transformation.
i
Assume that cia = 1.1.
i
Solutions to Exercises
Compiled by John C
.
6 7 .
.
Ion
What are the essential differences in martensite nucleation models based
(a) on changes at the core of a dislocation; (b) on dislocation strain field interaction? Discuss the advantages and disadvantages of both models in terms of the known characteristics of martensitic transformations.
8
6
.
6
9
.
Give possible reasons why the habit plane of martensite changes as a function of alloying content in steels and Fe-Ni alloys. What factors influence the retention of austenite in these alloys? What is the role of austenitic grain size in martensitic transformations? Is austenitic grain size important to the strength of martensite? What other factors are important to strength and roughness in technological
Chapter 1 1
1
.
Cp = 22 64 + 6.28 x HrTJ moP'K .
fT-C
Entropy increase AS =
- dT
,
hardened steels? 6
10 Suggest possible alloying and heat treatment procedures needed to design the following steels: (a) a quenched and tempered steel; (b) a dual phase steel; (c) a maraging steel; id) a TRIP steel. 11 How would you characterize the unique properties of alloys which can be utilized as 'memory metals'. How would you design a TiNi alloy for use as. e.g., a self-locking rivet? Give instructions on how it is to be
.
6
AC
f358 22.64 +:6.28 x lO T dT
_ 300-13SS -
MOO *
.
used.
/
H,L22 641n r+ 6 28 x HrT] .
.
40.83 J mol-'K'1 12 .
iquid Fe 1600 5-Fe y-Fe
9
i !
800 500a-Fe
o
l
£-Fe
-
300-
0
50
100
Pressure kbar ,
150
Solutions to exercises
Solutions to exercises
442
443
i
1
.
Liquid Fe
4
-Fe
a
y-Fe
Phases stable at low temperatures must have low enthalpies because the (-TS) term in the expression for G becomes negligible Phases .
stable at high temperatures, on the other hand have higher entropies to compensate for higher enthalpies. ,
y-Fe
15 .
J
At 800oC
At 1600oC
t 24
-
-
22
LL
2
3
5
5
Pressure kbar
a-Fe
At 5000C At 300oC
y-Fe
a-Fe
s-Fe E-Fe
4
110
120
Six distinguishable configurations. Schematic free energy-pressure curves for pure Fe.
Theoretical number of distinguishable ways of arranging two black balls and two white balls in a square is 1
.
3
From Equation 1.14 /dP\
(/VB + yVw)!
AH cq
6
2! 2!
TAV
Assuming AH and AV are independent of T and P f
the range of
1
interest, the equation may be rewritten as AP\
(2 + 2)!
.
6
Dividing both sides of Equation 1.30 by the number of moles of solution (nA + nB) gives
AH
dC7
TAV
{nA + nB)
where: AH = HL - Hss = 13050 J mol
A
Ha
("a + nB)
+ Mb
d«B
{nA + nB)
1
-
;
AK = VL - Ks = (8.0 - 7.6) x 10-6rm
!
,
:
.
The left-hand side of this equation is the free energy change per mole of solution and ean therefore be written dC.
i
T
= (1085 + 273) K.
Thus if AP is 10 kbar, i.e. 109 Nm-2, the change in the equilibrium melting temperature is given by the above equation as AT = 42 K
dn A
(«A + «b)
and
dn B
(«A + «B)
are the changes in the mole fractions of A and B, dXx and dXB. The above equation can therefore be written as
. i
1
444
'
Solutions to exercises
Solutions to exercises dG = M a +
b
17
b
.
Equation 1.31: G =
AXA + Ub b
Equation 1.39: G - AAGA + A'bGb + QXAXB + /?r(A'Aln,VA
dG
Thus - = nA
'
-
dXB
= -nA
f A pln
= AaGa
(1.6.1)
+ \iB
-
But using Equation 1.31
gives
-
e)
f XBGB + Q(X\XB + AiYA)
RT(XA\nXA -f A'BInA'B)
= XA[GA
G = \XaXa + I b b
x
445
+ fIA§ + /?rin,YA] +
b[Gb
'
r QA
-
x -r RT\nXB]
Comparison with Equation 1.31 and using AA + XB - 1 gives
=
XB
[iA = GA + Q(l - AA): + rin.Y B = GB + Q(.l - AB)2 + RT\nXH
x
.
G - Ha a
dG
giving
B
18 .
dG
or uA = G - xYB
(a) Atomic weight of Au = 197 Atomic weight of Ag = 108
dA- B
1
No. of moles ot Au = - = 0 076 .
From the figure nA = PR - X*
197
= PQ - OS
25
No. of moles of Au = - = 0 23! .
i
e
.
point S, the extrapolation of the tangent to point R on the
.
G-curve
108
represents the quantity |iA.
no. of moles of solution = 0 307 .
Equation 1.6.1 gives 0 07b .
(b) Mole fraction of Au
dG Ha
Hb
+
= 0 248 .
0 307 .
dXB 0 231 .
Mole fraction of Ag
Uv i e
.
.
= OS +- -
(c) Molar entropy of mixing. ASmix = -/ (A'AInAA -f ABlnArB} '
Thus jiB = OS + UV = TV e
.
.
.
.
But ITS = OT = 1 i
= 0 752 0 307
A5mix - -8.314(0.248-in 0.248 + 0.752 . In 0.752) = 4 66 J K
point V represents the quantity b-
.
mol "1
(d) Total entropy of mixing = Molar entropy of mixing x no
.
of moles of solution
= 4 66 x 0.307 .
= 1
43 J K"'
-
V CD
(e) Molar free energy change at 500oC = AGmix
R
0)
0)
-
LL
AG mix
s
-
rA5mix = -773 x 4.66 = -3.60 kJ mol"1
u
s
(0 I p
T
O A
X
B
au = GAu +
rin Au
= 0 + (8 314-773-In 0.248) .
-8 96 kJ mol .
RT{XA\nXA + XB In ArB) -
Solutions to exercises
Solutions to exercises
446
447
HAg = Gas + RT\nXAg = 0 + (8 314-773-In 0.752) .
= -1
.
83 kJ mor1
m
k e
(g) For a very small addition of Au
S 0)
T2-
dG' = HAu"d«Au(
E 0
T
.
P, nB constant)
n
.
f
9
h
a
a -
-
,
f
b
»c
d o
At 500oC, Au = -8.96 kJ mor1.
？
Avogadro's Number = 6.023 x 1023
Fe3C Graphite
1 eV = 1.6 x HT19 J -
.
8
96 x 103
.
-8.% kJ mol"1 =
-
.
-
1
6 x lO-9 x 6.023 x 10" eV at0m
i
Fe3C
Fe
Graphite
%C
.
= -0 1 eV atom"1
G-composition
and T-composition diagrams for the Fe-Fe3C and Fe-C systems (not to scale).
.
Adding one atom of Au changes the free energy of solution by -0.1 eV. 19 .
10
1
dC" = -nX dnA
.
Fe3C
dG = +\& dnA
8
At equilibrium dG" + dGp = 0 ie a + \i% dnA = 0 ncA i e \i% =
Graphite
-
1 = 1,
c
19
.
.
-
.
.
a b
Similarly for B, C, etc.
d
1
11
.
Equilibrium vacancy concentration XI = exp -
f S?
T = T2 e
AH,
A5V
9
CD
ul
RT
h
0) c 0) 03
AGV
= exp- -exp
f
1 eV = 1.6 x 10~i9 J R = 8.63 x HT5 eV atom"1 K-1 -
.
.
' .
.
Xt (933 K) = exp (2) - exp
n
= 3
.
/
\ \8.63 x lO-5 x 933/
58 x HT4
m
T-T3 k
(298 K) exp(2)-exp(
8
2 28 x 10 .
-
13
Q 8
63 x : 5 -
0
x
298)
L 12 .
Solutions to exercises
Solutions to exercises
448
449
2
Assume XSi
A
7V(Bi---
i
InXsi = In A - -
S
At 550oC (823 K): In 1.25 = \nA - 2/(8.314 x 823) At 450oC (723 K): ln0.46 = \nA - (2/(8.314 x 723) w
!
rE
.
!
-
A
Free energy of pure A
B
Free energy of pure B
I 1
hich can be solved to give
Q = 49.45 kJ moP1 A = 1721
Thus at 200oC (473 K) /
Xsi = 1721 -exp -
49450
\
\8.314 x 473/
= 0 006 atomic %
T
Gk G
.
According to the phase diagram, the solubility should be slightly under 0.01 atomic %. Reliable data is not available at such low
temperatures due to the long times required to reach equilibrium.
(d) E
A
1
13
.
A sketch of the relevant phase diagram and free energy curves is helpful in solving this problem. See p 449.
(a) Schematic phase diagram; (b) G-Tcurves for pure A; (c) G-Tcurves for pure B;
.
fd) Free energy curves for the A-B system at 7
ACA and ACb are as defined in (b) and (c).
Since A and B are mutually immiscible, the tangent to the liquid = X will intercept the curves for the A and B phases as shown, i.e. = GSA jib = G%. The liquid is assumed ideal, therefore from Fig. 1.12
B
curve GL at
,
AGA
Thus AGA AGB
-RTE\nX%
Finally therefore:
and -
E
AGb = ~RTE\nX%
But AGA and AGb can also be found from the relationships shown in
ASm(A)-{Tm(A) - rE) ASm(B)-(Tm(B) - 7E)
RTBlnX% = ASm(A)-(Tm(A) - TE) RTE\nX% = ASm(B)-(Tm(B) - TE)
-
or
Figs (b) and (c). -
314 rElnA = 8.4 (1500 - rE) 8 314 TEH(1 - Xl) = 8.4 (1300 - TE)
8
.
If Cp = Cp, Equation 1.17 gives AG = ~Ar m or
AG = ASm-AT
-
.
Solving these quations numerically gives
X% = 0.44 Xl = 0.56 TE = 826 K
"
F
.
Solutions to exercises
450
14
1
.
Solutions to exercises
If solid exists as a sphere of radius r within a liquid, then its free energy is increased by an amount
V
2lVm
.
s
Let the mole fractions of a, (3 and y in the final microstructure be X
i
Balance on A: 0.4 = 0.8 Xa + 0.1 Xp + 0.1 Xy Balance on B: 0.2 = 0.05 Xa + 0.7 Jfp + 0.2 Xy Balance on C: 0.4 = 0.15 Xa + 0.2 + 0.7 JfY
(from Equation 1.58)
Solving these equations gives: Xa = 0.43; Xp - 0.13; 1
16
.
.
e
G
.
7
0 44 .
From Equations 1.41 and 1.43 we have
GJ < GL i
<
X$ and Xy respectively.
r
where G] is the molar free energy of the sphere and GL is the molar free energy in the absence of interfaces. Growth of the sphere must lead to a reduction of the total free energy of the system, i.e. growth can occur when
451
liA = Ga + RTlnyAXA
- Gl >
2yVm
where GA is the free energy of pure A at temperature T and pressure
r
P
.
Suppose GA is known for a given temperature and pressure T0 and
See figure below.
/\) i
Growth occurs spontaneously with a decrease in free energy
e
.
.
GA(r0, Po) = Ga
From Equation 1.9 for 1 mole of A dGA = -5Adr + VmdP
0)
Thus if SA and Vw are independent of T and P, changing temperature
0}
G
S
from To to 7 and pressure from Pq to P will cause a total change in GA of
CO O
2
AGA = -5A(r-r0) +vm(P-Po)
mm
G
AT
L
Gs
Ga =
+ AG
= gSL + 5A(r0
r
-
r) + vm(p - p0)
and
Substituting Equation 1.17 for GL - Gi gives LAT T 1 m i
.
e
.
Ar>
increase.
r
27Vmrg
AT(r = 1 im) > 0.2 K AT{r = 1 nm) > 200 K .
15
-
T) + Vm(P - Po) + RT\nyAXA
The accuracy of this equation decreases as (T - Tq) and (P - P0)
2yVm
Substituting the numerical values given
1
= GSL + 5A(To
Composition = 40% A 20% By 40% C; ,
a = 80% A,
5% B 15% C; ,
p = 10% 70% B, 20% C; Y = 10% A, 20% B, 70% C.
i t
Solutions to exercises
Solutions to exercises
452
Similarly, if C = C; at x = /, the thickness of the sheet gives
Chapter 2 2
1
.
,
(a)
-
Jl = flC; + - C] - aC\
C]
Decarburizing
Sheet
Carburizing
gas
gas
n
453
i
e
.
.
7 - |«(C, - C:) + J (Ci - C \lL
14 .
The constants « and b can be determined from
-
D] = a + bC\
0 15 .
D
-
.
= a + hC
Carbon concentration
from which a = D
2
~
l WTj
D, - £>:
and 5 = Li - (-2
Thickness
Substitution of these expressions into the equation for / gives after simplification
f
(b) Under steady-state conditions, lux of carbon atoms into one side = flux out of the other side = J /= -
？
D-
.
+ D,\Ci - C.
Z)cdC
？
[dCl
Substituting: D, = 7.7 x 10 " irr s
fdC
I
i
D2 = 2.5 x 10"" mV 14
1). 15
.
Ci = - x 60 kg m-" 08 .
jcUj, / IcLcLu
0,4
_
7.7 x 10-"
0 32 .
0 15 .
0 8
(c) Assume that the diffusion coefficient varies linearly with carbon .
2 x Kr-1 m
concentration
..
.
x 60 kg m
.
y = 2.4 x 10~6 kg m-J2 ..-I s -
gives
/> = a +
.
where a and b are constants that can be determined from the data 22
given. Pick's first law then gives
.
.
1
2
/ = -(a + bC) dx
or J Jdx = -/ (a + bC) dC i e .
.
-Jx = aC +
bC2 -
j
- + d
where d is an integration constant.
d = -ac, -1 q S9
If we define C = C, at x = 0
Consider two adjacent (111) planes in an f.c.c. crystal. A vacancy in plane 1 can jump to one of three sites in plane 2. For the sake of
notations to exercises
Solutions to exercises 2
generality, let this number of jumps be designated P ( = 3). In all there
3 The activity along the bar is described by the following equation
.
are 12 possible sites (nearest neighbours).
If nx and are the numbers of vacancies m 2 in planes 1 and 2 respectively, the number of jumps from 1 to 2 will be given by ~
P 7V = - Tvnx
1
455
actIvlty
=
A»
__.ex
( x2 \
p(___j
where An = initial activity; m
"
-
s
D = diffusion constant; t = time;
where Vv is the jump frequency of the vacancies.
x = distance along the bar.
Lilcewise
Thus by plotting In (activity) vs jr a straight line of slope -(4D/)~ 1 is produced enabling D to be found since / is known. .
,
x ]xm
10
20
30
40
50
[im2
100
400
900
1600
2500
activity
83.8
66.4
42.0
23.6
8.74
In (activity)
4.43
4.20
3.74
3.16
2.17
2
Therefore following the same arguments as in Section 2.2.1 (p. 6) gives
x
where d is the perpendicular separation of the adjacent planes, i.e. we can write
4
-
In f.c.c. metals the jump distance a is given by a a
'
v
2
where a is the lattice parameter. For (111) planes
Putting P - 3 gives 2
a
2
v2
.
.
Dy = 76 a2rv
2000
x2
z
\im
.
.
-
1000
From the graph: slope = -8 66 x 10~4 jiirr2.
For (100) planes, adjacent planes are in fact (200) a
1
0
Z)v = 76 a2rv
\
,
"
Hence: -777-
8 66 x 10"4
-
.
I
Since t = 24 h
D = 3.34 x 10-15 n
.
s'1
\
!i
Solutions to exercises
Solutions to exercises
456
C = C + po sin
24 .
457
The first two terms of the series are
(t exp
"
4Cor . kx
For this equation to be a solution of Pick's second law, the following condition must be met
1 . 37rxl
C(x) = - nT + -slnTJ v
Plotting this sum for the range 0 < v < / gives the curve plotted opposite (b) If the surface concentration is effectively zero, the solution to the diffusion equation becomes .
5C
52C
57
Db &-
"
/tcxX
1
r
C(x, t)
C0S iyj
=
eXP " x7
52C
r.
=-p(1exp
.. ir
4C» V 71
/7u:\
.
-
1=0 2/ + 1
. exp
-.-
1
[-(2/
[
(2/ + 1) TLCj
+ l)VDr]
/=
J
The amplitude of the first term (A ) is obtained by putting jc = 1/2 and / = 0. i.e.
6C
52C rSX2 TT
x
bt But
T k
6C
Of 2
5
.
(Equation 2.21)
2D B
4C(,
A
The amplitude of the second term (Az) is obtained by putting x 1/6 and / = 1, i.e.
52C
Db
~
\-K2Dt]
i
4Co
2
5x2
(a)
r-97r:D/]
If A2 < 0.05 Ax
C(a-)
7i
2/ + r
(2/ + 1) 7U-
sin
4Co
/
\-9n2Dt]
exp rrn
'
3
where / = thickness of sheet,
which gives / > 0.0240
d = initial concentration.
4Ct)
r-7t2D/l
< 0.05 . --exp \ -p
J
K
D
(c) Assume that the time taken to remove 95% of all the hydrogen is so long that only the first term of the Fourier series is significant The hydrogen concentration at this stage will then be given by
Co
C{x, 0 = --sin y exp 1 i
.
CM
e
.
/2
.
I
as shown in the figure on page 458.
At the required time (tx) the shaded area in the figure will be 5% . of the area under the concentration line at / = 0 i.e. ,
V
C(xi, /,) dr = 0.05 C,,/ 0 31
0 2
4 X
4
4C0
/-7r2Df,\ ('
tvc
- -expl-sin y dx = 0.05 CJ
extrapolation of the tangents to the free energy curves at 1 and 2 to the corresponding sides of the free energy diagram, as shown above. All atoms diffuse so as to reduce their chemical potential. Therefore, A atoms will have a tendency to diffuse from a to P (\iaA >
2
which gives
459
Solutions to exercises
Solutions to exercises
458
= 0.282 t=0
\x%) and B atoms will have a tendency to diffuse from P to a
Co-
The resultant composition changes are indicated in the diagram.
T
Diffusion stops, and equilibrium is reached, when |iA = nil and \i% = ji . That this process results in a reduction in the total free energy of
c
the diffusion couple can be seen from the diagram below. The initial free energy Gj can be reduced to G? by a change in the compositions of the a and P phases to and A , the equilibrium compositions (provided < XBu]k < X%).
rr7777777777TT A 0 x
Note that this time is an order of magnitude larger than the time
derived in part (b). Consequently it is clearly justified to ignore all terms of the Fourier series but the first.
/-13400\
From Table 2.1. D = 0.1 exp i
e
.
.
P
RT
a CD c
Gi
Zl1\
2-1
D (20oC) = 4.08 x lO-4 mnrs ,
7
CD Z
.
f
Thus for / = 10 mm: f, = 19.2 h,
f
G2
for / = 100 mm: f, = 1920 h (80 days). 26 .
Mofar
free energy
A
XBulk
X°
B
xz P
1
2
2
7
.
Ha
Substituting into Darken's Equations (2.47) and (2.51) § Zn
/
5x
(equilibrium) (equilibrium)
Da
= XC
ll
DaZn
a
+ XZnD Cxi
we obtain
026 x lO"6 = (D n - Dacu) x 0.089 mm s
0
Ha
.
A
>
X
B
4
.
5 x 10~7
0
.
™ + 0.22 " 78 Dfin
.
From which
At the initial compositions 1 and 2 of a and P respectively the chemical potentials of A and B atoms in each phase can be found by
DaZn
5 1 x 10
Dr Cu
2
-
7
.
.
2 x UT7
2
mm
s
2 s
mm
"1
1
a
mnr2 s
"
1
460
Solutions to exercises
Solutions to exercises
The expected variation of Dzn, Dcu and Da are shown schematically
461
1 1
below
D
'
D2n 0
A
a
b
d
c
B
XB
%Zn Cu
a
t
Solubility limit for a
Since Zn has a lower melting point than Cu it diffuses faster of the
LJ
.
two, and since increasing the zinc content reduces the liquidus temperature, all diffusivities can be expected to increase with increasing
0
Zn concentration. 28 .
JC
-
GB
P
U
Hb(2) r
Ml) P+7
a+(3
He
G
i M2)
Ml
A
b
c
B
462
Solutions to exercises
Solutions to exercises
463
(i) a
t
a
U C
s
B
f = 3C
GO
b
P
p
L 2
~
r
a
t b
Bulk
A
o9
B
.
05 .
Bulk composition B
The total distance the interface moves 5, can be calculated in terms of ,
(ii)
the total couple thickness, L, by writing down an equation describing the conservation of B, i.e.
0
3
.
2
.
a
= 5.6 x lO-2
'
.
9 |y + sj + ()[~ - s] - 0.5L
.
a
Chapter 3 29
3
.
.
1
Considering only nearest neighbours, if a surface atom has B 'broken* bonds, it will have an excess energy of B. e/2, where £ is the bond energy.
a
For f.c.c. crystals, each atom has twelve nearest neighbours in the f = 0
limation and NLl Avogadro's number (no. of atoms per mole). The surface energy per surface atom is therefore given by
P
A
bulk, so that 8 = Ls/6/Va, where Ls is the molar latent heat of sub-
XB
7SV
B
B Ls = -./Va
per surface atom
.
If each surface atom is associated with a surface area A, the surface
!
a
o
1
energy is B >0
iV
12A Na
per unit area
Q
A can be calculated in terms of the lattice parameter a\
09 .
Solutions to exercises
464
Solutions to exercises
Each surface atom is connected to two nearest neighbours in the
*
and £ sv
{220} plane. Therefore it must be connected to ten others out of the plane. Since the atoms are symmetrically disposed about the {220}
8
(cos 9 - sin 9)
which gives | -r]
27
{200)
{111}
9 < 0
~ "
\ d9 /o.o
{hki}
,
465
2?
{220S
*
At 9 = 0 there is a cusp in the £sv - 9 curve with slopes
Ge© a
V2
a
*
A
a
V3
\'2
2
60©
a
.
3
For a two-dimensional rectangular crystal with sides of lengths /j and A and surface energies ji and 72 respectively the total surface energy is given by ,
aV2
_
\2
3
a
2
2
G = 2(/17, + /:Y2) B
3
4
0 33
Ysv
.
0
Vsv
-
58[
0 67 .
5
The equilibrium shape is given when the differential of G equals
*
zero. i.e.
dG = 2(/ldy, + y /, + /2d72 + y.d/,) = 0
rf
0 42
LA/aJ
.
Assuming that 71 and 72 are independent of length gives 0 59
a2/V;
.
[ 2W ] a
7id/, + y2dl2 = 0
a
But since the area of the crystal A = IJ2 dA = /2d/2 + hdli = 0
plane, there must be five bonds above the plane of the paper and five below (giving a total of 12) + It can be shown that in general for f c c
Giving
.
.
.
.
constant
metals
h
4
ii
For the simple cases above however. A can be calculated directly ,
from a sketch as shown ,
3
4
.
.
(a) By measuring, the misorientation 9 = 11. (b) By constructing a Burger- :ircuit around a dislocation, the
Burgers vector is found to be 1.53 mm in the photograph (i e .
32
Ew
.
i
.
e
.
£
sv
= (cos 0 + sin |0|)
~
( de /o
2fl-
(-sin
For a low-angle grain boundary, the spacing of the dislocations is given by
2a
(cos 9 + sin 9) ~
de
.
one bubble diameter).
e>0
b D == -
sin 9
9 + cos e)~ 2a
*
1.53
D
8 0 mm .
°
sin 11 =o
2a2
which is very close to the mean dislocation spacing in the boundary.
Solutions to exercises
466 3
5
.
Solutions- to exercises
Like all other natural processes, grain boundary migration always results in a reduction in total free energy.
467
(b) For nucleus growth, reduction in free energy due to annihilation of dislocations must be greater than or equal to the retarding force due to grain boundary curvature.
Equating this with the driving force across a curved boundary
Grain growth i
During the process of grain growth all grains have approximately the same, low dislocation density, which remains unchanged during the
e
.
.
1.96 x l(f
r
grain growth process.
'
.
.
27 r
1
Grain "boundaries move towards their centres of curvature in this
case, because atoms tend to migrate across the boundaries in the opposite direction (from the high pressure side to the low pressure side), in order to reduce their free energy, or chemical potential. The process also results in a reduction of the total number of grains by the growth of large grains at the expense of smaller ones. The net result is a reduction in the total grain boundary area and total grain boundary energy.
27
.
96 x 106
Thus the smallest diameter = 1.0 fim
3
7
.
From the phase diagrams, the limit of solid solubility of Fe in Al is 04 wt% Fe, whereas that of Mg in Al is 17.4 wt% Mg. If one element is able to dissolve another only to a small degree, the extent of grain boundary enrichment will be large. (See for example Fig. 3 28, p. 138.) Thus grain boundary enrichment of Fe in dilute Al-Fe alloys would be expected to be greater than that of Mg in Al-Mg alloys. 0
.
.
Recrystailization
In this case, grain boundary energy is insignificant in comparison with the difference in dislocation energy density between recrystallized grain and surrounding deformed matrix. The small increase in total grain boundary energy that accompanies growth of a recrystailization nucleus is more than compensated for by the reduction in total dislocation energy. The boundaries of recrystailization nuclei can therefore migrate away from their centres of curvature.
3
.
6
3
8
.
See Fig. 3.35, p. 145.
If d(t < dp, then in general the dislocation spacing (D) will span n atom planes in the P phase and (n + 1) planes in the a phase, i.e.
D = nd$ = (n + 1) d, From the definition of 8 we have
(-ve)
(a) The pulling force acting on the boundary is equivalent to the free energy difference per unit volume of material.
dv = (l + S)da
2
If the dislocations have an energy of j- J m
and the
Substitution into the first equation gives
dislocation density is 1016 m~2, then the free energy per unit
n(l + 5) da = (n + 1) d(x
volume, G, is given by
1
10-x(0.28x , »x
G,10
4
Y
i e .
1%MJ|n.3
,
Thus the pulling force per unit area of boundary is 1.96 MN m~2.
.
n = o
and D = 5
"
7
Solutions to exercises
468 39
Solutions to exercises 1 26 - 1.43
2*2
.
469
.
Hence zone misfit
x 100%
1 43 .
2*1 -
71
11.89%.
When the misfit is less than 5%
strain energy effects are less important than interfacial energy effects and spherical zones minimize the total free energy. However when the misfit is greater than 5% the small increase in interfacial energy caused by choosing a disc shape is more than compensated by the reduction in coherency strain ,
72
,
72
,
energy.
The edges of the plate exert a force on the periphery of the broad
Thus the zones in Al-Fe alloys would be expected to be discshaped.
face equal to 72 . 4 . 2t2
This force acts over an area equal to (2X2)2
i
3
.
AP
Y2 ' 4 '
12
272
2 =
(2v2)2
x2
Assuming that the matrix is elastically isotropic, that both Al and Mg atoms have equal elastic moduli, and taking a value of 1/3 for Poisson's ratio, the total elastic strain energy ACS is given by:
AGS - 4 jiS2!/
|i = shear modulus of matrix; 5 = unconstrained misfit; V = volume of an Al atom.
\ 2x1 1 60 - 1.43 .
5
0 119 .
1 43 .
2x 2
V
= 4/3 * tt (1 43 x 10~!0)3 = 1.225 x lO"29 m3 \xAl = 25 GPa = 25 x lO9 Nm"2 .
AGS = 4 x 25 x 109 x (0.119)2 x 1.225 x 10~29 J atom = 1
2 -72 '2xx + 2-7, - 2x2
1
'2 2
_
, Yi h -
1 eV
3
.
3
11
Atomic radius of Al = 1.43 A Atomic radius of Fe = 1.26 A
1
1 .
.
6 x 10~19 J, thus
735 x lO"20
1
6 x 10~19
eV atom
-
1
.
1 eV atom"1
.
It is also implicitly assumed that individual Mg atoms are separated by large distances, so that each atom can be considered in isolation,
X2
See Section 3.4.1 (p. 143).
-
1000
0
.
10
.
kJ mol
1
AGS
(see also Exercise 3 3)
Xi
735 x HT20 x 6.023 x 1023
10.5 kJ m-r' Y2
From the Wulff theorem (p. 115) for an equilibrium plate shape: - =
.
AG s
The area of each edge is 2xx . 2x2 AF =
735 x HT20 J atom"1
.
In 1 mol there are 6.023 x lO23 atoms
The periphery of each edge is acted on by a force of magnitude
2y2-2xl +
i
-
i
e
.
.
dilute solutions.
The use of Equation 3.39 is also based on the assumption that the matrix surrounding a single atom is a continuum. 3
13
.
See Section 3.4.4. (p. 160).
3
Solutions to exercises
Solutions to exercises
470
crystal itions A are shifted to B the atoms above the glide plane in pos
14 When a Shockley partial dislocation passes through an
.
f
c c
.
.
3 16 .
.
positions, B into C positions, etc. i
e
.
.
A
B
A
v
A
B
C
A
C B
C
C
X
A
A
A
C
B
B
A
1
1
Plane B
B
B
B
A
A
A
A
A
C
C
C
C
C
B
B
B
B
B
1
8
B
B
B
B
B
C
C
C
C
C
C
A
A
A
A
A
A
B
B
B
B
- A-
-
Fee
/////////
B , '
C
C
B
B
B
A
A
C
C
c
C -
B
B
B
A
A
A
A
C
C
C
C
C
B
B
B
B
B
B
A
A
A
A
A
A
Twin
C
C
C
C
C
B
Fee
B
B
B
B
A
471
If a single atom in crystal I attempts to jump into a crystal II position a ring of dislocation and an unstable A upon A situation results. A Shockley partial dislocation in every {111} slip plane creates a glissile interface between two twinned crystals:
The above series of diagrams shows the twinning process. 3 15 .
A A
~
-
t
h
B "
-
'
C
i
A i
t
T
B
,D
I
E
F
C i
Let the interface CD move with a velocity v perpendicular to the
A
I
C
J. i
B
JL I
J
B
r
Consider unit area of interface perpendicular to ooth BC and CD. Mass flow perpendicular to BC = u x h. Mass flow perpendicular to CD = v x /. From the conservation of mass: u x h = v x /. u x h -
11
C
i
interface.
v
A .
I
C
B !
!
A -
X
A
.
Note, however, that as a result of the shape change produced by the transformation large coherency stresses will be associated with the interface (see Fig. 3.62a). Similar coherency stresses will arise as a result of the f.c.c./h.c.p. interface in Fig. 3.61. Strictly speaking. Fig. 3.60 is an incorrect representation of the stacking sequence that results from the passage of the partial dislocations. In layer 10, for example, there will not be a sudden change across the extra half-plane of A to B or B to C, but '
'
472
Solutions to exercises
Solutions to exercises
rather a gradual change associated with long range strain fields in
3 20 .
both crystals. 3
.
17
473
r2.
Solid/vapour interfaces and solid/liquid interfaces in non-metals are faceted and therefore migrate by ledge mechanisms .
Solid/liquid interfaces in metals are diffuse and migration occurs by random atom jumps. 3
.
18
See-Section 3.3.4 p. 130. ,
3 19 .
X
X,p
Suppose the alloy had reached equilibrium at a temperature T[ and consists of long plate-like precipitates. The bulk alloy composition is Xih the equilibrium concentrations at 7, and 7 are Xx and respectively, where 7% is the temperature to which the alloy is heated.
A
B
X
(ii) Interface control
(i) Diffusion control
From equations 1.41 and 1 43 we can write .
'
hb
= GB + RT In Yi*t
i
Hfc = Gb + RT\nycXc r
.
Aii =
a
- (is = RTlnycXc
/ = 0
a
X2Xo\
For ideal solutions: y; = y = 1 For dilute solutions (X « 1): y, = yc = constant (Henry's Law)
L
c
Distance
}
f = f1
such that in both cases X2
Aji'b = RT In
X
Xc
This can also be written
A k = RTln If the supersaturation is small
X
o
,
i.e.
(X,- - Xe) « Xe, then = RT X0
f = t3 > '2
Solutions to exercises
Solutions to exercises
474
475
Substituting Q - 1.6 x 10~29 m3 and y - 0.177 Jm : gives
(iii) Mixed control: similar to diffusion control except the interface
~
concentration in the a-matrix will be less than A , the equilibrium concentration at T2-
'
AGr - (5.435 x l(r20)nl 3 For 1 mm3, n{i - 6.25 x 1019 atoms = 10 atoms, nr - 9 x 1013 clusters mm 3: and when nQ = 60 atoms nx - 3 clusters mm-3; when n = 100 atoms nr = 4 x 10"8 clusters mm 3: or. alternatively. 1 cluster in 2.5 x 107 mm3 (251)
Therefore when n
Chapter 4
"
c
,
4
.
1
*
-47ir3 3
"
AGV + 47cr2YsL
c
,
.
Differentiating this equation with respect to r. dAGr
4
3
.
As the undercooling (AT) is increased there is an increasing contribution from AG in the equation ,
V
47i/-::*AGv + 87rr/SL
-
dr
4
AGr = --7cr3AG + 47rr2YS[ v
At the critical radius, / *, this expression is equal to zero "
0
= -4Trr
*2
'AG
v
whereas the interfacial energy is independent of AT Consequently
+ 8nr*ySL
.
for a given r. AG decreases with increasing AT, and the 'maximum' r
r
*
=
2ysl
cluster size increases somewhat.
AGy
4
.
4
From Equation 4
In order to calculate the critical value of AG, AG* at this radius,
13
.
oCoeXp{
the value of r* is substituted into the original equation where
ILlkT
7 = 7m- bJ
From which the following values are obtained: 167rysL 3(AGV)2
=
4
2
.
AT K
From Equation 4.10. at the equilibrium melting temperature Tm
iVh
om
m"3 s""1
180
07
200
8 x 106 1 x 1012
7 x lO"7
.
220
iVhom cm-3 s-1
8
1 x 10A
At the equilibrium melting temperature AGV = 0. so that Equation
Note the large change in N over the small temperature range (see
4 4 becomes
Fig. 4.6).
.
AGr(r= Tm) = 4Ttr2ySL For a cluster containing nQ atoms with an atomic volume Q we have
4
5
.
AG* = i-F*-AG
For homogeneous nucleation it has been shown (see 4.1) that ,
47173 3
ncQ.
v
2TSL
* _
AGV
Therefore the expression for AGr becomes
Thus for a spherical nucleus
'
3QAic\2/3-r
V
4n7-*3
327tysL 3(AGV)3
Solutions to exercises
476 1
AG* - --
Solutions to exercises
16Try3SL
-AGv
All
Heterogeneous nucleus
2
3AG;
77777
L
This is identical to that derived in 4.1, and so the equation holds for homogeneous nucleation. For heterogeneous nucleation, it can be shown that
/
s
Size of
r
> = 2ysL
homogeneous
AGV
nucleus at same AT
The volume of a spherical cap on a flat mould surface is given by V
(same r')
\
(2 + cose)(l - cos9)2
\
,
nr
3
Thus v*
=
/2ysl\3 (2 + cos6)(l - cos e): \AGv/ 3
where 0 is the
'
wetting angle.
Substituting into the given equation
4 yIl AG* = - V*-AGV = -k-(2 + cose)(l - cose) i
2
The wetting angle between nucleus and mould wall (8) is fixed by the balance of surface tension forces (Equation 4.14). The activation energy barrier (AGhet) depends on the shape of the nucleus as determined by the angles a and 0. From Equation 4.23, for a given undercooling (AT), AGV and r* are constant, such that the following equalities apply .
7
j AGy
S
Writing the normal free energy equation for heterogeneous nucleation in terms of the wetting angle 0 and the cap radius r
AGhet = j- AGv + 4nr2YSLj
(2 + cos0)(l - cos0)2
i
e
.
.
het
hom
AGh*et
S
_ ~
AGh*hum
4
But from Equations 4.19 and 4.17 we have
AGh, he! AGh* horn
Volume of the heterogeneous nucleus Volume of a sphere with the same nucleus/ liquid interfacial radius
It can be seen that the shape factor {S) will decrease as a decreases and on cooling below Tm the critical value of AG will be reached at progressively lower values of AT, i.e. nucleation becomes easier. When a =S 90 - 0, S = 0 and there is no nucleation energy barrier. (It can be seen that a = 90 - 0 gives a planar solid/liquid interface, i e r = x even for a negligibly small nucleus volume.) Once nucleation has occurred, the nucleus can grow until it reaches the edge of the conical crevice. However, further growth into the liquid requires the solid/liquid interface radius to pass through a minimum of R (the maximum radius of the cone). This requires an undercooling given by ,
*
167ty|L (2 + cos0)(l - cos0)2
AG =W
4
which is identical to that obtained using AG* = tV *AG
v
46
See Section 4.1.3.
4
Consider a cone-shaped crevice with semiangle a as shown below:
.
t
'
>
-
.
7
.
.
<
2ysl R
h
LbT m
r
i
.
I i
e
.
Ar
2YsiTm RL
478 4
.
8
Solutions to exercises
Solutions to exercises
For conical crevices with a < 90 - 0 the solid/liquid interface can maintain a negative radius of curvature which stabilizes the solid above the equilibrium melting temperature (rm):
1 m
479
2
V
V 6
L
(1 - 6)
S
S
1 m
L
7777,
77777 Thus
G(II) = Cs(l - 8) + CL8 4- ySL + yLV (8 > 0) G(II) - Gs + 5(GL - Gs) + YSL + 7Lv (8 > 0)
S
LAT
At an undercooling AT below 7m, GL - Gs = -*
m
LAT
i
e
.
.
G(U) - Gs + --5 + 7sl + 7lv / rm
As the temperature is raised above rm the solid will melt back into the crevice to maintain equilibrium with a radius given by
LAT
or
G(II) = G(I) - Ay + - --5 Tm .
r
where Ay =
LAT
where { - AT) is now the superheat above r 4
4
.
9 10
.
- 7SL - 7Lv
This is shown in the figure below: m
.
If the situation described above is realized in practice it would explain the observed phenomena.
G(li
(a) The values of the three interfacial energies are as follows:
Gradient
LA 7
el
Solid-liquid = 0.132 J m"2; Liquid-vapour = 1.128 J m"2; Solid-vapour = 1.400 J mT2.
Ay
Thus the sum of the solid-liquid and liquid-vapour interfacial free energies is less than the solid-vapour free energy and there is no increase of free energy in the early stages of melting Therefore, it would be expected that a thin layer of liquid should form on the surface below the melting point because the dif,
.
,
ference in free energies could be used to convert solid into liquid (b) Imagine the system I below. The free energy of this system is
0
do
6 max
b
.
given by:
G(I) = Gs + y
sv
System II contains a liquid layer of thickness 5 and solid reduced to a height (1 - 8). (The difference in molar volume between liquid and solid has been ignored.)
Note that as 5 -> 0. G(II) G(I), which means that in practice ySL + Ylv ~> Ysv a result of an interaction between the SiL and LIV interfaces as they approach to within atomic dimensions of each other.
The optimum liquid layer thickness (5o) will be that giving a minimum free energy as shown. We cannot calculate this value
Solutions to exercises
Solutions to exercises
480
without a knowledge of the above interaction. However, it is reasonable to assume that the minimum will occur at a separation of a few atom diameters, provided 5max in the above diagram is at least a few atom diameters. 5max is defined by G(I) = G(II)
1
i
.
e
.
G(I) - Ay +
G(I)
7m
-5
/
-Ke"
undercoolings we have max
nuclei m 2 s
oc exp LAT
Alternatively, AT
where k is approximately constant. Each time a cap is nucleated, it should grow rapidly across the interface to advance a distance h. It seems reasonable to suppose therefore that the growth rate will be proportional to N, i.e.
L5 max
If Smax = 10 nm (say), then AT = 16 K. It seems therefore that surface melting is theoretically possible a few degrees below 7m. 11
1
Ayrm
5 max
.
\
But. AG 3c AT,, the undercooling at the interface, so for small
,
4
481
v x exp
AT;
(a) Repeated surface nucleation (see Section 4.2.2, p. 198). (b) Very roughly. Equation 4.28 can be seen to be reasonable as
I
-
follows:
i;
Firstly, it is reasonable to suppose that the distance between successive turns of the spiral (L) will be linearly related to the
r
\
*
ii
r
h
L
t
S
minimum radius at the centre (r ) Thus we have .
L
r* * AT~]
Secondly, for small undercoolings, the lateral velocity of the steps (w) should be proportional to the driving force, which in turn is proportional to AT,
Suppose the edge of the cap nucleus is associated with an energy e (J m 1) Formation of such a cap will cause a free energy
u
ATj
-
.
Thus the velocity normal to the interface v is given by
change given by
uh
AG = -Kr2hAGv + lure The critical cap radius r* is given by
v = dr
= 0
e
i
.
e
.
where
4
r* =
.
12
r .
i
ATf
is the step height.
Equilibrium solidification (see Figs, 4.19 and 4.20)
MGV
From Fig. 4.19 the lever rule gives the mole fraction solid (/s) at T2
and
as
AG*
ne
2
XL - Xn
hAG V s
The rate at which caps nucleate on the surface should be proportional to exp I-I
XL - Xs
(Xs/k) - X0 (Xs/k) - Xs
kX0 s
1 - (1 - *)/s
482
Solutions to exercises
Solutions to exercises
This expression relates the composition of the solid forming at the interface at T2 to the fraction already solidified. For the case shown in Fig. 4.1°, it will be roughly as shown below:
483
No diffusion in solid perfect mixing in liquid (see Fig. 4.21). ,
Again, we have 2 ~~ T$
Xjy _
7\ - 7"}
X; - k-X
.
is now given by Equation 4.33 such that
Xo -
t t2-t3
.
i-k(i-fsy
k-»
_
Tx-Ts
(l-k)
where Tx > T2 > TE. For the phase diagram in Fig 4.21a. the following variation is therefore obtained (k - 0.47. the exact form of the curve depends on k, of course) .
0
4
The temperature of the interface (7%) as a function of the fraction solidified can be obtained using the following relationship which is apparent from Fig. 4.19 T2 - T3 r, - t3
To
X{) - Xs xi} - kXo
T3-
Substituting for Xs gives
0
This will be a curved line roughly as shown below for the case described for Fig. 4.19 (* - 0.47).
1
No diffusion in solid, no stirring in liquid (see Fig. 4.22)
t Q)
T2
Initial transient
<5 Q
T3-
.
E 52
Final transient
Steady state
CD CD
CD
T3 0
1
0
fs
1
i
484 4
Solutions to exercises
Solutions to exercises
13
Diagram (a), above is a typical phase diagram for k > 1. (In this
No diffusion in solid, complete mixing in liquid
.
485
,
case, A- = 3.) The variation of composition along the bar can be calculated using Equation 4.33 i a)
e
.
.
Xs = kX(>(l -/s) <*-i)
L
The result for k = 3 is shown in diagram (b) /s is proportional to distance along the bar. Note that the final composition to solidify is
ri
.
pure solvent (Xs = 0).
No diffusion in solid no stirring in liquid. ,
5 CO J
f
S
Steady state
0
4
0
4
.
14
During steady-state growth the concentration profile in the liquid must be such that the rate at which solute diffuses down the concen-
tration gradient away from the interface is balanced by the rate at which solute is rejected from the solidifying liquid i.e.
(b)
,
-
DCL = v(CL - Cs)
Assuming the molar volume is independent of composition this ,
becomec
DdXL
-
v
dx
at the interface
The concentration profile in the liquid is given by
exp - m\
s
dXi -
:
0 0
1
= Xn I -X
o[
k II d exn exp
-
(D/ (D/v)
Solutions to exercises
486
Solutions to exercises
(d) For an Al-2 wt% Cu alloy;
Substituting this expression into the solute equation
Interface temperature = 620 40C
v -
487
D--(Xo-XO
3 x 10
Diffusion layer thickness
Since = Xq/Ic at the interface, the expressions are equivalent, and the profile satisfies the solute balance.
4
re = 6 x 10
6 x 10
.
700"
9
(653.2 - 620.4)
Temperature gradient
4 15
5 x 10
-
-
4
m
54.7 K mm
16 Scheil equation: XL = X0fik-D
.
Since it is assumed that the soiidus and liquidus lines are straight
©
,
k is
constant over the solidification range and may be calculated usine ,
OS
a3 600-
max
and XE as follows
Q
-
Xs
0)
k = tt at a given temperature At the eutectic temperature
s = Xmax and X{ = X
,
500 0
5
15
10
25
20
30
35
X solute
For an Al-0.5 wt% Cu alloy: 33-
(a) Interface temperature in the steady state is given by the soiidus temperature for the composition concerned, Interface temperature = 650, TC 6
(b) Diffusion layer thickness is equivalent to the characteristic width of the concentration profile. 3 x lO"9 Thickness = v
i
6 x lO"4 m
5 x 10~6
(c) A planar interface is only staM: if there is no zone of constitutional undercooling ahead of it. Under steady-state growth, con-
5 65.
20 .
sideration of the temperature and concentration profiles in the liquid ahead of the interface gives that the critical gradient, Ti, can be expressed as follows liquidus temp at
where L
soiidus temp at X0
DN
-
y
-
0 34' .
T
0
0 38 .
0 97 .
Distance along bar
The above plot may be constructed by considering the composition of the initial solid formed (kX0) the position at which the solid has the compositions X0 and X and the eutectic composition XR. ,
max,
ThUS
=
(658.3 - 650.1)
6 x 10-
,
Cm
1
-
Initial solid formed = kX0
.
= 0 17 x 2 wt% Cu .
13.7 K mm
-
i
= 0 34 wt% Cu .
488
Solutions to exercises
Solutions to exercises
The volume fraction of liquid remaining, /L when the solid
4
.
17
deposited has a composition Xq is found from the Scheil equation.
489
Cells grow in the direction of maximum temperature gradient, which is upstream in a convection current.
4 18 .
Thus when
=
XL =
and the Scheil equation becomes L
T,-
k
a
d
c
1
(-0 83) .
b
0 17 .
9
e
/l = 0.12. hence/s (position along bar) = 0.88 Similarly, when Xs = X
max
i
5
.
65 wt% Cu, the Scheil equation
becomes 5 65 .
9 X / "0"83)
0 17
0 25 .
.
0
j_
\
.
Assume equilibrium conditions between <) and L
175 65x 2y '0-83 .
= 0 03 .
2
T-p
Hence /s (position along bar) = 0.97 h
L
From the information given, X = 33 wt% Cu for positions along the
9
bar between 0.97 and L
c .
b
(b) From the diagram, the fraction solidifying as a eutectic, fE = 0.03. (c) For an Al-0.5 wt% Cu alloy solidified under the same conditions. the fraction forming as eutectic may be found from the Scheil equation as before by putting equal to X
e
miix f
xL = x0f<£-" Temperature at which all 6 disappeared
f S _ y ,(A-1) y
max
4
v r(k-l)
k ~xofL _
"
5 65
17
It can be shown that the growth rate of a lamellar eutectic v. is given by the following equation v = kDATi
.
0
19
.
U :)X/e -
.
/e
/0.17 x 0.5\ \
5.65
0 006 .
/
1/0.83
where k D ATf, X
= = =
proportionality constant; liquid diffusivity; interface undercooling; lamellar spacing; minimum possible value of X
.
490
Solutions to exercises
Solutions to exercises
Differentiating a second time
(i) When the undercooling is fixed, k, D and ATI, may be combined to form a constant c. thus
'
a
d AZi -
-
-
A7n -
dv
-c
dA
r
A"
d2v
2c
Differentiating a second time; - = dA~
A"
6ca* A
A To
d>.-
dA"
AT
2ATi) ijji ii J.1 «
d:Ar()
2a*\\
A/p /
£
OA
f)
_
Ia
2Ari) To / J1
dA2
Hence - = -r- at the max. or min. urowth rate. "
dA
6a
'
/
« \;
Substituting X - 2a*
rj-.
2ca* A"
A
2a
d Ar(i
d>.2
X
c
a
d2A7
2cX
- = -H
dAT,, r
;
A 7,)
Differentiating this equation with respect to X. V
491
Thus
a
d Ar,,
\4X*
1_\
4> 2/
:
_
A7n / J
6
a 14;. :
\
l(v. : ' _
is positive
dA
Hence undercooling is a minimum when /. = 2/.''
Substituting this value into the equation of the second differential d2v
2c
6ca*
4
.
20
The total change in molar free energy when liquid transforms into lamella a + (3 with a spacing a is given by Equation 4 37. i.e. .
_
dx3
=
>7 " "x12c
AG(a) = -AG( ) +
6ca*
A
4
16a * -c
8/,
Thus when /. = 2a*. the growth rate is a maximum.
The equilibrium eutectic temperature TE is defined bv X = * and G(x) =0. We can define a metastable equilibrium eutectic temperature at (7"E - ArE) such that at this temperature there is no change in free energy when L a + (3 with a spacing a. i.e. at TE - ATE ,
(ii) When the growth rate is fixed, the original equation may be
AG{X) = 0.
Also from Equation 4.38 at an undercooling of ArE
rewritten as follows
AHATe
AC( )
a = A To . r where
T,
Finally, then, combining these equations gives v
2ynfiVmTE
a
Li
Thus:
a
1
AT,)
\
AHX AH
Substituting: Yap - 0.4 J m-". rE = 1000-K..-p = 8 x 10x J m
k2
'
-
Differentiating with respect to /. -
a
Ar5
dATi)
1
2X*
dl
X2
X
dX
10
~
a \a2
6
-
AT,
X
3
i
r2
dAT.
gives
a3 /
e
.
.
for a - 0.2 \xm. ATE = 5 K X = 1.0 [im. ATB = 1 K
v . jn
Solutions to exercises
492
Solutions to exercises
Note that if these eutectics grow at the optimum spacing of 2X* the total undercooling at the interface during growth (Ar0) will be given by Equation 4.39 such that for A. = 0.2 urn.
X* = 0.1 \im.
and X = 1.0 (im,
= 0.5 \xm.
A To = 10 K
493
Chapter 5 5 1 .
AG,, = Rrf ln (a) By direct substitution into the above equation
AT{) = 2 K
AGU = 420.3 J mor1
4 21 .
h
(b) Applying the lever rule to the system at equilibrium (X - X )
U>-d -J
Mole fraction of precipitate = j-
- = 0 08 .
(Xp - XQ)
Assuming the molar volume is independent of composition P
P
P
a
will also be the volume fraction
,
this
.
(c)
50 nm
Rod-like eutectic
Lamellar eutectic
For a lamellar eutectic the total interfacial area per unit volume of eutectic is given by: 2/X. irrespective of volume fraction of p. For the rod eutectic, considering rods of unit length, and diameter d the area of a/(3 interface per unit volume of eutectic is given by
Assuming a regular cubic array with a particle spacing of 50 nm the number of particles per cubic metre of alloy =
.
1
.
nd
X
(50 x w y
2nd
3/2
X2V3
For the rod eutectic to have the minimum interfacial energy, then 2nd
X
2
< T' X '
.
e.
d
x3
A
71
21
Let all the particles be of equal volume and spherical in shape with a radius r. Then the total volume of particles in 1 m3 of alloy =
7 < ,
8 *10
_
8 x 1021 x 7tr3
d depends on the volume fraction of p. (/)
Equating this with the volume fraction of precipitate
f~ 4 /
8 x 1021 x nr = 0.08 m3
2
v3
From which / < /c = - 2n
4 22
See Sections 4.4 and 4.5.
4 23
See Section 4.5.
.
.
0 28. .
r = 13.4 nm.
Thus in 1 m3 of alloy the total interfacial area = 8 x 1021 x 47U-2 = 1.8 x 107 m2
I:
Solutions to exercises
Solutions to exercises
494
(d) If Tap = 200 mJ m
2
495
From Equation ] .40
"
GB GB Mb GA HX = GA
total interfacial energy = 200 x 1.8 x 107 mJ m 3 alloy = 3 6 x 106 J m"3 alloy "
.
= 36 J mol
"
1
+ + + +
RT}nXq + D(l - Z,,)2 KTlnXe + na - Xe)2 RT\n(l - X0) + QJCo2 RT\n (1 - Xe) + nx2
36 9%
(e) The fraction remaining as interfacial energy
Combining the above equations gives
420.3
ir
-
(f) When the precipitate spacing is 1 im; 1
No of particles per m3
(1 x 10~6)3
53 .
(a) AGn
RTln - per mole of precipitate
1 x 1018m"3
Thus for a precipitate with X{)
Using the same method as in (c), the particle radius is found to Thus in 1 m3 of alloy the total interfacial area =
(b) Assuming that the nucleus is spherical with a radius
7
1 x 10i8 x 4n x (2.67 x 10 )2 = 8
orientation, the total free energy change associated with nucieation may be defined as
96 x 105 m2
Total interfacial energy =1.8 x 10 J irT3 alloy .
and ignor-
ing strain energy effects and the variation of y with interface
.
= 1
1 andZe - 0.02 at 600 K:
.
AGn = 8.0 kJ mol
be 267 nm.
~
0
4
AG = --7W3-ACV + 4nr2y
8 J mol"1
Fraction remaining as interfacial energy = 0.4%
where AGv is the free energy released per unit volume. Differentiation of this equation yields the critical radius r*
52 .
m
r
G
0 50 nm .
AG n
a
(c) The mean precipitate radius for a particle spacing of 50 nm was calculated as 13.4 nm = 27 r*. For a 1 jam dispersion the precipitate radius, 267 nm = 534 r*.
Go
03
E3
54 .
Mi
Gr
CQ
a
O
P A
X
,
B
US
AG0 = G0 - Gf
G0 = XQv.l + (1 - Z0)nO Gf = Zon| + (1 - x0) \i%
Xe
Xo
496
Solutions to exercises
Solutions to exercises
From Equation 1.68
n£ = GB + = Gb
i
5
f)
.
497
(a)
rinyoZo
V
+ RT\njcXc
where y0 and ye are the activity coefficients for alloy compositions X{) and Xe respectively
I
AGn
0
e
Co
dti YO O yeXe
For ideal solutions y0 = ye = 1
For dilute solutions yo = Ye = constant (Henry's Law) In both cases i
AG n
RT\n
t
t
(b) 5
5
.
(a) Consider equilibrium of forces at the edge of the precipitate:
a -
X
full
L
v
8 x
Using the simplified approach, above, the carbon concentration gradient in the austenite, - may be expressed as due
For unit area of interface Yaa
L
= 2ya(3COSe
For unit area of interface to advance a distance dr, a volume of
9 = cos"17""
= 53 1
material 1. dx must be converted from y containing CY to a containing Ca moles of carbon per unit volume, i.e. (Cv - Ca)dbc
°
.
moles of carbon must be rejected by diffusion through the y. . . The flux of carbon through unit area in time dt is given by V
(b) The shape factor 5(0) is defined as 5(8) = ~(2 + cos8)(l - cos0)2 = 0 208 .
D{dCtdx) dt, where D is the diffusion coefficient. Equating the two expressions gives
(Cy - Ca) dx
dt
Solutions to exercises
Solutions to exercises
498
* =
dt
d(\dx )
(D ~ ?v
l
\;
L
)*s
-
(C7 - Ca) where / is the volume fraction of austenite
Thus using the simple concentration profile obtained earlier
dr
,
:
X
'
.
L
.
3 D *max = (1 - f}.;>),
UCy - Co) 2
2(C0 - Ca)x
D
(Cy - Co)
(The same answer is obtained for any polyhedron.) Approximately, f-, is given by
Substituting for L in the rate equation
D(C, - Cp)2
dx
.
J (CY-Cn:)
The width of the diffusion zone L may be found by noting that conservation of solute requires the two shaded areas in the diagram to be equal (Co - Ca)x
499
_
0
XfL
Xy
- Xa
"
dr
2(Co - Ca)(CY - Ca)x
Y
Assuming that the molar volume is constant, the concentrations may be replaced by mole fractions (X = CV ). Integration of the rate equation gives the half-thickness of the boundary
In the present case /7 *max
slabs as X
= 0
.
43. such that for D - 300 jim;
= 36 5 fim .
This value will be approached more slowly than predicted by the (Xy-X
V
parabolic equation, as shown schematically in the diagram below.
Dt)
(xt) - xay*{xy - xa)™ 40-
(c) The mole fractions in the above equation can be replaced approximately by weight percentages. For ferrite precipitation from austenite in an Fe-0.15 wt% C alloy at 800 0C we have ,
"1
O? .
xn
0 15; .
giving x
1
.
Parabolic equation
\ Real variation (schematic) 20-
.
3 x HT12
% 304
i
:
0 02;
Dl
Maximum half-thickness
"
*
m
49 x 10""6
"
t
s
- I
1/2
10-
(d) The previous derivation of x(t) only applies for short times. At longer times the diffusion fields of adjacent slabs begin to overlap reducing the growth rate. The lever rule can be used to calculate the maximum half-thickness that is approached for long times. Assume the grains are spherical with diameter D. When the transformation is complete the half-thickness of the ferritic slabs Omax) is given by
0 0
100
200
300
400
500
600
700
Time (S)
The exact variation would require a more exact solution to the
Solutions to exercises
Solutions to exercises
500
diffusion problem. However, the approximate treatment leading to the parabolic equation should be applicable for short times. 57
501
59 .
v
.
E
a
0)
u
X
Nucleation or growth rate
Consider unit area of interface perpendicular to the diagram: Civilian transformations that are induced by an increase in temperature show increasing nucleation and growth rates with increasing superheat above the equilibrium temperature (Tc). This is because both driving force and atom mobility (diffusivity) increase with increasing AT.
Mass flow in the direction of u = u x h: Mass flow in the direction of v"= v x X. From the conservation of mass: u x h = v x X u x h v
5
.
8
5
.
10
(a)
G = XAGA + XBGB + QXAXB + RKXA\nXA + XBlnXB) GA = GB = 0 gives: G = QXaXq + RT(XA\nXA + XBlnXB)
/= 1 - exp(-tfO
dG = ClXAdXB + nXBdXA
At short times this equation becomes
+ RT[dXA + dXB + \nXAdXA + \nXBdXB}
n
/= Kt
but
(a) Pearlitic nodules grow with a constant velocity, v. The voiume fraction transformed after a short time t is given by
XA + XB - 1 -
.
d A + dXB = 0 dG
/4tcv3\ 3
47t(v03
/
.
_
d e
-
b) + RT(\nXB - \nXA)
d2G
e
.
= Q( A
. -
2Q + RT
K =
-r 3 3d
.
n = 3
{ + A) xB
dZg d2G
RT
dXi
XAX]B
-
2Q
(b) For short times, slabs growing in from the cube walls will give
(b) This system has a symmetrical miscibilitv gap with a maximum at . /6v\
i2-vt 6d-
f i e .
d3
_
Ui
=
XA = XB = 0.5 for which t
d2G ART - in
.
6v
K = a
.
n = 1.
It can be seen that as T increases
d2G
d
changes from negative to
Solutions to exercises
502
Solutions to exercises
positive values. The maximum of the solubilitv gap (T = Tc)
(d) The locus of the chemical spinodal is given by
d2G
corresponds to
503
0
dA-B .
2R
e
.
RT
2O = 0
-
dG
A A A'b
(c) Equating -to zero in the equations gives dX B
T
= 4
-
Q(XA - XB) + RT(\nXQ - \nXA)
0
Putting Q = 2RTC gives T _
B(1
b)
-
This is also shown in the figure.
2(1 - 2XB) 5 11 .
.
G o + AA ) = G(Xn) + - (AX) +
-T1 + . . .
G(X{) - AX) = G(X,) + (-AX) +
This equation can be used to plot the coordinates of the miscibility gap as shown below:
!-p + . . .
Total free energy of an alloy with parts of composition (Xt) + AX) and (Xu - AX) is given by -
G(XI) + AX)
,
G(Xn - OX)
Miscibility
lnr,Y
,
(kv
.
(AX)2
= G(X0) +
gap r
Original free energy = G(A',)) 1 d2G
Change in free energy 05
2
2 dX
-
.
Chemical
spinodal
5
12
.
Equation 5.50 gives the minimum thermodynamically possible wavelength
m
in
as
IK
-
Iv E'V
0 0
0 5 .
is a positive constant, while d2G/dJt'2 varies with com-
position X& as shown below:
Solutions to exercises
Solutions to exercises
504
505
Chapter 6
d2G dx2
6 1 .
0 05 .
B
At 7> T0 Chemical
spinodal
Coherent
spinodal
G mm
"
G
0) 03 0) 0)
At T = "r0 1
05 .
e
Thus Amin = * at the coherent spinodal, but decreases as XB increases towards 0.5, as shown schematically above. The wavelength
that forms in practice will be determined by a combination of thermodynamic and kinetic effects, but qualitatively it will vary in the same G
way as A.mm. 5
13
.
(a) Massive transformations are classified as civilian nucleation and growth transformations which are interface controlled. This is because massive transformations do not involve long-range diffusion, but are controlled by the rate at which atoms can cross the parent/product interface (see also Section 5.9). (b) Precipitation reactions can occur at any temperature below that marking the solubility limit, whereas massive transformations cannot occur until lower temperatures at least lower than r0 (Fig. 5.74). Massive transformations therefore occur at lower temperatures than precipitation reactions. However, at low temperatures diffusion is slow, especially the long-range diffusion required for precipitation. Massive transformations have the advantage that only short-range atom jumps across the parent/ product interface are needed. Thus it is possible for massive transformations to achieve higher growth rates than precipitation reactions despite the lower driving force.
'
A
MT=M5(
AG7 '
G
B
X
%Ni
For an alloy of composition X, at T > r0, the free energy curve for a lies above that for y, thus austenite is stable at this composition and temperature, and the martensitic transformation is unable to occur. At a temperature T = T0 the a and a' free energy curve coincides with that for y and so at this temperature and composition both the ,
Solutions to exercises
506
Solutions to exercises
martensite and austenite have equal free energy, and there is no
6
.
4
driving force for the martensitic transformation. At a temperature T = Ms the y free energy curve lies above that for a, therefore y is thermodynamically unstable, and there is a driving force for the martensitic transformation proportional to the length AB. The significance of the Ms temperature is that it is the maximum temperature for which the driving force is sufficient to cause the martensitic transformation. No such driving force is present at temperatures above Ms. At the equilibrium temperature Tq, AG for the transformation is zero, thus '
AGy~a = AHy-a - TcfiS = 0 '
AH f~a at
Tn, AS
To
.
For small undercoolings AH and AS may be considered to be independent of temperature, thus the free energy change may be expressed in terms of the undercooling as follows: AGy-a = AHy-a
'
(Tp - Ms) To
at the Ms temperature. The driving force for the martensitic transformation has been shown to be proportional to the undercooling (r0 - A/s), where T{] is
the temperature at which austenite and martensite have the same free energy, and Ms is the temperature at which martensite starts to form. In the Fe-C system both To and Ms fall with increasing carbon content, with an equal and linear rate. Thus the difference (r0 - Ms)
6
.
5
507
The habit plane of martensite is a common plane between martensite and the phase from which it forms which is undistorted and unrotated during transformation. Thus all directions and angular separations in the plane are unchanged during the rransformation. The martensitic habit plane may be measured using X-ray diffraction and constructing pole figures. The figures are analysed and the plane index may be determined by measuring the positions of diffraction spots from martensite crystals produced from austenite crystals. The main reason for the scatter in the measurement of habit planes is that the martensite lattice is not perfectly coherent with the parent lattice, and so a strain is inevitably caused at the interface. This may act to distort the habit plane somewhat. Internal stress formed during the transformation depends on transformation conditions. Habit plane scatter has been observed to increase when the austenite has been strained plastically prior to transformation, indicating that prior deformation of the austenite is an important factor. Another reason for the scatter is that during the formation of twinned martensite, the twin width may be varied to obtain adjacent twin widths with very low coherency energies. Experimental studies have shown that the lowest energy troughs are very shallow and quite extensive, enabling the production of habit planes which may vary by several degrees in a given alloy.
The key to the phenomenological approach to martensitic transformations is to postulate an additional distortion which reduces the elongation of the expansion axis of the austenite crystal structure to zero. This second deformation can occur in the form of dislocation
slip or twinning as shown below:
remains constant for different carbon contents, which means that the
s
driving force must remain constant. 6
2
.
Z
7
See Section '-.3.1 (p. 398).
i
J nucleus
i
7 Z
t
Slip
1
-
2y c s
a
#
167n(S/2)2 (ACV)2
T
J Twinning
Substitution of the values given gives AG* - 3.0 x lO"18 J nucleus"1 0 23 nm
c
a
.
*
= 8 5 nm .
Austenite
Martensite
Solutions to exercises
Solutions to exercises
508
Dislocation glide or twinning of the martensite reduces the strain of the surrounding austenite. The transformation shear is shown as S. Both types of shear have been observed under transmission electron
509
content which means that the austenite is not as uniformly or as efficiently eliminated as with lath martensites. Plate martensite is
formed by a burst mechanism, this factor contributing to the fact that
the habit plane changes to {225}, and to {259} with even higher
microscopy.
carbon content.
Similar arguments may be used to explain the change in habit plane with increasing Ni content in FeNi alloys since Ni acts in a similar way to C. lowering the Ms temperature and influencing martensite
66 .
,
S5
morphology and amount of retained austenite. C
The amount of retained austenite is also influenced by the austenitizing temperature since this influences the amount of dissolved iron
carbide. The quenching rate is also important an oil quench will produce more retained austenite than a water quench. ,
a
a
,,
.
V2
69
See Section 6.4 5
6 10
See Section 6 7
6 11
See Section 6.7 4
.
Martensite
Austenite
i
.
.
Assuming that ay = 3.56 A and aa = 2.86 A. and that da for martensite is equal to 1.1, the movements of atoms in the c and a directions may be calculated
}
aa = 2.86 A
. ca = 3.15 A
56 A
.\ - 7 = 2.52 A
a
y
= 3
.
.
Vertical movement of atoms = 3.56 - 3.15 A = 0 41 A .
o
Horizontal movement of atoms = 2.86 - 2.52 A = 0
.
34 A
Thus by vector addition, the maximum movement is found to be 0 6 7 .
6
8
.
.
53 A
See Sections 6.32 and 6.33.
The habit plane of martensite is found to change with carbon and nickel contents in FeC and FeNi alloys respectively. This may be explained by considering the nature and the method of formation of the martensite which is dependent on alloy content. In low-carbon steels the Ms temperature is high and martensite
forms with a lath morphology growing along a {111} plane." Growth occurs by the nucleation and glide of transformation dislocations. However, as the carbon content is increased the morphology changes to a plate structure which forms in isolation. The degree of twinning is higher in this type of martensite. An important difference in this process is that the Ms temperature is lowered with increasing alloy
.
.
.
.
.
.
Index relaxation time. 72 self-diffusion, 75. 78. 79 short-circuits, 98
Gibbs-Duhem relationship. 54
Gibbs free energy, 1 of grain boundaries. 122 of interphase interfaces, 146 of mixing, 12 of solid/liquid interfaces. 171
steady-state. 69 substitutional, 61, 75
Index
surface. 98 X
Activation energy, 55, 67, 172 barrier, 56 interstitial diffusion. 68 substitutional diffusion, 75
Activity, 21 coefficient, 22
Age hardening, alloys. 291
i
1
aluminium-copper alloys, 307 Antiphase domains, 363 Arrhenius rate equation, 56 Avrami equations, 290 Bain deformation, 391
Bainite, 334 lower, 337
upper. 334 Binary phase diagrams. 33 Broken-bond model. 114
of surfaces, 113
tracer. 94
of twin boundaries, 123
Diffusional transformations in solids, 263 Diffusionless transformations. 382
Gibbs-Thomson effect, 46 Glissile interfaces, 163. 172, 409. 413
Common tangent construction. 31. 50 Components of a system, 1 single-component systems, 4 two-component systems, 11 Constitutional supercooling, 215
Discontinuous precipitation, 322
Gradient energy, 311
Dislocations in martensite transformation, 401
Grain boundaries. 116
Constrained misfit, 158
transformation, 409 Divorced eutectic, 229
twin boundaries, 122
Cold cracking, 376 Columnar zone, 235
up-hill, 60, 96. 308 vacancy, 79
Down-hill diffusion, 60, 309
Controlled transformation steels. 430
Driving force for precipitate nucleation
Cooperative growth, 222 Coring, 229 Correlation factor. 75 Critical nucleus size for martensite nucleation. 400
for precipitation, 267. 272
268
Enthalpy of formation of borides, 424, 425 of carbides, 424. 425 of nitrides. 424. 425 of vacancies. 43, 76
Entropy. 1. 6
Carbides, 422
Darken's equations, 88
of fusion. 11
Carbon equivalent, 376
Dendrites cellular, 219 columnar. 235
of mixing, 13. 14 of vacancy formation. 43
Cellular precipitation. 288, 322 Cellular solidification. 214 Cellular transformations, 288 Cementite, 422
crystallography. 206 equiaxed, 234, 236 secondary-arm spacing, 221
Chemical potential. 16
thermal, 206
gradient, 60
Chemical spinodal, 309 Chill zone, 234 Civilian transformations, 172
Classification of phase transformations, 173
Clausius-Clapeyron equation, 9 Coarsening
of grain size, 131 of particle dispersions, 314 Coherency, 143 loss, 161 strains. 155, 310. 398 Coherent
interfaces, 143
tip velocity, 207 Diffuse solid/liquid interfaces, 169, 198 Diffusion. 60 atomic mechanisms, 61
controlled growth. 105, 173, 175, 279 controlled lengthening of plates and needles, 283
dilute substitutional alloys, 91 dislocation-assisted, 102
Epsilon carbide, 417, 421 Epsilon martensite, 402 Equiaxed zone, 236 Equilibrium, 1 freezing range, 216 heterogeneous systems, 28 shape of crystal, 115 shape of precipitates, 149, 154, 179
,
low-angle, 116 special, 122 Grain boundary aliotriomorphs, 317 energy, 117 junctions, 124 migration, 130 mobility, 135 segregation, 138 Grain coarsening, 131. 140 Grain growth. 131, 139 abnormal, 142
during tempering of steel, 426
Growth fedges. 179 199, 285
Guinier-Preston (GP) zones, 149, 291 equilibrium shape of. 158 Habit plane, 153
of Widmanstatten plates 150, 152, ,
317
of martensite laths and plates
Henry's law, 22 Heterogeneous nucleation (see
.
grain boundary, 98
Fibrous precipitation, 349
390.
Hardenability, 338
impurity effects, 229 Euteetojd transformations, 263, 326 Extensive thermodynamic properties, 4 .
,
396
Heat flow, 203. 239 Helical dislocations, 303
Ferrite, 317
multiphase systems, 103 nonsteady state, 69
in nucleation. 271
solidification, 208 Eutectic solidification, 222
down-hill, 60, 309 Pick's laws, 65, 71, 88 interstitial, 61, 63
high angle, 118 in cellular precipitation. 322
in precipitate nucleation. 274 Shockley partials, 164
Continuous casting, 238 Continuous growth process, 198
CCT diagrams, 346
f
ternary effects. 96
pressure effects. 7 temperature effects, 4 Gibbs phase rule, 36
spinodal, 312
for solidification, 187. 193, 196
Carburization, 73
511
Nucleation, heterogeneous) Homogeneous nucleation (see Nucleation, homogeneous) Homogenization, 71
Pick's law of diffusion, 65, 71, 88 First order transformations, 359
Ideal solutions, 13
Free energy-composition diagrams, 15
Inoculants, 196
Ingot structure, 233
v
-
Index
512
Intensive thermodynamic properties, 4 Interdiffusion coefficient. 88 Interface
coherence, 143 .
175,
285
Order
long-range order parameter 358 short-range order parameter 24 ,
513
Raouh's law, 22 Recalescence. 346
Mf temperature, 383, 386 V/ temperature. 383, 386
Order-disorder transformations 263. 358
Recrystallization 138, 288 426
nucleation, 397
Orientation relationship
Retained austenite 383, 426 Reversion 301
,
S
,
144 of some martensites 390 .
role of grain size. 416
Recovery during tempering
426
,
.
,
.
Massive transformations, 263, 288, 349
migration. 171 mobility, 172
,
Overageing, 306
Richard's rule 11 ,
Mechanical properties
reaction, KS6
of age hardening alloys. 294
stability during solidification, 203
of controlled transformation steels, 434
Interfaces. 110 coherent, 143
complex semicoherent. 148 effect on equilibrium, 44 free energy of, 110
of titanium alloys, 372 Metastable equilibrium, 2 Metastable (transition) phases, 2, 292
interphase, 142
solid/liquid, 168. 197 solid/vapour, 112 Intermetailic compounds. 27 Interphase precipitation, 349
382
Misfit parameters, 157 Mobility
compounds, 27 diffusion, 61, 63
,
of twin boundaries, 136
sites in cubic crystals, 385 Invariant plane strain, 391
Mushy zone, 236
Kinetics, 55
of grain growth. 139 of phase transformations. 287 Kirkendall effect, 89
Kurd]umov-Sachs orientation relationship, 148, 317 394 ,
Nishiyama-Wasserman orientation relationship, 148. 317. 394 Nitinol, 431
Non-equilibrium lever rule, 212 Nucleation, heterogeneous. 192 activation energy barrier, 193, 195, 272
Latent heat
in liquids, 185
of fusion, 7, 112 170
in solids, 271
.
of melting, 7, 112
,
170
of sublimation 112, 170 of vaporization 112 Lateral growth, 178 198, 285 Lath martensite, 410 ,
,
,
Laves phases, 27 Ledge mechanism 178, 198, 285 Local equilibrium 97, 103, 177 210, ,
,
224, 279
of martensite, 400 on dislocations, 274
on grain boundaries, 271 rate of, 194, 276
vacancy-assisted, 275 Nucleation, homogeneous activation energy barrier, 187, 266
in liquids, 185, 186 in solids, 265 of martensite, 397 rate of, 191, 267
Long-range order, 358 Lower bainite 337 ,
151
Second-order transformation 361
Pearlite, 288 326
Secondary hardening in high-speed tool
,
,
growth, 330
steels, 423
in off-eutectoid alloys
,
333
Peritectic solidification 231
420
,
Phase diagrams, 33
Seif-diffusion, 75 "
Al-Cu, 291 binarv, 33
activation energy of, 76 experimental data, 78
Cu-Zn,353
Semicoherent interfaces 145
eutectic, 36. 51
Shape of inclusions and precipitates coherent precipitates 155
,
.
Fe-Cr-C, 432. 433 Fe-Cr-Ni, 250. 257
grain boundary effects, 154 incoherent precipitates and
Fe-Mo-C, 418
inclusions 159 ,
Fe-Cr-Mo-W-V-C 253
interfacial energy effects. 149
,
Mg-Al. 326
misfit strain effects 155
ternary, 48
electron, 28
plate-like precipitates. 160 Shape-memory. 431 Shockley partial dislocations Short-range order, 24
intermediate, 26
Shrinkage in ingots and castings
Laves, 27
Site saturation, 288 Solid solutions
,
Ti-Ni/436 Phases, 1
,
effect of external stresses 415 ,
Off-eutectic alloys, 229 Off-eutectoid alloys, 333
164
,
metastable, 2, 292
237
binary. 11, 52
ordered, 24, 35 transition. 292
carbon in iron, 285
Plate martensite, 412
free energy of, 11
Polymorphic transformations 263 Precipitate coarsening, 314 growth, 279, 283 Precipitate-free zones 304 Precipitation, 263 in Al-Ag alloys, 302 in Al-Cu alloys, 291 of a from p brass. 349 ,
,
of ferrite from austenite, 317
ideal, 13
interstitial, 24 ordered, 23, 24
quasi-chemical model, 18 real, 23
regular. 18, 41 solubility as a function of temperature, 41 Solidification, 185
alloy, 208
Pre-martensitic phenomena. 416
carbon steels, 249
Pro-eutectoid ferrite 317 Pro-eutectoid cementite 322
castings, 233 driving force for, 10
Quenched and tempered steels, 428
eutectic, 222 fusion welds, 243
Quenched-in vacancies, 303
high-speed steels, 251
,
Martensite, 382 ,
Segregation in ingots and castings 237 Segregation of carbon duirno temperins .
,
crystallography 389
Schaeffler diagram, 257 Scheil equations, 212
Pasty zone, 236
Fe-C,250. 318
of atoms, 92
of glissile interfaces. 172 of grain boundaries, 135 of interphase interfaces 172
Interstitial
,
,
.
,
coherent, 312 incoherent, 313 Misfit dislocations. 145
irrational, 149, 167 semicoherent, 145
Partially coherent precipitates Particle coarsening 314
nucleation. 327
Microstructure, 110
Military transformations, 173 Miscibility gap, 33, 308
incoherent, 147
tit
growth, 409 habit planes, 389, 390, 396 ;
controlled growth, 106. 173
S
Index
.
.
.
.
rem
Index
514
recrystallization 138 Thermal activation 56, 66, 172
ingots, 233
,
low-alloy steels, 249 peritectic, 231 rapid, 249 shrinkage, 237 single-phase alloys, 208
,
Thermodynamics, 1 Ti-6V-4A1 alloys, 366 Tie-lines, 50
Torque term, 126
stainless steel weld metal, 256
Transformation shears, 337, 383 Transformation dislocations, 409
unidirectional, 208
Solubility product, 426
Transition phases, 292
Solute drag, 138 Spinodal decomposition, 308 Spiral growth, 201
Twin boundaries, 122 in solidification 202
Stabilization of austenite, 415
Unconstrained misfit, 155
,
Stacking faults, 167 in martensitic transformations. 41)2
,
Up-hill diffusion, 60, 96, 308 Upper bainite, 334
404
in precipitate nucleation. 273, 276
.
303
Concentration, 43
Stirling's approximation. 14
diffusion, 79
Substitutional diffusion, 75 Surface nucleation, 200 Surface tension. Ill
formation enthalpy 43. 76 formation energy 43, 76 formation entropy 43 jump frequency. 79 quenched-in, 303 Valency compounds 28 ,
,
,
TTT diagrams, 287, 301 339 Temper embrittlement. 427 .
,
Tempering of ferrous martensites Ternary alloys. 48
.
diffusion in, 96 Texture deformation, 138
Vacancy
417
Weldability, 372
Widmanstatten side-plates
,
317
Widmanstatten structures. 153. 279 318 ,
Wulff construction 115 ,
