Page 94
This problem involves plotting the freeenergy of an ideal solid solution against the mole fraction, N A
between N A = 0 and 1.0. The basic equation involved in obtaining the above gure is
G, = NAGA+NBG93+RT(NA1nNA+N31nNB)
where G0 = 7 000 J/mol, G

Page 50
Now solve for p
where do is the intercept of the straight line, through the data in Fig. 5,35, with the stress axis. The
value of arc, is 39 kg/mm2 and k = 2.55-10"l kg2-cm. Accordingly for the 60 kg/rm-n2 stress we
mm
have
_ 6039 2_ 9 3_ 13 3
g

Page 48
area, A5, or
P
of = " = _100_ = 1.273 .107 Pa. = 12.73 MPa
A.- (1: - (.01?)
4
al area minus the nal area divided by the initial
(c) The reduction in area, RA, is dened as the initi
0-7 A:
areaor
A.-A
where RA = 0.3, Ai = 7.85 - 10'5 m, thus h 3

Page 43
muss Fri Jun 07 xa-Jtzli Isul
5.14 (a) The diagram accompanying this problem shows a 111 standard projection of a fee
crystal in which the standard stereographic triangles are outlined. Assuming that point a in the gure
represents the orientatio

Page 45
5.16 For the case of tensile deformation considered in Prob. 5.14, determine the indices of the
primary, conjugate and cross-slip systems as well as those for the critical plane.
Solution:
(a) The primary system is (111) [011].
(b) The conjugate s

Page 52
6.2 According to quantitative metallography, N1, the average number of grain-boundary
intercepts per unit length of a line laid over a microstructure is directly related to S, Wrea
f as:
per unit volume, by the relation
A~f f
Su=2NI
(3) Determine

Page 47
Solving for v and substituting the given values of p, b and I) gives
e 0.00167 -12 -1
=_=0.734.10 m-s
1 p~b 1014-0248-10-9
-7
10_T=4.032-1012 m-s'12
10 0243-10
&1
111 =
5.20 Necking of a tensile specimen begins at a true strain of 0.20. The corr

Page 72
next gure shows the required plot.
run museum-I
N
1.0
0.4
Relative graln boundary energy, Im / I
0.2 0 8
D
an 5 1D 8202530354045505550557075803590
Angle of h1tin degrees
This plot is quite similar to that in Fig. 6.8. The data is, however, skewe

/
12.2, A diffusion couple, made by weldin a thin one-centimeter uare slab of pure metal A to a
similar slab of pure metal B, was given a diffusion anneal at an elevated temperature and then cooled
to room temperature. On chemically analyzing successive l

8.2 (a)Using the 283K data, given ain Sec. 8.5 for the Drouard, Washburn and Parker
zinc single crystal, determine the value of A in the rate equation, Eq. 8.2.
(b) Next determine the temperature at which the crystal should recover one-fourth
of its yield

6.1 (a) Given a small angle tilt boundary, whose angle of tilt is 0.1 deg., find the spacing
between dislocations in the boundary if the Burgers vector of the dislocations is 0.33 nm.
(b) On the assumption that the dislocations conform to the conditions i

7.1 Gold has a meltng point of 1,063 and its latent heat of fusion is 12,700 J/mol.
Determine the entropy change due to the freezing of one mole of gold and indicate
whether it is positve or negatve.
7.8 Compute the equilibrium concentraton of vacancies i

6.1 (a) Given a small angle tilt boundary, whose angle of tilt is 0.1 deg., find the spacing
between dislocations in the boundary if the Burgers vector of the dislocations is 0.33 nm.
(b) On the assumption that the dislocations conform to the conditions i

Chapter. 3
3.9 Do you except a correlation between bonding energy and melting points of
metals? Justify your answer.
Solution.
,. .,
.(bonding energy)
(melting point)
3.10 The lattice energy of an ionic solid, U, is the amount of energy required to
separa

7.1 Gold has a melting point of 1,063 and its latent heat of fusion is 12,700 J/mol.
Determine the entropy change due to the freezing of one mole of gold and indicate
whether it is positive or negative.
7.8 Compute the equilibrium concentration of vacanci

IM [a] How rnan'g.r eduivalent cfw_111: 115 :3 slip svsterns are there In the fcc lattice?
[bl Indentifv each svstem bv writing out its slip plane and slip direction indices.
Solution:
[a] The various slip svsterns are listed below.
(111J[1Io] (TI1)[110]

5.5 Defamation twins are also able to form along the cfw_111 planes of fee crystals, as a re5ult of the
application of a shear stress across this type of plane. In twinning. the shear directions are
< 211 as.
cfw_a Prove. using Eo. 5.3, that the [1Z1]ancl

/ 12.2., A diffusion couple, made by welding a thin one-centimeter mug-g slab of pure metal A to a
similarFsl/ab of pure metal B, was given a diffusion anneal at an elevated temperature and then cooled
to room temperature. On chemically analyzing successi

Page 74
end points. Thus, the upper left point correspdnds to 1 / T = 1 / 283 = 0.00353 and 111(1/7') = ln(1 / 0.007)
= 4.96 while the lower right point corresponds to l/T = 0.00492 and ln(,1/ 1') = - 1.88. The slope is
accordingly
Q 4.96 ( 1.88) _ 6.84
5

Page 80
PROBLEMS CHAPTER 9
9.1 At room temperature the stable crystal structure of iron is bcc. However, above 1183 K it
becomes fee. The iron fcc crystal structure is able to dissolve a much larger concentration of carbon
than is the bcc structure. A pri

Page 82
9.3 An approximate equation givingthe atom fraction of carbon soluble in aust ' e, or the ne
form of iron, when the carbon comes from cementite, is
28 960 a j" *"t 7
Cice=1-165'm(-ii;r Li; 1;) 4K )
. J .J "
Where Q is in J / mole, R in J / K mole

Page 84
Solution:
The constants for this problem, as given in the text, are
b=2.48-101m A=1.84-1o29 Pa-m4 k=1.381-10"23
The basic equation for the time to traverse the solute path becomes, on substituting the values of the
parameters;
k 1.381-10-23 T
i

Page 88
PROBLEMS CHAPTER 10
10.1 List and identify the phases in Fig. 2.27.
Solution:
The phases are
(1) An intermetallic phase, Nb3Si3.
(2) An intermetallic phase, Nb3Si.
(.3) A Nb rich. solid solution.
_"
10.2 The three elements zirconium, titanium, and

Page 92
10.7 Compute the activity coefcients corresponding to the four activities involved in Probs. 10.4
and 10.5.
Solution:
The activity coefcient is dened as the ratio of the activity to the mole fraction; that is
7A T and 7B = TB:
Thus for the (A) all

Page 86
The above values should be substituted into the CottrellBilby equation that gives f, the fraction of
solute precipitated. This is Eq. 9.32 which reads as follows
2
5
f = apcfw_Ak11).t .
The answer to the problem is f = 0.039 or 3.9 percent.
_,4.-

Page 96
PROBLEMS CHAPTER 11
11.1
GAUSS Fri Dc! 25 09:03:21 I991
Temperature, C
0
A Weight percent component B
The points a, b, c, d, and e on the 60 percent component B line give some temperatures through which
a slowly cooled alloy containing 60 percen

Page 98
\/ (c) Just above the eutectic temperature at 779C the structure will consist of primary solidfigendtes
_of composition? percent Cu and 8 percent silver in a, liquid matrix of the eutectic composition, 28.1
percent copper and 71.9 percent silver

Page 90
GB = 693+RTIMB
Therefore,
G = NA091+ N3033+ RT(NA1naA+ NBlnaB)
=70,000 J/mol, R=8.314 J/Kmol, T=1000 K, GA: 0.50 and
where GE, = 50,000 J/mol, G33
ee energy equation for the solution results in
a3 = 0.90. Inserting these values into the fr
G, = 0.

Page 56
and with x = 2 and y = 1 this yields
2=2z+12-3=7
(b) The angle 9 is determined with the aid of Eq. 6.21
9 = 2tan_1(y']:1/i) = 2tan_1(k-)= 2.40.9= 81.8
The cfw_111 plane in a fcc metal has a sixfold symmetry. If we subtract 60 from this value of

Page 70
to determine the value of the constant A. On Page 233 it is stated that Q = 83,140 J /mol, and at a
temperature of 273 K, the yield point recovers 1/4 its original value in a period of 5 min or 300 s.
Inserting these data in Eq. 8.2 results in
83,

Page 58
PROBLEMS CHAPTER 7
7.1 Gold has a melting point of 1063C and its latent heat of fusion is 12,700 J /mol. determine
the entropy change due to the freezing of one mole of gold and indicate Whether it is positive or
negative.
Solution:
First convert

Page 64
have
ASm - 2(8.37) cfw_0.5m 0.5 + 0.5ln.5] = 11.53 J/K for the two moles.
(6)
AU = A6) + AW = 0
(1) Now compute the change in free energy AG:
AG=AQTAS = TAS
AG = 293(11.53) = 3,436 J
This change in the free energy is large enough to be signicant.