... High-strength; high-temperature.
(free- flowing) + 16 Zn + 25 Cd
Silver; general-purpose 38 Ag + 20 Cu 605–650 High-strength; high-temperature.
(pasty) + 22 Zn + 20 Cd
Case studies in phase diagrams ... hot-water layout, you will already have had some direct experience of
this system.
22 Engineering Materials2
Fig. 2. 7. Many metals are made up of
two
phases. This figure shows some of the shapes ... (Fig. 2. 2c). But if A atoms prefer to have A neigh-
bours, or B atoms prefer B neighbours, the solution can cluster (Fig. 2. 2d); and when A
atoms prefer B neighbours the solution can order (Fig. 2. 2e).
Many...
... ∆H = −334 kJ kg
−1
. For water at 27 2 K, with T
m
− T = 1 K, we find that W
f
=
1 .22 kJ kg
−1
(or 22 J mol
−1
). 1 kg of water at 27 2 K thus has 1 .22 kJ of free work avail-
able to make it turn ... Conservation of volume gives
4
3
4
3
4
3
3
3
1
3
2
3
πππ
rrr =+
. (5.30)
Combining eqns (5 .29 ) and (5.30) gives
∆A =
4
1
3
2
323
1
2
2
2
πγ
[( ) ( )].
/
rr rr+−+
(5.31)
For r
1
/r
2
in the range ... coarsening process? If we put r
1
= r
2
/2
in eqn. (5.31) we get ∆A = −4
πγ
(−0.17
r
2
2
). If
γ
= 0.5 J m
2
and r
2
= 10
−7
m our two
precipitates give us a free work of 10
−14
J, or about...
... the vol-
ume of the nucleus. For 0 ഛ
θ
ഛ 90° these are:
solid–liquid area = 2
π
r
2
(1 − cos
θ
); (7.4)
catalyst–solid area =
π
r
2
(1 − cos
2
θ
); (7.5)
nucleus volume =
2
3
1
3
2
1
2
3
3
πr
... (7.1) we can write
W
f
= 2
π
r
2
(1 − cos
θ
)
γ
SL
+
π
r
2
(1 − cos
2
θ
)
γ
CS
−
π
r
2
(1 − cos
2
θ
)
γ
CL
−−+
−
cos cos
( )
.
2
3
1
3
2
1
2
3
3
πr
H
TT
T
m
m
θθ
∆
(7.7)
Note ... the interfacial energy terms to reduce
to
2
π
r
2
(1 − cos
θ
)
γ
SL
+
π
r
2
(1 − cos
2
θ
)(–
γ
SL
cos
θ
),
or
2
π
r
2
{1 −
3
2
cos
θ
+
1
2
cos
3
θ
}
γ
SL
. (7.9)
The complete result...
... 2. 1 41–160 150 25 0
Ti alloys 4.5 120 170– 128 0 27 2. 4 1.1 38 28 0 400–600
(Steels) (7.9) (21 0) (22 0–1600) 27 1.8 0.75 28 20 0 (400–600)
* See Chapter 25 and Fig. 25 .7 for more information about these ... strength
E
/
r*E
1 /2
/
r*E
1/3
/
r* s
y
/
r*
Creep
r
(Mg m
−
3
) modulus
s
y
(MPa) temperature
E
(GPa) (°C)
Al alloys 2. 7 71 25 –600 26 3.1 1.5 9 22 0 150 25 0
Mg alloys 1.7 45 70 27 0 25 4.0 2. 1 41–160 150 25 0
Ti ... reading
M. F. Ashby and D. R. H. Jones, Engineering Materials I, 2nd edition, Butterworth-Heinemann,
1996, Chapters 7 (Case study 2) , 10, 12 (Case study 2) , 27 .
Further reading
I. J. Polmear, Light Alloys,...
... 0 .20 0.55
B 0.40 0.60 1 .20 0.30 1.50
C 0.36 0.70 1.50 0 .25 1.50
D 0.40 0.60 1 .20 0.15 1.50
E 0.41 0.85 0.50 0 .25 0.55
F 0.40 0.65 0.75 0 .25 0.85
G 0.40 0.60 0.65 0.55 2. 55
116 Engineering Materials ... Jones, Engineering Materials I, 2nd edition, Butterworth-Heinemann,
1996, Chapters 21 , 22 , 23 and 24 .
Further reading
K. J. Pascoe, An Introduction to the Properties of Engineering Materials, Van ... started the crack which led to the Alexander Keilland
failure.)
Example 2: Pressure vessel steel to A 387 grade 22 class 2
A 387 22 (2) is a creep-resistant steel which can be used at about 450°C. It...
... concretes 20 9
Fig. 20 .2. (a) The hardening of Portland cement. The setting reaction (eqn. 20 .5) is followed by the
hardening reactions (eqns 20 .6 and 20 .7). Each is associated with the evolution ... 20 .7). Each is associated with the evolution of heat (b).
2C
2
S + 4H → C
3
S
2
H
3
+ CH + heat (20 .6)
2C
3
S + 6H → C
3
S
2
H
3
+ 3CH + heat. (20 .7)
d
Tobomorite gel.
Portland cement is stronger than ... any
cement are, in this nomenclature
Lime CaO = C
Alumina Al
2
O
3
= A
Silica SiO
2
= S
Water H
2
O = H.
20 0 Engineering Materials2
Fig. 19.9. Forming methods for glass: pressing, rolling, float-moulding...