Tutorial 3 mats-404

Faculty of Engineering and Materials Science
Introduction to Materials Engineering (MATS-404)
Tutorials 3
Dipl. -Ing. Ismael Abdel MAKSOUD
Engr. Yasser Sheasha

Exercise 1: Lattice defects
In Figure 1 different lattice defects are displayed.
Classify these lattice defects accordingly. Explain by which processes/mechanisms the lattice
defects presented in Figure 1 can be created?
Figure 1: Different lattice defects

Answer Exercise 1: Lattice defects
Substitutional atom
Interstitial atom
Self/matrix
Interstitial atom
Vacancy
s ta c k in g f a u lt a n ti p h a s e b o u n d a r y p a rtic
(p re c ip it
d is p e rs
tw in b o u n d a r y e d g e d is lo c a tio n s e m i p a
c o h e re n
b o u n
s ta c k in g f a u lt a n ti p h a s e b o u n d a r y p a
(p re c i
d is p
tw in b o u n d a r y e d g e d is lo c a tio n s e m i
c o h e re
b o u
s ta c k in g f a u lt a n ti p h a s e b o u n d a r y p a rtic le
(p re c ip ita t io n ,
d is p e rs io n )

Substitutional atom
Interstitial atom
Vacancy
Self interstitial atom
Substitutional atom:
Interstitial atom
Vacancy
Self interstitial atom

Stacking fault:
Instead of ABCABCABC => stacking order ABCABAABC
Stacking faults develop during crystal growth or because of
interaction with dislocations.
Twin boundary:
Twin boundaries are characterized by a symmetric arrangement of the
atoms with reference to a vertical plane.

Particle (precipitation, dispersion):
Field of lattice with changed structure.
Precipitations are formed in the material because of
thermodynamic equilibrium (formation of a new phase via heat
treatment of a meta-stable solid solution).
Dispersions are extra phase of small particles distributed in
the material.
Edge dislocation:
An edge dislocation is characterized by an
additional half-lattice plane.
It is caused by vacancy diffusion.

Exercise 2: Vacancies
i) State the correlation between the concentration of vacancies cv and temperature. By the aid of
a diagram (plot), describe this relationship qualitatively.
ii) State the order of magnitude of cv in metallic materials at their melting temperature.
iii) Which of the two materials Al and W has at 20°C the higher concentration of vacancies?
*
....
63,000
660 933
f
m
For Al
JH
mol
T C K
∆ =
= ° →
*
.....
300,000
3,350 3,623
f
m
For W
JH
mol
T C K
∆ =
= ° →

Answer Exercise 2: Vacancy concentration
200 400 600 800 1000 1200 1400
-1,0x10
-6
0,0
1,0x10
-6
2,0x10
-6
3,0x10
-6
4,0x10
-6
5,0x10
-6
6,0x10
-6
7,0x10
-6
8,0x10
-6
c
Temperatur T [K]
*
0 exp 8.31B
v
H Jc c with R mol
KR T
 ∆
= × − = × 
Basic equation:

( ) ( )
( )
( )
( ) ( ) ( )
* *
0 0
*
0
*
0
*
4
exp exp
exp
exp
1 1
exp , 10
f f
v v m
m
f
v
v m f
m
f
v v m v m
m
H H
c T c c T c
R T R T
H
c
R Tc T
c T H
c
R T
H
c T c T c T
R T T
−
   ∆ ∆
= × − = × −   
× ×   
 ∆
× − 
× =
 ∆
× − 
× 
 ∆  
= × − ≈  ÷
  

*
63,000
660 933
f
m
JH
mol
T C K
∆ =
= ° →
*
300,000
3,350 3,623
f
m
JH
mol
T C K
∆ =
= ° →
( )
( ) 12
4
10220
293
1
933
1
31,8
000.63
exp1020
−
−
⋅≈°⇒














−⋅=°
Cc
KKmol
K
J
Cc
Al
L
LAl:
W: ( )
( ) 54
4
10720
293
1
3623
1
31,8
000.300
exp1020
−
−
⋅≈°⇒














−⋅=°
Cc
KKmol
K
J
Cc
W
L
L
AL: W:
( ) ( )CcCc W
L
Al
L °>>° 2020

Exercise 3: Correlation between mass, atomic and volume percent
i)The homogeneous copper-based alloys CuSn5 and CuAl5 contain 5 weight - % of tin or
aluminium respectively. Determine the concentration of the alloying elements in atomic - %.
ii) Determine the chemical compositions in atomic - % and in weight - % for the following
intermetallic and intermediate compounds:
a) Titanium aluminide TiAl
b) Iron carbide Fe3C
c) Chromium carbide Cr23C6
- The relative atomic masses for the elements in exercise (i) and (ii) are:
Ar (Ti) = 47.90 Ar (Fe) = 55.85 Ar (Al) = 26.98 Ar (Cr) = 52.00
Ar (C) = 12.01 Ar (Cu) = 63.55 Ar (Sn) = 118.67

Answer Exercise 3: Correlation between mass, atomic and volume
percent
(i) The homogeneous copper alloys CuSn 5 and CuAl 5
contain 5 weight - % of tin and aluminium respectively.
Determine the alloy concentration in atomic - %.
The relative atomic masses are:
Ar (Cu)= 63.55 Ar (Sn) = 118.67 Ar (Al) = 26.98

Formulae
100 . %A
A
A B
m
c wt
m m
= × −
+
Weight concentration of element A:
atomic concentration of element A:
)(
)(
1
%100'
BAc
AAc
atomic
c
rA
rB
A
⋅
⋅
+
−
=

Solution (i)
' 100 % 100 %
2.74 %
( ) 95 118.67
11
5 63.55( )
Sn
Cu r
Sn r
atomic atomic
c atomic
c A Sn
c A Cu
− −
= = = −
× ×
++
××
CuSn 5 CuAl 5
cSn= 5 wt.-% cAl= 5 wt.-%
' 100 % 100 %
11.03 %
( ) 95 26.98
11
5 63.55( )
Al
Cu r
Al r
atomic atomic
c atomic
c A Al
c A Cu
− −
= = = −
× ×
++
××

(ii) Determine the chemical compositions in atomic -% and in weight - %
for the following intermetallic and intermediate compounds:
a) titanium aluminide TiAl
b) iron carbide Fe3C
c) chromium carbide Cr23C6
The relative atomic masses are
Ar (Ti) = 47.90 Ar (Fe) = 55.85 Ar (Cr) = 52.00
Ar (Al) = 26.98 Ar (C) = 12.01

Solution (ii)
A B Ar(A) Ar(B)
c`A
[atomic-%]
c`B
[atomic-%]
cA
[Wt.-%]
cB
[Wt.-%]
TiAl Ti Al 47.9 26.98 50 50 63.97 36.03
Fe3C Fe C 55.85 12.01 75 25 93.31 6.69
Cr23C6 Cr C 52 12.01 79.3 20.7 94.31 5.69

Exercise 4: Different alloy materials

i)The diagrams in Figure 2a and b depict alloys of different kinds. Ascribe to each of the
diagrams their proper designations.
ii) What are the main differences between these two kinds of alloys?
Figure 2. (a) (b)

Figure 2. (a) (b)
Answer Exercise 4: Different alloy materials

i)Figure 2a represents the structure of a heterogeneous alloy while Figure 2b represents the
structure of a homogeneous alloy.
ii) Heterogeneous alloys have grains which are chemically and structurally different.
Homogeneous alloys are solid solutions in which the atoms of combining species are in structura
unity (i.e. atoms of the solute are completely miscible in the lattice of the solvent.

Exercise 5: The phase rule
i)Write the expression of the phase rule and define the variables in this expression.
ii)Explain the significance of the reduced phase rule

Answer Exercise 5: The phase rule
i)Write the expression of the phase rule and define the variables in this expression.
The phase rule is given by: F + P = C + 2
This phase rule is helpful for the physical characterisation of a “thermodynamic” system.
Where C is the number of components in P number of phases having F degrees of freedom.
- The components are the chemical elements of the system. Note that compounds with fixed
chemical composition are treated as one chemical element in the phase rule.
-The phases are areas of materials which are physically and chemically different from each othe
and are separated by interfaces (phase interfaces or boundaries)
-The degree of freedom represents the state variables of a system which can be changed
independently without changing the number of phases. Established state variables are: Pressure
Temperature and the concentration of the chemical species or components present in the system
ii) Explain the significance of the reduced phase rule.
The reduced phase rule provide a reasonable basis for describing real equilibrium system, as is
commonly encountered in technical practice. Since most investigations in materials science are
carried out at standard pressure i.e. p = const. = 1 atm = 1.013 bar.
Since in this case the degree of freedom of pressure is already occupied, thus the result degree
of freedom would be: F* = F - 1. Hence the „reduced phase rule“ is: (F* + 1) + P = C + 2
or F* + P = C + 1

Exercise 6: Change of state in materials

i)Describe, using an appropriate diagram, the transformation of materials from the liquid solid
 gaseous state. (consider the roles of temperature and enthalpy in your answer)
Figure 3. Relationship between temperature T and enthalpy H for a metal without phase transition in the
solid state

Figure 3. Relationship between temperature T and enthalpy H for a metal without phase transition in the
solid state
Answer Exercise 6: Change of state in materials

i)Figure 3 illustrate the transformation of a materials from the solid to the liquid state.

Exercise 7: Construction of phase diagrams
i)Consider the temperature / time relation associated with a two component system AB where
component A has a melting temperature (TM,A) relatively higher than that of B (TM,B). The profile in
Figure 4 shows the influence of concentration of both components on the T – t diagram.
For a system where the concentration of both components are varied accordingly, you are required
to use the thermal analysis information in Figure 5 (next slide) to construct a binary phase diagram
Figure 4. Temperature – time plot for a two component system (p is constant)

Exercise 7: Construction of phase diagrams
i)You are required to use the thermal analysis information below to construct a binary phase
diagram.
Figure 5. Temperature – time profile for a
two component system.

Answer Exercise 7: Construction of phase diagrams
Determining phase diagrams
procedure:
1. Registration of the T, t-curves for sufficient amount of cB with adequately
slow „cooling in equilibrium“
2. Transition of hold points and so called knee points in the T, cB-diagram
(→ phase diagram)
3.Connection of all liquidus and solidus points including melting points of the pure
components A and B by a single line each
------------------------------------------------------------------------------------------------
Liquidus line and solidus line:
• are lines of equilibrium!
• they mark the boundaries of phase fields
⇒ area of melting temperature “M“
(1 phase)
⇒ area of melting temperature and
solid solution “M” + “α” (2 phases)
⇒ area of solid solution “α” (1 phase)

Answer Exercise 7: Construction of phase diagrams
Construction of a binary phase diagram

Tutorial 3 mats-404

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