كانزاسازى Kanza sazi metallurgy كانزاسازى ئەندازیاری کانزاسازی Metallurgy كانزاسازى KanzaSazi كانزاسازى علم المعادن

2
Republic of Iraq
Kurdistan Regional Government
Metallurgy
‫كانساشازى‬

3
Types of Materials
Question: answer by true or false and then rewrite the false statements in
the correct form.
1. The branch of metallurgy that deals with the structure of metals and
alloys and its effect on their physical and mechanical properties is
termed as process metallurgy.
Answer: false …the branch of metallurgy that deals with the structure
of metals and alloys and its effect on their physical and mechanical
properties is termed as physical metallurgy. Page 1
the branch ‫لق‬ deals with ‫َتةوة‬‫ل‬‫دةكؤ‬ structure
‫ثةيكةر‬
alloys ‫ِشتةكاى‬‫ر‬‫دا‬ effect ‫كاريطةرى‬ is termed ‫ناونراوة‬ Atoms ‫طةرديمة‬
2. Materials are generally classified as metals and polymers.
Answer: false… Materials are generally classified as metals, ceramics and
polymers. Page 1
3. Metallurgy is a science that deals with metals and polymers.
Answer: false… metallurgy is a science that deals with metals alone. Page
1
4. polymers are extremely strong

4
Answer: false… polymers are weak and extremely flexible. Page 1
5. Thermoset plastics soften when heated and hardened when cooled.
Answer: false… Thermo plastics soften when heated and hardened when
cooled. Page 2
6. Metals are generally good electrical and thermal conductors due to
the covalent bonded atoms.
Answer: false… Metals are generally good electrical and thermal
conductors due to the metallic bonded atoms. Page 2
7. Thermo plastics are those polymers that can only be heated once to
set into permanent shape.
Answer: false… Thermoset plastics are those polymers that can only be
heated once to set into permanent shape. Page2
8. Isotopes of an element are formed when the number of protons is
not fixed for all atoms.
Answer: false… isotopes of an element are formed when the number
of protons is not fixed for all atoms. Page3 elements‫تومخةكاى‬
9. Atoms of some elements may have different number of protons in
their nucleus producing that are called isotropics.

5
Answer: false… Atoms of some elements may have different number
of neutrons in their nucleus producing that are called isotopes.
Page 3
10. Composite materials
are usually designed to display the combination of the component
materials.
Answer: true Page 3
11. physical and
mechanical properties of engineering materials are derived from the
type of the interatomic bonding
Answer: true Page 4
12. Secondary bonds are
some times called intermolecular bonds.
Answer: true Page4
Generally ‫بةطشتى‬ electrical‫كارةبايى‬ thermal‫طةرمى‬ conductor
‫طةيةنةر‬ due to ‫بةيؤى‬ covalent bond ‫ياوبةشى‬ ‫بةندى‬ extremely ‫ِى‬‫ر‬‫ئةوثة‬
flexible ِ‫ري‬‫ج‬ properties ‫شيفةتةكاى‬ once ‫جار‬ ‫يةك‬ permanent ‫يةميشةيى‬
component materials ‫َهةرةكاى‬‫ي‬ً‫َك‬‫ي‬‫ث‬ ‫مادة‬
13. Hydrogen bonds are
strong and stable interatomic bonding.

6
Answer: false….hydrogen bonds are strong and stable intermolecular
bonding. Page7
14. Anisotropy
represents a state in which the properties are the same in all
directions.
Answer: false… isotropy represents a state in which the properties are
the same in all directions. Page8
15. Crystalline solids are
isotopics because of random arrangement of atoms in all three
dimensions.
Answer: false… Crystalline solids are anisotopics because of random
arrangement of atoms in all three dimensions. Page 8
16. There are only two
atoms contained along [111] directions for BCC crystal structure.
Answer: true Page11
Fix ِ‫ر‬‫نةطؤ‬ stable ‫َطري‬‫ي‬‫ج‬ same ‫يةماى‬ direction ‫ِاشتة‬‫ر‬‫ئا‬
Question: fill in the following blank with a suitable word or term.

7
1. Extraction metallurgy is the science that deals with the………of metals
from earth’s crust.
Answer: mining. Page1
2. The science that deals with the mining of metals from earth’s crust is
called…………..
Answer: Extraction metallurgy. Page1
3. Thermoset plastics are those polymers which when set into
permanent shape by heating, reheat will not……..
Answer: soften it. Page2
4. Ceramic materials are characterized by……...electrical conductivity.
Answer: poor. Page2
5. AL2O3 and SiO2 are considered as …………..
Answer: ceramic materials. Page2
6. Most of polymers are extremely……..
Answer: flexible. Page2

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7. For the formation of ionic bonding, two kinds of atoms must exist
one is………and the other is………..
Answer: Losing electron, receiving electron. Page5
8. Hydrogen bonding occurs between molecules in which is…..bonded
to fluorine, oxygen.
Answer: covalently Page7
9. Materials are the same in all direction are called………..
Answer: isotopic. Page8
10. All metals are solid at
room temperature except…………
Answer: mercury.
Earth’s crust ‫زةوى‬ ‫َى‬‫م‬‫َك‬‫ي‬‫تو‬ mining ‫ثوختةكردى‬ soften ‫بونةوة‬ ‫نةرم‬
formation ‫َكًاتو‬‫ي‬‫ث‬ occurs ‫ِودةدات‬‫ر‬ except ‫َجطةلة‬‫ي‬‫ب‬
mercury ‫جيوة‬ are considered ‫وادانراوة‬ as ‫وةكو‬
Question: Explain each of the followings.
1. Ceramic materials are poor electrical and thermal conductors.
Answer: because of ionic bond. Page2 and page 5

9
2. What is deriving force behind the bonding mechanism between two
atoms and consequently the creation of all kinds of materials?
Answer: the bonding mechanism between atoms derives from the desire
of atoms to remain in or revert to a stable condition, thus, this stability
depends on an atoms ability to maintain eight electrons in its outer most
rings, it can achieve stability by one of the following ways:
1) Gaining electrons [to form an ionic bond]
2) Sharing electron [to form a covalent bond]
3) losing electrons [to form a metallic bond] Page 4
3. What is the main objective in designing of composite materials?
Answer: A composite is designed to display the combination of the best
characteristic of each of the component materials. Page3
4. Is it true that all mechanical properties of engineering materials depend
on the type of interatomic bonding. Discuss it.
Answer: The type of interatomic bonding affects mechanical and other
physical properties of engineering materials which depend on the valence
electron of the outer most rings around the nucleus, since the atoms
composing a solid material are bonded to each other y means of these
outer most valence electrons. Page 4
5. iquefaction of some inert gases at subzero temperature.
Answer: dispersion bond causes inert gases to liquefy. Page 6

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6. Metals are good conductors of electricity.
Answer: the movement of free electrons means that metallic bonded
materials have good thermal and electrical conduction. Page 6
7. Polymeric materials have low densities.
Answer: polymer’s bonds are non-crystalline but they are direction bond.
Page 6
8. Non-crystalline solids are usually referred to be anisotropic.
Answer: because of random arrangement of atoms or molecules, non-
crystalline solids are usually referred to be anisotropic.
9. The attraction that exist between water molecules.
Answer: hydrogen bonding occurs in which hydrogen (H) is covalently
bonded to oxygen in H2O,in this case the single electron of hydrogen is
shared with the other atom, thus, the hydrogen end of the bond is
essentially a positively charged bare proton as shown below:

11
Page7
referred ‫َكراوة‬‫ي‬‫ث‬ ‫ئاماذةى‬ random ‫ِةمةكى‬‫ر‬‫ية‬ arrangement ‫ِيسبوى‬‫ر‬
Liquefaction ‫شــمــبـبونةوة‬ charge ‫بارطة‬ Inert ‫شـصـت‬ subzero
‫َرشفر‬‫ي‬‫ذ‬ temperature‫ثمةيطةرمى‬ density‫ِي‬‫ر‬‫ض‬ consequently ‫يةروةيا‬
creation ‫دروشتبو‬ kinds ‫جؤرةكاى‬ desire ‫خواشنت‬ remain in
‫مانةوة‬ revert ‫ِنةوة‬‫ر‬‫طة‬ stable condition ‫َطريي‬‫ي‬‫ج‬ ‫باري‬ ability ‫توانا‬
maintain ‫ثاراشنت‬ outer most rings ‫دةرةوة‬ ‫بةرطي‬ achieve ‫دةيك‬‫ات‬
form‫َهاى‬‫ي‬ً‫َك‬‫ي‬‫ث‬
Question: What makes the metals to be different from non metallic
material?
Answer: metallic bonding makes the metals to be different from non
metallic material.

12
Question: Distinguish between dispersion and hydrogen bonds.
Answer: dispersion bond is temporary and very weak interatomic or
intermolecular bonding between atoms or moleculars that electrically not
symmetric as shown below.
But hydrogen bond is the strongest secondary intermolecular bonding
type, it is especial case of polar molecular bonding . it occurs between
molecules in which hydrogen (H) is covalently bonded to fluorine (F)

13
,asin HF, oxygen as inH2O …..etc is such case like this one , the single
electron of hydrogen is shared with the other atom, thus, the hydrogen
end of the bond is essentially a positively charged bare proton as shown
below:
Unit cells
the correct form.
1. the number of atoms per one FCC unit cell is 6atoms/unit cell.
Answer: false… the number of atoms per one FCC unit cell is 4atoms/unit
cell. Page 12
2. In BCC crystal structures the lattice constant (a) can be expressed as
a=2D/√3 , where D is the atomic diameter in angstrom.
Answer: true Page12
3. Allotropy is also called polymorphism and could happen due to
temperature and pressure.

14
Answer: true Page13
4. In HCP crystals structures the lattice constant (a) is related to the
atomic radius R through the equation…………..
Answer: a=2R page 13
5. Crystals of high coordination number CN are said to be closely packed
crystals.
Answer: true Page14
6. Miller indices of planes are represented as [hkl].
Answer: false… Miller indices of planes are represented as (hkl). Page16
7. The plane (1010)in HCP crystals is called basal plane.
Answer: false… the plane (0001) in HCP crystals is called basal plane.
Page 19
8. Atomic packing factor APF is defined as the number of atoms per unit
area.
Answer: false…planer density is defined as the number of atoms per unit
area.

15
page 20
9. The plane (110) is the only slip plane with in FCC crystal structure.
Answer: false… The plane (111) is the only slip plane with in FCC crystal
structure. Page 20
1. Plane………..possesses the highest degree of atomic packing within a
BCC unit cell.
Answer: (110). Page 20
2. The crystallographic direction………is the close packed direction in
FCC crystal structure.
Unit cell Close packed plane
(slip plane)
Close packed direction
(slip direction)
FCC
BCC
HCP
(111)
(110)
(0001)
[110]
[111]
a1,a2,a3

16
Answer: [110] Page 20
3. Miller index of the basal plane of HCP crystals is written as ……...
Answer: (0001). Page 20
4. During X-ray diffraction analysis of BCC crystals constructive
diffraction occurs when the sum of (h+k+l) is……….
Answer: even page 24
5. X-ray diffraction in BCC crystal structures occurs constructively at
planes in which the sum (h+k+l) must be even.
Answer: true Page 24

19
Home exercise: show that APF for BCC=0.68
Answer:
Where: n=2atom/unit cell
( ) ( )
(
√
)
Page 14

20
Home exercise: show that APF for HCP=0.74
Answer:
Where: n=6atom/unit cell
Total volume of the unit cell=area of the base * height
Total volume of the unit cell=area of the hexagon * c
Total volume of the unit cell=3asin60 * c
But : c = 1.633 a
Therefore Total volume of the unit cell= 3asin60 *1.633 a
Total volume of the unit cell= 3asin60 *1.633
But : a=2R

21
( )
page 14
Question: Niobium Nb has an atomic radius of 0.143nm and a density of 8.57
g/cm3.determine whether it has an FCC or BCC crystal structure. Atomic
weight on Nb is 92.91 gr/mole.
Solution:
‫كةمادةكة‬ ‫َني‬‫ي‬‫وايدادةن‬ ‫َك‬‫ي‬‫بكةيو.جار‬ ‫شيكار‬ ‫ثرشيارةكة‬ ‫كةدوجا‬ ‫َويصتة‬‫ي‬‫ث‬ ‫ثرشيارانة‬ ‫جؤرة‬ ‫ئةم‬ ‫دانةوةى‬ ‫َام‬‫ل‬‫بؤوة‬
FCC‫كة‬ ‫َني‬‫ي‬‫وايدادةن‬ ‫َكىرت‬‫ي‬‫وةجار‬BCC‫دةدؤزيهةوة.لةكام‬ ‫ئاظؤطاردرؤ‬ ‫ذمارةى‬ ‫جارةكةدا‬ ‫يةردو‬ ‫َت.وةلة‬‫ي‬‫ب‬
‫ئةوا‬ .‫ِاشتةقيهةوة‬‫ر‬ ‫ئاظؤطاردرؤى‬ ‫لة‬ ‫بوو‬ ‫نسيك‬ ‫نرخةكة‬ ‫بارةياندا‬.‫جؤرةيانة‬ ‫ئةو‬ ‫مادةكة‬
We assume that the material has FCC crystal structure:
( √ )

22
Therefore the material is not FCC
We assume that the material has BCC crystal structure:
(
√
)
Therefore the material has BCC crystal structure. (Answer) Page 15

23
Question: write the Miller indices of the shaded crystallographic planes
shown in following:

24
shown in following:

26
shown in following:

28
Question: write the Miller indices of the crystallographic direactions shown
in following:

29
Question: determine the Miller indices for the following crystallographic of
planes and directions.
Solution:
Plane F (112)
Plane E (001)
_
Direction D[ 2 0 1]
_
Direction C[0 1 1]

30
Page 16
Question: determine the Miller indices for the following crystallographic of
directions.
Solution :
Direction OA [100]
Direction OB[110]
Direction OC[111]
Direction OD[210]
_
Direction OE[0 1 0]
Direction OF[112]
_
Direction CB[001]

31
Planer density
Home exercise: determine the planer density for the following planes within
an BCC unit cell assuming atomic radius is 1.24A .plane (001)
Solution:
No of atoms in the plane = 1 atom/plane
Area of the plane = a * a but
√
√ √

32
Page 20
Home exercise: determine the planer density for the following planes within
an FCC unit cell assuming atomic radius is 1.25A plane (111).
Solution :

33
No of atoms in the plane =1/2 +1.5 =2 atom/ plane
√
√
Sin 60=h/L therefore h =L sin60 thus h=√ a*sin60
√
√ (√ )
Area of the plane = a*a*sin60
√
( √ )( √ )

34
page20
Question: determine the planer density for the following planes within an
BCC unit cell assuming atomic radius is 1.24A plane (111).
Solution :

35
√
√
Sin 60=h/L therefore h =L sin60 thus h=√ a*sin60
√
√ (√ )
Area of the plane = a*a*sin60
√

36
(
√
) (
√
)
Question : determine the planer density for the following planes within an
FCC unit cell assuming atomic radius is 1.25A, plane (110).
Solution :

37
√
√
√
√
√ ( √ )( √ ) √
√
√

38
Question : determine the planer density for the following planes within
an FCC unit cell assuming atomic radius is 1.25A, plane (002).
Solution :
No of atoms in the plane =2atom/plane
√
( √ )( √ )

39
Home exercise: compute the planer density for plane (0001) in HCP crystals
assuming R=1.53A
Solution:
No of atoms in the plane=2+1=3atoms/plane
But : a=2R

40
Page 20 _
Home exercise: compute the planer density for plane (1010) in HCP crystals
assuming R=1.53A
Solution:
No of atoms in the plane=1=3atom/plane
Area of the plane= length *width
Area of the plane = c*a But a=2 R and c=1.633 a
Area of the plane =1.633 a * a
Area of the plane= 1.633(2R)(2R)=6.532(R)(R)

41
page 20
Linear density
Home exercise: compute the linear density for direaction[111] in FCC crystals
assuming R=1.53A
Solution:
linear density=1633986.92 atom/mm page 21
Home exercise: compute the linear density for direction [110] in FCC crystals
assuming R=1.53A
Solution:

42
linear density=3267973.85 atom/mm page 21
Volumetric density
Home exercise: prove that ρv =1/( √ ) fro HCP crystals .
Solution:
Total volume of the unit cell=area of the base * height
Total volume of the unit cell=area of the hexagon * c
Total volume of the unit cell=3 sin60 * c
But :c = 1.633 a
Therefore Total volume of the unit cell= 3 sin60 *1.633 a
Total volume of the unit cell= 3 sin60 *1.633
But : a=2R

43
Total volume of the unit cell= 3 sin60 *1.633
Total volume of the unit cell= 24 sin60 *1.633 =33.94
√
page 22
X-ray diffraction
Solution:
(A)we assume that the material has BCC crystal structure:
1. n=1(first order reflection)
if θ=21.66 , the plane is (110)
n.λ =2 sin θ
1.54=2 sin21.66 therefore =2.08Ǻ

44
√ √
therefore = 2.941Ǻ
2. if θ =31.47 , the plane is (200)
n.λ =2 sin θ
1.54=2 sin31.47 therefore =1.474 Ǻ
√ √
Therefore = 2.94Ǻ
3. if θ=39.74 ,the plane is (211)
n.λ =2 sin θ
1.54=2 sin39.74 therefore =1.2 Ǻ
√ √
=2.93Ǻ
4. if θ=47.58,the plane is (220)
n.λ =2 sin θ
1.54=2 sin47.58 therefore =1.04 Ǻ

45
√ √
=2.95Ǻ
n.λ =2 sin θ
1.54=2 sin55.63 therefore =0.932 Ǻ
√ √
=2.949Ǻ
n.λ =2 sin θ
1.54=2 sin 64.71 therefore =0.85 Ǻ
√ √
=2.95Ǻ
(B) we assume that the material has FCC crystal structure:

46
if θ=21.66 , the plane is (111)
n.λ =2 sin θ
√ √
=3.613Ǻ
2. if θ =31.47 , the plane is (200)
n.λ =2 sin θ
1.54=2 sin31.47 therefore =1.474Ǻ
√ √
therefore = 2.94 Ǻ
a1is not equal to a2 therefore it is not FCC .
its crystal structure is BCC
Results:
1. The Miller indices of the reflecting planes:
Bragg’s reflection in BCC system occurs at planes
(110)(200)(211)(220)(310)(222).

47
2. The structure of the crystal:
Therefore a1=a2=a3=a4=a5=a6 so that the material has BCC crystal
structure.
3. Lattice parameter =a=2.94Ǻ
4. the radius of atoms :
a=4R/√3 therefore 2.94=4R/√3 so that R=1.27Ǻ

48
Solution:
1. The Miller indices of the reflecting planes:
Bragg’s reflection in BCC system occurs at planes (11)(200)(220)(311)(222).
1) if 2θ=43 so that θ=21.5 , the plane is (111)
n.λ =2 sin θ
2) if 2θ =51 so that θ=25.5 , the plane is (200)
n.λ =2 sin θ
3) if 2θ=74 so that θ=37, the plane is (220)
n.λ =2 sin θ
4) if 2θ=90 so that θ=45, the plane is (220)
n.λ =2 sin θ

49
1.54=2 sin45 therefore =1.088Ǻ
√ √
therefore = 3.63 Ǻ
but √
3.63= √ so thut R=1.28Ǻ *atomic radius+
the atomic diameter of the copper=2R=2*1.28Ǻ=2.57Ǻ
Mechanical properties and solid solution
the correct form.
1. Hardness refers to metal ability to resist deformation under tension.
Answer: false… tensile refers to metal ability to resist deformation under
tension. Page 27
2. The impact test is performed to evaluate the tensile strength of
metals.
Answer: false…the tensile is performed to evaluate the tensile strength
of metals. Page 27

50
3. Tensile strength of metals decreases with an increase in
temperature.
Answer: true Page
4. Creep is the phenomena by which a metal elongate under the action
of both stress and high temperature.
Answer:
5. Fatigue is the failure that takes place in member’s structures
operation at elevated temperature.
Answer:
6. Solidification of a molten metal is possible when the free energy of
the liquid state GL is higher than the free energy of the solid state
Gs.
Answer: true Page28
7. Heterogeneous nucleation of the new phase occurs randomly in the
parent phase depending on the degree of super cooling only.
Answer: false… Homogeneous nucleation of the new phase occurs
randomly in the parent phase depending on the degree of super cooling
only. Page29

51
8. Solid solution of pure substance is an exothermic reaction.
Answer: true Page29
9. The critical radius of embryos is unaffected by temperature.
Answer: false… The critical radius of embryos is affected by
temperature. Page 30
10. During the process of metals solidification
nucleation centers are formed in several locations in the existing
liquid. These nucleation centers are called seeds or embryos all of
these embryos are capable to grow into crystal.
Answer: false…that embryos having R greater than Rc are capable to
grow into crystal. Page 30
11. During the process of metals solidification
nucleation rate RN is reduced as degree of super cooling DT is
reduced which lead to higher rates of grain growth.
Answer: true…….. page30
12. Eutectic is a term used to describe alloys of high
melting point.
Answer: false… Eutectic is a term used to describe alloys of low melting
temperature. Page 30

52
13. Nucleating centers or embryos are assumed to have
a spherical shape of a well-defined diameter.
Answer: true Page30
14. RN increases as the degree of super cooling is
decreased and that RG reaches its maximum point and that RN falls
to zero at very high DT. page 30
Answer: false… RN increases as the degree of super cooling is
increased and that RG reaches its maximum point and that RN falls
to zero at very high DT. page 30
15. The critical radius of embryos decreases as the
degree of super cooling increases.
Answer: true Page 31
16. Metals are stronger when they possess finer
microstructure.
Answer: true Page32
17. High solidification rates are achieved when the
molten metal is poured into metallic molds.
Answer: true Page33

53
18. Thermal equilibrium diagram shows the relationship
between composition – temperature and the structure of alloys.
Answer: true Page35
19. Pure metals are usually utilized in engineering
practice when high ductility, high electrical conductivity of high
corrosion resistance is required.
Answer: true Page35
20. All alloy system exhibit complete solid solubility.
Answer: false…the two metals completely or partial or totally in soluble
in each other. Page 35
21. Bismuth Bi and cadmium Cd are not soluble in each
other in the solid state, therefore they form a(Eutectic) structure
which consist of alternative layers of α and β phase.
Page 36
Answer:
22. All alloys are soluble in each other in their liquid
state except Fe-AL, Cu-Pt and W-Cu.
Answer: false… All alloys are soluble in each other in their liquid state
except Fe-Pb, Cu-Pb and W-Cu. Page 37

54
23. Some alloy systems are not soluble in each other in
the liquid state due to the difference in there crystal structure.
Answer: false… Some alloy systems are not soluble in each other in the
liquid state due to:
(1) The large difference in melting temperature between two metals.
(2) The large difference in specific weight (density) between two metals.
Page 37
24. Metal A is said to substitutional dissolve in metal B
when atoms of metal A occupied lattice sites that would normally be
occupied by atoms of metal B.
Answer: true Page37
25. Metal A dissolve interstitially in metal B when the
relative difference between the atomic diameter of both is less than
15%.
Answer: false… Metal A dissolve substitutionly in metal B when the
relative difference between the atomic diameter of both is less than 15%.
Page38
26. Carbon atoms dissolve substitution in the solid state.
Answer: true Page39

55
27. High solidification rates are achieved when Titanium
Ti is added to the molten metal.
Answer: false… High solidification rates are achieved whenFe3C is added
to the molten metal.
1. Stiffness can be defined as the ability of material to………..deformation.
Answer: resist. Page 27
2. The radius of any embryos must be……….than that called the critical
radius of embryos Rc to grow into crystal.
Answer: greater. Page 30
3. The process by which a solid material transforms to a gas without
passing through the liquid state is called………….
Answer: sublimation.
4. During the process of metals solidification, the critical radius of
embryos Rc is related to the degree of super cooling DT by the
equation………….
Answer: Rc=2γsl * TE /(DH.DT). Page31

56
5. The term (eutectic) means……………..
Answer: able to easily melted. Page36
6. Pure metals usually solidify at ………….temperature.
Answer: constant. Page40
7. Electron compounds possess metallic properties because of the
…….....that exist between the atoms forming it.
Answer: metallic bonding. Page 47
Question: Prove the change in Gibbs free energy is directly proportional
to the degree of super cooling during the process of metal solidification.
Answer:
Substitution of equation (2) into equation (1)

57
( )
َ‫ي‬‫بةذ‬ ‫دةكةيو‬‫ر‬‫ياوبةش‬ ‫ةى‬
‫كةرتةكةدا‬ ‫لةشةرةى‬‫ياوبةشدةكةيو‬
1. Rapid cooling rate during metal solidification, results in finer grains.
Answer: the critical radius of embryos is reduced as the degree of super
cooling is increased and hence a large number of embryos will have a chance
to grow into crystals and so, the final structure will compose of finer grains.
Page32
2. Low cooling rates of casting, results in cause large grained size.

58
Answer: the critical radius of embryos is increased as the degree of super
cooling is decreased and hence a few numbers of embryos will have a chance
to grow into crystals and so, the final structure will compose of large grains.
Page32
3. Using of single crystals of certain alloys in the manufacture of gas turbine
and jet engine blades.
Answer: Page
Question: differentiate between the following:
1.casting and ingots.
Answer: Ingots: simple block of cast metal are called ingots which different
from castings in that they have to be rolled, forged or extruded in order to
produce semi finished products. Page33
2. Embryos and crystals.
Answer: Embryos are assumed to be tiny particles of a spherical shape
identical in structure to the solid phase crystal and consist of a few numbers
of atoms in the form of a cluster and their radius is less or equal to critical
radius of embryos .
But Crystals: are embryos that have grown into crystal because their radius
were the greater than critical radius of embryos. Page32
3.Strength and hardness .
Answer: strength refers to metal ability to withstand or support external
loads without rupture.

59
Hardness: refers strength refers to metal ability to resist perpetration or
abrasion by other metals. Page 27
4. Valance and electron compound.
Answer: Valence Compounds: [fixed composition]
These are compounds of electropositive and electronegative elements, the
interatomic bonding may be ionic or covalent therefore they are usually
hard, brittle, poor electrical conductors and having high melting
temperatures such as NaCl.Mg2Sn, Fe3C and Fe2N.
Electron compounds:[variable composition]
These compounds are formed between two metals that have different
number of valence electrons such as CuZn, Cu3AL and Mg2Pb.The
interatomic bonding is metallic therefore they possess metallic properties.
Page 41
Question: Describe each of the solidification sequence in metals by detail
sketch only. Page 32
Answer:

60
Question: Two important steps discuss them fully.
Answer: solidification of metals proceeds through two important steps:
1. Nucleation of the new solid phase [The formation of embryos or seeds].
2. Growth of the solid phase.
Nucleation of the new solid phase can either be:
a. Homogeneous Nucleation: the solid phase nucleates randomly in the
parent liquid phase.
b. Heterogeneous Nucleation: the solid phase nucleates preferentially at
specific sites in the parent liquid phase like grain boundaries, foreign
particles, mold walls…etc. page 29
Answer: There are many factors that controls the extend of solid solubility
in metal form which:
1. The size factor.
2. Crystal structure.

61
3. Valence electron.
4. Chemical reactivity or affinity. Page 38
States of solid solubility:
1. Al-Cu system is partial solid solubility.
2. Bi-Cd system is negligible solid solubility.
3. Fe-C system is partial solid solubility.
4. Ni-Cu system is complete solid solubility.
5. Ag-Au system is complete solid solubility. Page 35 and 36
Thermal equilibrium diagram (phase diagram)
Example:

63
The change in microstructure that take place when an alloy containing
70% Pb cools slowly to room temperature is described with clear on X-
axis.
Cooling curve:

64
Answer:
35% Si cools slowly to room temperature is described with clear on X-axis.

65
Iron carbon diagram
Question: Sketch and fully label the Iron-Iron Carbide phase diagram and
then show position of the following alloys on the diagram and indicate what
microstructural change take place in each alloy while slowly heating them up
to melting;
1. Alloy no.1 containing 0.4%C.
2. Alloy no.2 containing0.8%C.
3. Alloy no.3 containing1.6%C.
Answer:

66
0.4%C while slowly heating to melting temperature is described with clear
on X-axis.
on Y-axis.
on Z-axis.
the correct form.
1. Maximum solubility of carbon in FCC austenite is about 0.02% at 1147
C.
Answer: false… Maximum solubility of carbon in FCC austenite is about
2.14% at 1147 C. page 50
2. Ferrite that is percent in the pearlite is called eutectoid ferrite.

67
Answer: true Page
3. In the peritectic reaction, liquid and one solid phase transform to a
second solid phase.
Answer: true Page51
4. Nickel is what makes stainless steel stainless.
Answer:
5. Nodular graphite gray cast iron is so named because of the presence of
MnS globules in the microstructure.
Answer: false… Nodular graphite gray cast iron is so named because of
the presence of Mn or S globules in the microstructure.
6. Galvanized steels are also termed stainless steels.
Answer: true
7. austenitic stainless steel grade 316L has superior resistance to
intergranular corrosion.
Answer:
8. White cast irons are produced either by rapid cooling rates or low
silicon content.
Answer: false… White cast irons are produced either by rapid cooling
rates and low silicon content. Page 60
9. Martensite has a BCT crystal structure and is the strongest of the
common phase found in steels.
Answer:

68
10. High speed steels HSS can keep their not hadness at temperatures as
high as 600 C.
Answer: true Page60
11. α –iron exhibit a BCC crystal structure.
Answer: true Page50
12. The crystal structure of pearlite phase is orthorhombic.
Answer: false… The crystal structure of cementite phase is orthorhombic.
Page50
13. The solubility of carbon in α-iron decreases on heating above 723 C
due to the formation of γ-iron.
Answer: true Page49
1. The crystal structure of Fe3C is…………..
Answer: Orthorhombic. Page 50
2. In the iron-carbon system eutectic reaction take place at………%C
and……………C.
Answer: 4.3%C and 1147C.
3. The solubility of carbon in ferrite is ……….than that in austenite.
Answer: less.

69
4. The steels that contain carbon in the range of……………is called
Hypereutectoid steel.
Answer: 0.8 – 2.1 %C. page 48
5. Cast iron are generally…………. Materials.
Answer: hard and brittle. Page 49
6. Gray cast iron usually contains more than ………….% Silicon.
Answer: 1 – 3 . page 61
7. Pearlite is eutectoid microstructure of steel which contain about
………..% carbon and consist of lamellae of ………and………
in …………..ratio.
Answer: 0.8 , ferrite , cementite, 7:1 in. page 50
8. The steels that contain lower than 0.8% carbon are called…………
Answer: Hypoeutectoid steels. Page 48

70
1. Carbon has a limited solubility in BCC iron.
Answer:
2. Using of mild steel in the manufacture of sheets and sections for
structural works.
Answer: Because they possess good weldability characteristics.
Page54
3. Using of Gray cast iron in the manufacture of hevy mashines body.
Answer:
4. Using of HSS in the manufacture of cutting tools.
Answer: because its extremely hard and retains hardness up to 600 C
during cutting operation. Page 63
5. Steel gets harder as te carbon content is increased.

71
Answer: due to the fact that increasing carbon content leads to the
formation of further cementite which is very hard and brittle phase.
Page 56
6. Gray cast iron gets stronger as it alloyed with Mg or Ce.
Answer: because it causes the graphite to solidify as roughly spherical
nodules. page 61
7. Stainless steels loss its corrosion resistance as the carbon content is
increased.
Answer: When Cr is present in amounts in excess of 12% the steel
becomes highly resistance to corrosion, owning to the protective film of
chromium oxide that forms on the metal surface, but when carbon
increases the Cr is decreases. Page 63
8. The dark appearance of the fracture surface of gray cast iron.
Answer: due to existence of carbon in form of free graphite and not as
Fe3C. page 61
9. Silicon is usually added to alloys intended for casting.
Answer: imports fluidity to steel parts intended for casting. Page 57
10. Nodular graphite gray cast iron are tougher and softer than flackes
graphite gray cast iron.
Answer: Gray cast iron alloyed with magnesium or cerium before casting
which causes the graphite to solidify as roughly spherical nodules. Page
61

72
11. The corrosion resistance of martensitic stainless steels is lower than
that of ferritic and austenitic types.
Answer: due to the relatively high carbon contents compared to ferritic
an austenitic stainless steels. Page 64
12. Tempered martensite is hard, strong and tough.
Answer: alloy elements render material harder, stronger and tougher.
13. Silicon content of high and tool carbon steels is usually limited to be
low %.
Answer: because silicon increases the tendency for cementite
decomposition into free graphite and ferrite in high and tool carbon
steels, therefore Si should not exceed 0.3 % . page 57
14. White cast iron products are usually harder than gray cast iron
products.
Answer: due to existence of carbon in the form of cementite Fe3C which
it is hard and brittle phase. Page 60
15. Steels containing sulphur in an excess amounts should be treated with
manganese.
Answer: Because the presence of sulphur element in the structure of steel
leads to the formation of iron sulphide FeS that precipitate at the grain
boundaries of the steel. FeS is a hard and brittle phase of low melting
temperature , therefore, FeS melts at the hot working temperature of steel
leading to a phenomena known as (Hot shortness) by which the part is
cracked and fractures during the hot working process.

73
In order to nullify the effect of sulphur in steel , Manganese is added to the
molten steel before casting.Mn reacrts with sulphur to form manganese
sulphide (MnS) in preference to FeS. MnS is insoluble in the molten steel and
some is lost in the slug, the remainder is present as fairly large globules
distributed throughout the structure of steel. MnS is also plastic at the hot
working temperature, so that the tendency of hot shortness is removed.
Page 58
16. Cast iron products are hard and brittle.
Answer: Cast iron products are hard and brittle due to the high carbon
content which leads to the formation of the cementite which is very hard and
brittle phase.
17. Excess of sulphur in steel cause hot shortness.
Or sulphur content of steel should be kept as low as possible.
Answer: Because the presence of sulphur element in the structure of steel
leads to the formation of iron sulphide FeS that precipitate at the grain
boundaries of the steel. FeS is a hard and brittle phase of low melting
temperature , therefore, FeS melts at the hot working temperature of steel
leading to a phenomena known as (Hot shortness) by which the part is
cracked and fractures during the hot working process.
18. The carbon content of ferritic and austenitic stainless steels must be kept
below 0.1% .

74
Answer: to obtain high corrosion resistance of ferritic and austenitic
stainless steels because Cr decreases as carbon increases.
19. Austenitic stainless steel is more corrosion resistance than ferritic
stainless steel.
Answer: when Cr is present in amounts in excess of 12% the steel becomes
highly resistance to corrosion owing to the protective film of chromium oxide
that forms on the metal surface, and austenitic stainless steels contain 18%
Cr. Page 63
Question: differentiate between the following:
1. White and gray cast iron, in term of carbon content; silicon content;
microstructure; fracture appearance; mechanical properties and application.
Answer: White Cast Iron possesses the following properties:
Carbon content is above 2.1%.
Silicon content is less than 0.1%.
Its microstructure is pearlite and cementite.
Its fracture surface has a white appearance.
Its mechanical properties are extremely hard and brittle.
Its applications are used to a very hard surface and wear-resistance surface
such as in rollers in rolling of mills and railway wheels.

75
But Gray Cast Iron possesses the following properties:
Carbon content is above 2.1%.
Silicon content is between1-3.
Its microstructure is ?-iron and free graphite.
Its fracture surface has a gray appearance.
Its mechanical properties are strong and tough.
Its applications are used in manufacture of valves, pump bodies, crankshafts,
gears, engine block and heavy machines.
2. Iron, steel and cast iron in term of carbon content, microstructure,
mechanical properties and application.
Answer: Iron possesses the following properties:
Carbon content is less than 0.005%.
Its microstructure is ?-iron.
Its mechanical properties are soft, ductile, resists corrosion and fatigue
failure.
Its applications are used for making bolts, pipes, and tubes.
Steel possesses the following properties:
Carbon content is between 0.005- 2.1%.

76
Its microstructure is ?-iron and Fe3C.
Its mechanical properties are
Its applications are used for making
Cast iron possesses the following properties:
Carbon content is between 2.1-6.7%.
Its microstructure is ?-iron and Fe3C.
Its mechanical properties are hard and brittle.
Its applications are used for making
3. Eutectic and Eutectoid reaction.
Answer: A eutectoid reaction ia at 0.8% C and 723C
A eutectic reaction is at 4.3%C and 1147C

77
4. Cold and hot shortness.
Answer:
Cold shortness is a phenomena that occurs during cold working of steel parts
when steel is solidify into insolated bands which known as ghost bands.
Hot shortness: is a phenomenon that occurs during hot working processes
due to the melting of FeS at hot working temperature of steel.
5. Black heart and white heart malleable cast iron.
Answer: Black heart malleable cast iron can be obtained by black heart
process which white iron castings are heated in a neutral atmosphere at a
temperature of about 900C for between 2and 3days, and then cooled very
slowly. As a result of this process, Fe3C in the white iron structure breaks
down into ferrite and roughly spherical aggregate of graphite in the form of
rosettes.
White heart malleable cast iron can be obtained by white heart process
which white iron castings into boxes containing ferric oxide Fe2O3 and
heating to temperature of 900C for 2-5 days. Because the castings are in
contact with ferric oxide, carbon is oxidized away from the surface, and the
resulting structure is composited of firrite at the edge of the casting and
ferrite, pearlite and some graphite nodules at the center.

78
Question: Briefly describe method for producing black heart malleable cast
iron.
Answer: : Black heart malleable cast iron can be obtained by black heart
process which white iron castings are heated in a neutral atmosphere at a
temperature of about 900C for between 2and 3days, and then cooled very
slowly. As a result of this process, Fe3C in the white iron structure breaks
down into ferrite and roughly spherical aggregate of graphite in the form of
rosettes.
Question: Briefly describe method for producing nodular graphite gray cast
iron.
Answer: gray cast iron alloyed with magnesium or cerium before casting
which causes the graphite to solidify as roughly spherical nodules.
Question: Briefly discuss the allotropic transformation of iron and its
technological significance.
Answer:

79
Question: The microstructure of an iron-carbon alloy consists of
proeutectoid ferrite and pearlite, the mass fractions of these two micro
constituents are 0.286 and 0.714, respectively. Determine the concentration
of carbon in this alloy.
Solution:‫ِي‬‫ر‬‫ب‬ ‫ئةوةى‬ ‫لةبةر‬ ‫كة‬ ‫ئةوةية‬ ‫تةنًا‬ ‫ثرشيارة‬ ‫ئةم‬ ‫فكرةى‬ (proeutectoid ferrite and
pearlite) ‫ب‬ ‫َت‬‫ي‬‫دةب‬ ‫كةوانة‬ ‫دراوة‬‫لةناوضةى‬ ‫ئةمة‬ ‫كة‬ ‫سانني‬ (Steel) (‫َواى‬‫ي‬‫لةن‬ ‫واتة‬ ‫داية‬0.005-0.8 %
C)‫.داية‬

80
Effect of alloy elements on the Fe- Fe3C diagram
(a) The most ferrite stabilizers are Cr, Si, Mo, W, V, Ti, All these elements
have the same crystal structure as ferrite (BCC).

81
Therefore, the stability of ferrite increases, when Cr, Si, Mo, W, V, Ti added to
steel parts.
(b) The most austenite stabilizers are Al, Mn, Ni, All these elements have the
same crystal structure as austenite (FCC).
Therefore, the stability of austenite increases, when Al, Mn, Ni added to steel
parts.
(c) The most graphite stabilizers are Co, Mg, All these elements have the
same crystal structure as graphite (HCP).
Therefore, the stability of graphite increases, when Co, Mg added to steel
parts.
(d) The most carbide stabilizers are W, Cr.
Therefore, the stability of carbide increases, when W, Cr added to steel parts.
Question: show the effect of the following elements on the stability of
ferrite, austenite, carbide and graphite when added to steel parts.(1.Cr, 2.Ni,
3.Ti, 4.W, 5.Si, 6.Mn, 7.Al, 8.Mo, 9.Co, 10.Mg, 11.V)
Answer:

82
The stability of ferrite increases, when Cr, Si, Mo, W, V, Ti added to steel
parts.
The stability of austenite increases, when Al, Mn, Ni added to steel parts.
The stability of graphite increases, when Co, Mg added to steel parts.
Crystal imperfection and deformation of metals
Question: answer by true or false.
1. Strain hardening is the phenomenon by which a metal get stronger due to
cold working.
Answer: true page 81
2. Critical resolved shear stress represents the amount of friction between
atomic layers within the crystal.
Answer: true page70
3. Block slip theory of plastic deformation of metals was able to explain why
metals strengthen as they plastically deformed.
Answer: false……… Block slip theory of plastic deformation of metals was
not able to explain why metals strengthen as they plastically deformed.
Page 68

83
4. Hardening heat treatment is carried out to increase the ductility of steel
parts.
Answer: false …… Hardening heat treatment is carried out to increase the
hardness of steel parts.
Crystal structure Slip plane Slip directions Slip system
FCC
BCC
HCP
4(111)
6(110) 1(0001)
basal plane
3[110]
2[111]
3-a
12
12
3
Question: explain each of the following.
1. Ductile materials fracture usually at angle of 45.
Answer: if ? = ?, under this condition, the applied shear stress reaches its
maximum value, and therefore, plastiv deformation will be initiated first in
that slip planes oriented at 45 since the applied shear stress is a maximum
and consequently ductile material have FCC crystal structure which have four
slip planes oriented at 45 (111) .
2. FCC metals are typically more ductile than BCC and HCP metals.
Answer: Metal of FCC crystal structures have the highest number of close
packed plane within one unit cell each of which have three close packed

84
directions (slip direction)and consequently the highest slip system. Page
72
3. HCP metals are typically more brittle than BCC and FCC metals.
Answer: : Metal of HCP crystal structures have the lowermost number of
close packed plane within one unit cell which it is basal plane (0001) which
have three close packed directions (slip direction)and consequently the
lowermost slip system. Page 72
4. BCC metals are typically stronger than FCC and HCP metals.
Answer: Because phenomenon of slip takes place on slips planes which they
are close packed planes while slip plane in metals of BCC crystal structures is
plane (110) which is in reality not a close packed plane but it has the highest
atomic density plane among others in the BCC system, so that a slip system
of BCC is not a true system. Page 72
5. High loading rates may cause ductile material to fracture in a brittle
manner?
Answer: High speeds of loading or rapidly varying loads may cause metal to
fracture in a brittle manner because this type of loading leads to occur a
rapid propagation of crack with negligible plastic deformation as shown
below:

85
6. Recrystallization temperature Tr decreases as the degree of cold working
increase.
Answer: The degree of prior cold working should be greater than a certain
amount known as critical cold working CCW for the recrystallization to take
place. As the degree of cold working is increased the recrystallization
temperature is reduced as shown below: page 77

86
7. Plastic deformation of metals due to the presence of dislocation.
Answer: Dislocation theory of plastic deformation is based on the existence
of small linear imperfection in the crystal lattice known as Dislocations which
are responsible on the plastic behavior of metals. After long term of
experiments on the plastic behavior of metals, it is observed that the plastic
deformation is in fact being due to step by step dislocations movement
within the crystal structure. Page 69
8. linear crystal imperfections are very beneficial in the process of plastically
deformation of metals?
Or
Do you believe that there is a relationship between the mechanical
properties of metals and the ease with which dislocations can move within
the metals? Explain
Answer: Dislocation theory of plastic deformation is based on the existence
of small linear imperfection in the crystal lattice known as Dislocations which
are responsible on the plastic behavior of metals. Page 69
9. Metals gets harder as they cold worked?
Or
Metals gets harder as the plastically deformed?

87
Answer: This phenomenon is explained on the bases of dislocation theory
of plastic deformation which attribute this behavior of metals to the
interaction between dislocation strain fields because its density is greatly
increased in the metal forming that is known as a forest of dislocations and
because of this, the motion of dislocations is hindered by the presence of
other dislocations which leads to higher strength and hardness.
Page 74
Questions: Cite the difference between the following:
1.Block slip theory and dislocation theory and dislocation theory of plastic
deformation of metals.
Answer: Block slip theory accounted for many of the observed phenomena
but possessed a number of draw back, one of which being that the
theoretical strength of metals calculated on the bases of this theory was
about a thousand times greater than the experimentally observed strength.
Second, block slip theory failed on explaining why do metals become
stronger as they plastically deformed ?
But dislocation theory could explain the mechanism by which metals are
plastically deformed and it can solve the draw backs of block slip theory.
Page 68
2. Elastic and plastic deformations

88
Answer: Elastic deformation is the temporary deformation of metals under
action of an external load and the metal will return back to original state
after the applied load has been released. Such as in polymer materials.
But plastic deformation is permanent deformation of metals under action of
an external load without fracture, such as in metals. Page
68
3. Mechanical and thermal twins.
Answer: mechanical twins is deformations of metals due to mechanical
energy , such as deformations of machine structures.
But thermal twins is deformations of metals due to thermal energy, such as
in heat treatments.
4. Frankel and Schottky crystal imperfection.
Answer: page 65

89
5. Brittle and ductile fracture.
Answer: ductile fracture is the fracture which take place by a slow
propagation of crack with appreciable plastic deformation as evidenced by
necking, such as in steels, as shown below;
Brittle fracture is the fracture which take place by a rapid propagation of
crack with negligible plastic deformation, such as in glass, as shown below:
page 75

90
6. Crystallization and recrystallization.
Answer: crystallization is the formation of crystal from embryos that
have radius R greater than Rc during metal solidification.
Recrystallization is the formation of new set of strain free grains and
equiaxed grains that have low dislocation densities and they are
characteristic or the precold worked condition. The new grains form as a
very small nuclei at position of high strain energy in the parent metal
matrix and grow until they completely replace the parent metal. Page
77
7. Hot and cold work.
Answer: hot work is plastic deformation of metals at a temperature
which is higher than recrystallization temperature.
Cold work work is plastic deformation of metals at a temperature which is
lower than recrystallization temperature.

91
8. Dislocation can move easily in grains compare to grain boundaries?
Answer: grain boundaries can act as an effective barriers to dislocation
motion. page 81
9. Reducing grain size results in an increase in the strength of metals.
Answer: When The grain size is reduced, the number of grain boundaries
will be increase and grain boundaries can act as an effective barriers to
dislocation motion. Because of this, a metal with small grains will tend to
be stronger than the same metal with large grains at ordinary
temperature . page 81

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