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

Nov 01, 2014

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Engineering

Kurd Engineer

كانزاسازى
Kanza sazi
metallurgy
كانزاسازى
ئەندازیاری کانزاسازی
Metallurgy
كانزاسازى
KanzaSazi
كانزاسازى
علم المعادن
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Page 1: كانزاسازى Kanza sazi metallurgy كانزاسازى ئەندازیاری کانزاسازی Metallurgy كانزاسازى KanzaSazi كانزاسازى علم المعادن
Page 2: كانزاسازى Kanza sazi metallurgy كانزاسازى ئەندازیاری کانزاسازی Metallurgy كانزاسازى KanzaSazi كانزاسازى علم المعادن

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Republic of Iraq

Kurdistan Regional Government

Metallurgy

كانساشازى

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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 دةكؤلَتةوة structure ثةيكةر

alloys دارِشتةكاى 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

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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 elements تومخةكاى

9. Atoms of some elements may have different number of protons in

their nucleus producing that are called isotropics.

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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

ئةوثةرِى extremely بةندى ياوبةشى covalent bond بةيؤى due to طةيةنةرflexible ِجري properties شيفةتةكاى once يةك جار permanent يةميشةيى

component materials مادة ثيَكًيَهةرةكاى

13. Hydrogen bonds are

strong and stable interatomic bonding.

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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 ِنةطؤر stable جيَطري same يةماى direction ئارِاشتة

Question: fill in the following blank with a suitable word or term.

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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 تويَكمَى زةوى mining ثوختةكردى soften نةرم بونةوة formation ثيَكًاتو occurs رِودةدات except بيَجطةلة 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

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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:

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Page7

referred ئاماذةى ثيَكراوة random يةرِةمةكى arrangement رِيسبوى Liquefaction شــمــبـبونةوة charge بارطة Inert شـصـت subzero يةروةيا consequently ضرِيdensity ثمةيطةرمىtemperature ذيَرشفر

creation دروشتبو kinds جؤرةكاى desire خواشنت remain in

توانا ability باري جيَطريي stable condition طةرِنةوة revert مانةوة maintain ثاراشنت outer most rings بةرطي دةرةوة achieve ات دةيك

formثيَكًيَهاى

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.

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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)

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,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

Question: answer by true or false and then rewrite the false statements in

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.

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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.

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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

Question: fill in the following blank with a suitable word or term.

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

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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

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Home exercise: page 13

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Home exercise: show that APF for BCC=0.68

Answer:

Where: n=2atom/unit cell

(

)

(

)

(

√ )

Page 14

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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

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(

)

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:

بؤوةلَام دانةوةى ئةم جؤرة ثرشيارانة ثيَويصتة كةدوجا ثرشيارةكة شيكار بكةيو.جاريَك وايدادةنيَني كةمادةكة FCC وةجاريَكىرت وايدادةنيَني كةBCC بيَت.وةلة يةردو جارةكةدا ذمارةى ئاظؤطاردرؤ دةدؤزيهةوة.لةكام

مادةكة ئةو جؤرةيانة. بارةياندا نرخةكة نسيك بوو لة ئاظؤطاردرؤى رِاشتةقيهةوة. ئةوا

We assume that the material has FCC crystal structure:

( √ )

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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

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Question: write the Miller indices of the shaded crystallographic planes

shown in following:

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Question: write the Miller indices of the shaded crystallographic planes

shown in following:

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Question: write the Miller indices of the shaded crystallographic planes

shown in following:

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Question: write the Miller indices of the crystallographic direactions shown

in following:

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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]

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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]

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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

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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 :

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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

( √ )( √ )

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page20

Question: determine the planer density for the following planes within an

BCC unit cell assuming atomic radius is 1.24A plane (111).

Solution :

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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

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(

√ ) (

√ )

Question : determine the planer density for the following planes within an

FCC unit cell assuming atomic radius is 1.25A, plane (110).

Solution :

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No of atoms in the plane =1/2 +1.5 =2 atom/ plane

√ ( √ )( √ ) √

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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

( √ )( √ )

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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

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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)

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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:

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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

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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

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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

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=2.95Ǻ

5. if θ=55.63,the plane is (310)

n.λ =2 sin θ

1.54=2 sin55.63 therefore =0.932 Ǻ

=2.949Ǻ

6. if θ=64.71,the plane is (222)

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:

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1. n=1(first order reflection)

if θ=21.66 , the plane is (111)

n.λ =2 sin θ

1.54=2 sin21.66 therefore =2.08Ǻ

=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).

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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

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Solution:

1. The Miller indices of the reflecting planes:

Bragg’s reflection in BCC system occurs at planes (11)(200)(220)(311)(222).

2. n=1(first order reflection)

1) if 2θ=43 so that θ=21.5 , the plane is (111)

n.λ =2 sin θ

1.54=2 sin21.5 therefore =2.1Ǻ

2) if 2θ =51 so that θ=25.5 , the plane is (200)

n.λ =2 sin θ

1.54=2 sin25.5 therefore =1.788Ǻ

3) if 2θ=74 so that θ=37, the plane is (220)

n.λ =2 sin θ

1.54=2 sin25.5 therefore =1.27Ǻ

4) if 2θ=90 so that θ=45, the plane is (220)

n.λ =2 sin θ

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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

Question: answer by true or false and then rewrite the false statements in

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

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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

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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

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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

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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

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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

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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.

Question: fill in the following blank with a suitable word or term.

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

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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)

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(

)

ةى ياوبةشردةكةيو بةذيَ

ياوبةشدةكةيو لةشةرةى كةرتةكةدا

Question: Explain each of the followings.

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.

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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.

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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:

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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.

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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:

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Answer:

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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:

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Answer:

The change in microstructure that take place when an alloy containing

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

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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:

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The change in microstructure that take place when an alloy containing

0.4%C while slowly heating to melting temperature is described with clear

on X-axis.

The change in microstructure that take place when an alloy containing

0.8%C while slowly heating to melting temperature is described with clear

on Y-axis.

The change in microstructure that take place when an alloy containing

1.6%C while slowly heating to melting temperature is described with clear

on Z-axis.

Question: answer by true or false and then rewrite the false statements in

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.

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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:

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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

Question: fill in the following blank with a suitable word or term.

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.

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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

Question: Explain each of the followings.

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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.

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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

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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.

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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% .

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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.

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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%.

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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

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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.

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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:

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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:فكرةى ئةم ثرشيارة تةنًا ئةوةية كة لةبةر ئةوةى برِي (proeutectoid ferrite and

pearlite) سانني كة ئةمة لةناوضةىدراوة كةوانة دةبيَت ب (Steel) 0.8 - 0.005داية واتة لةنيَواى) %

C)داية.

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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).

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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:

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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

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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

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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:

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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

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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?

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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

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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

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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

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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.

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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