mle1101 ay1213 sem2 detailed tutorial solutions

8/22/2019 MLE1101 AY1213 Sem2 Detailed Tutorial Solutions

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MLE1101AY12/13 Semester 2

Tutorial 1-6 Detailed Tutorial Solutions

Written by: Ms Kong Hui Zi :3

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Determinetheelectronconfigurationsforasiliconatom(Z=14)

andagermaniumatom(Z=32).

Explainwhythesetwoelementsdisplaysimilar characteristics.

Tutorial1Question2(ElectronicConfig.):

e configurationofSi(Z =14): 1s22s22p63s23p2

e configurationofGe (Z=32): 1s22s22p63s23p64s23d104p2

or 1s22s22p63s23p63d104s24p2

Bothelementshaveavalenceelectronstructureoftheformxs2xp2

wherex is3 forSiand4 forGe,

(bothbelongtoGp.4inPeriodictable)

soverysimilarcharacteristics.

5

1s

2s 2p

3s 3p 3d

4s 4p 4d 4f

5s 5p 5d 5f

6s 6p 6d

7s 7p

1s22s22p63s23p64s23d104p6

36electrons

In a commercial xray generato

(Cu) or tungsten (W) is expose

energy electrons. These electron

metal atoms. When the metal athey emit xrays of characteris

example, a tungsten atom st

may lose one of its K shell

another electron, probably from

into the vacant site in the K s

occurs in tungsten, a tungsten K

Tutorial1Question4(E

34 8

15

9

6.63 10 J s 3 10 m/s9.30 10 J

1m0.02138nm

10 nm

hcE

AtungstenK

xrayhasawavelength of0.02138nm.

Whatisitsenergy?Whatisitsfrequency?

Tutorial1Question4(Contd):

8

19

9

3.00 10 m/ s1.40 10 Hz

1m0.02138nm10 nm

c

7

[LNChap2p.6]

Tutorial1Question5(EDescribetheelectrontransferp

formationoftheioniccompoun

Electropositive : Li:1s22s1

Electronegative: O:1s22s22p4

Li++Li++O2 Li2O

Li LiO

++ + ++

[LN: Noble gas:s2p6 ,Metals:1 #of


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Tutorial1Question6(IonicBonding):IftheattractiveforcebetweenapairofSr2+ andO2 ionsis1.29x

108NandtheionicradiusoftheO2 ionis0.132nm,calculatethe

ionicradiusoftheSr2+ ion innanometers.

21 2

net a tt ract ive repulsive 2 1

04n

Z Z e nbF F F

a a

2192

1 20

12 2 2 80 attractive

10

2 2 1.6 10 C

4 4 8.85 10 C / N m 1.29 10 N

2.672 10 m 0.2672nm

Z Z ea

F

0 0 0.2672nm 0.132nm 0.135nma r R r a R9

o = Permeabilityoffreespace,b&nareconstants

[LNChap2p.2223]

Tutorial1Question7(CUseschematicdiagramstodepictt

ineachofthefollowingmaterials:

(b)C2H6 (ethane)

(c)C2H3Cl (VinylChloride)

#ofcovalentbonds # ofvalenceelectNo.ofvalenceelectrons: H:1 O:6

c.f. C2HCl3 (Chloroform

8O:1s22s22p4

Usetheconceptofsecondarybondstrengthtopredictwhichmemberofeachpairof

materialsbelowhasahigher meltingtemp.

Matls Typeof 2o bond Size Ans: m.p. (oC)

(a) C2H4 WeakVan der Waals Similar Lower 169C

C2H2F2 Permanent dipolefromhighly electrove F(EN=4.1)

Similar Higher 144or

165C

(b) H2O Permanentdipolew moreelectrove O(EN=3.5)

+HBonding

Similar Higher 0C

H2S Permanentdipolewithelectrove S(EN=2.4)

Similar Lower 85.5C

(c) C3H8 organicmoleculescomposedofcarbonand

hydrogen V.d.W. only

Smaller (orshorter)

moleculeLower 189.7C

C12H26 organicmoleculescomposedofcarbonand

hydrogen V.d.W. only

Largermolecule

canformalarger

numberofdipoles

Higher 9.6C

11

Tutorial1Question9(SecondaryBonds): SecondaryBondin

Aspecialcase:hydrogenbonding

(a) It occurs between molecules in which h

(F), oxygen (O) and nitrogen (N).

(b) For each HF, HO and HN bond, the si

other atom. Thus, the hydrogen end of th

bare proton, which is capable of a strong at

adjacent molecule.

H Cl H Clsecondarybonding

Fluctuating

dipole

Permanent

dipole+ +


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Chap.3Cyrstal Structure1. CrystalStructure,SpaceLattice,UnitCell,Amorphous

2. 14Bravais Lattices:Cubic,Tetragonal,Orthorhombic,Rhombohedral,Hexagonal,Monoclin

3.

4. AtomicPackingFactor Interplanar spacing

5. AtomsPositions

6. CrystallographicDirtn:Start&End,End start(orwalk),I

7. MillerIndices:Origin,Intercept,Inverse,Integer,Bracket,

8. VolumeDensity(g/cm3 ), PlanarDensity(Ctr),LinearAto

9. Polymorphismor

Allotropy

10. XRayDiffraction:

hkla d

2

2 2 2 2

2sin

4h k l

a

2 sinhkl

d

2

2

sin

sin

A

B

3

3

4

3n r

a A A

A

A

A

A

AB

B BC

CC

Stacking:ABAB

vs ABCABC

BCC 2atoms

CN=8

0.68

FCC 4atoms

CN=12

0.74

HCP 6atoms

CN=12

0.74

4 3R a 4 2R a 2 c

R aa

Massu.c./Volumeu.c. Equiv.#ofatomctr intersected

SelectedArea

Equiv.#ofat

S

Integer#ofwavelength BCC:h+k+l =even,FCC:(h,k,l)allevenorallodd

HCP(0001) FCC(111)


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Thed130 interplanar spacinginaBCCelementis0.1587nm.

(a) Whatisitslatticeconstanta?

(b) Whatistheatomicradiusoftheelement?

(c) Whatcouldthiselement be?

Tutorial2Question5:

2 2 2 2 2 2

0.1587 1 3 0 0.502hkla d h k l nm (a)

3 0.50234 3 0.2174

4 4

nmaR a R nm (b)

(c) Lattice

Constant

Crystalstructure

Atomicradius

Qn 5 Ans: 0.502 BCC 0.2174

Barium,Ba 0.5019 BCC 0.217

Lead,Pb 0.49502 FCC 0.175

Potassium, K 0.5344 BCC 0.238

Silicon,Si 0.54282 Diamond 0.1171

Question8:ThelatticeconstantforBCCtantalumat20

16.6g/cm3.Calculateavalueforitsatomic

3

Atominumber of atoms

mV a

3

7

216.6

0.33026 10 6.023 10

Atomic mass

180.08 /Atomic mass g mol

18 1 2

8 Number of atoms

Calculatetheplanaratomicdensityinatomspersquaremillimetreforthe

(111)crystalplaneinFCCgold,whichhasalatticeconstantof0.40788nm.

1 1sin

2 2Area of Triangle bh ab

21 3

2 2 sin 602 2

oArea of a a a

1 1

3 3 22 6Number of atoms

2

6

13 2

2 2

3 0.40788 10

1.39 10 /

Number of AtomsPlanar Density

Area mm

atoms mm

3

a a

(111)

Question9:

1

6

1

2

1

4

12 12

Number of atoms

62.33 10

Number of Linear Density

Length

ato

2 2

6

7

2

2 0.3039 10

4.3 10

Face diagonal a a a

mm

Calculate the linear atomic density in at

direction in BCC vanadium, which has a lat

Draw[110]inBCCunitcell

Question10:


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5

Cubic structure

XRD

2A

2BXRDpattern

0.75

0.5

3

4

2

4

2

2 2 2 2

2sin

4h k l

a 2 2 2

hkla d h k l

2 sinhkld

No. 2 sin sin21 38.60o 19.30o 0.3305 0.109

2 55.71o 27.86o 0.4672 0.218

Anxraydiffractometer (Wavelength ofth

recorderchartforanelementthathaseithe

showeddiffractionpeaksatthefollowing2

38.60,55.71,69.70,82.55,95.00and10

(a) Determinethecrystalstructureofthe

(b) Determinethelatticeconstantofthee

(c) Identifytheelement.

Question11(Contd):

(a)

2 2 2

0.15405

2sin 2sin 19.3

na h k l

(b)

2

2 2 2 2

2sin

4h k l

a

7

Anxraydiffractometer (Wavelength oftheincomingradiation=0.15405nm)

recorderchartforanelementthathaseithertheBCCortheFCCcrystalstructure

showeddiffractionpeaksatthefollowing2 angles:

38.60,55.71,69.70,82.55,95.00and107.67.

(a) Determinethecrystalstructureoftheelement.

(b) Determinethelatticeconstantoftheelement.

(c) Identifytheelement.

Question11(Contd):

Lattice Constant(nm) CrystalstructureQn 11Ans: 0.3296 BCC

Magnesium, Mg 0.32094 HCP

Niobium,Nb 0.33007 BCC

Tantalum,Ta 0.33026 BCC

Zirconium,Zr 0.32312 HCP

(0.30to0.34)(c)

Coordinates

x, y, z

(0, 1,0)


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Drawdirectionvectorsinunitcubesforthefollowingcubicdirections:

9

Question1:

111

113

121

110(a) (b) (c) (d)

y

x

z

(0,0,1)

(0,1,0)(0,0,0)x

(b)1. Start&Endpositions:

2. End start(orwalk):

3. Integer:

4. Sqr Bracket:

1, 1, 0 0, 0, 0 1, 1, 0

1 1 0

1 1 0

1 1 0

1 1 0

x

x

1, 0, 0 0,1, 0 1, 1, 0

x

Drawdirectionvectorsinunitcubesforth

Question1(Contd):

111

(a) (b

(d 121

(c)

y

x

z

1

3

11

Question2:

Whataretheindicesofthedirectionsshownintheunitcube:

(e)1. Start&Endpositions:

2. End start(orwalk):

3. Integer:

4. Sqr Bracket:

(1,1,0)

1,0,1

4

1,0,1

4

1 3

, 0,1 1,1, 0 , 1, 14 4

3

4, 1 4, 1 4 3, 4, 44

(1,1,0)

3 4 4

Question2(Contd):

Whataretheindicesofthedirectionsshow

(g)

11,0,

4

10,1,

2


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13

Question2(Contd):

Whataretheindicesofthedirectionsshownintheunitcubebelow:

Subtractingorigincoordinates

fromemergencecoordinates

(e)

(f)

(g)

(h)

x y z

Origin: 1 1 1

Intercept: 1 1

Inverse: 1 1 0

Integer: 1 1 0

Bracket:

x

Origin: 0

I nt ercept : 1

Inverse: 1

Integer: 1

Bracket: 1

1 1 0 h k Family

15

Question3:

(a)1. Origin

2. Intercept

3. Inverse

4. Integer

5. Bracket

x y z

Origin: 1 0 1

Intercept: 1

Inverse: 1 0 3

Integer: 1 0 3

Bracket: 1 0 3

y

x

z

x

13

(1,0,1)

WhataretheMillerindicesofthecubiccrystallographicplanesshownbelow:

Question3(Contd):

W

c

(0,0,0) x y

x

z

1 10

4 4

2 1 8 3 51

3 4 12 12

5 1 1 12 30

12 4 4 5 5

y ax b

a b b

a a

y x x


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17

Question3(Contd):

x

x

(0,0,0)

Top

view

xx y

x

z

3 30

4 4

1 3 4 9 51

3 4 12 12

5 3 3 12 90

12 4 4 5 5

y ax b

a b b

a a

y x x

(c) x y z

Origin: 0 1 0

Intercept:

Inverse: X 9 X 9 0 X 9

Integer: 5 12 0

Bracket: 5 12 0

3

4

4

3

9

55

9

(0,1,0)

Question3(Contd):

(b) x y z

Origin: 0 1 1

Intercept: 1 1

Invers e: 1 X2 1 X2 X2

Integer: 2 2 3

Bracket: 2 2 3

2

3

WhataretheMillerindicesofthecubic

crystallographicplanesshownbelow:

3

2

(d)

Orig

Inte

Inve

Inte

Brac

19

(a)

Question4:

y

x

z

DrawinunitcubesthecrystalplanesthathavethefollowingMillerindices:

111 213 121 102(a) (b) (c) (d)

(a) x y z

Origin: 0 1 1

Intercept:

Inverse: 1 1 1

Integer:

Bracket: 1 1 1

x(0,1,1)

y

x

z

x

x

x

x

Originat

(b)

Question4(Contd):

y

x

z

Drawinunitcubesthecrystalplanesthat

111 102(a) (b) (c)

(0,0,1)

y

x

z

x

x

x x

x

x

1

2

Originat


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21

(c)

Question4: DrawinunitcubesthecrystalplanesthathavethefollowingMillerindices:

111 213 121 102(a) (b) (c) (d)

(c) x y z

Origin: 0 1 1

Inverse: 1 1

Bracket: 1 2 1

1

2

y

x

z

(0,1,1)

y

x

z

x

x

1

2

x

x

x

(d)

y

x

z

(0,0,1)

y

x

z

x

x12 x

x

x

1

3

(d) x y z

Origin: 0 0 1

Inverse: 1

Bracket:

1

2

1

3

2 1 3

OriginatOriginat

1

1

x

y

Intercept:

x=1

y=1

Intercep

a1 a2 a31 1 1 1

23

x

xx

DrawthehexagonalcrystalplaneswhoseMillerBravais indicesare:

Question6:

1 0 1 1(a) (b) (c) 0 1 1 1 1 2 1 0

(a) a1 a2 a3 c

Origin: 0 0 0 0

Intercept:

Inverse: 1 1 1

Integer:

Bracket:

c

a1

a2

a3

a3

1 0 1 1

(b)

(c)

c

c

(b)

Orig

Inte

Inve

Inte

Brac

(c)

Orig

Inte

Inve

Inte

Brac

Question6(Contd): DrawthehMillerBrav


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DetermineMillerBravais indicesofthehexagonalcrystalplanesinthefigues below:

25

Question7(Contd):

(b) a1 a2 a3 c

Origin: 0 0 0 1

Intercept: 1 1 1

Inverse: 1 1 0 1

Integer: 1 1 0 1

Bracket: 1 1 0 1

(b) a1 a2 a3 c

Origin: 0 0 0 0

Intercept: 1 1 1

Inverse: 1 1 0 1

Integer: 1 1 0 1

Bracket: 1 1 0 1

c

a1

a1

a2

DetermineMillerBravais indicesofthehexag

c

Question7:

(c)

Origin

Interc

Invers

Intege

Bracke

27

AreaofCD=a 0cos 30

cV HC

a

a

60o

A

B

CArea of

cV HCP

Equilateral

Triangle

VolumeofHCPunitcell:


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Chap.4Solidification&Imperfections

12nN d

VG

r

2

*1. HomogeneousNucleation:

2. HeterogeneousNucleation:

3. Growth,GrainRefiner,GrainBoundaries(polycrystalline),

SingleCrystal Creepresistance

4. Substitutional solidsolution:Substi.Solvent,distort,Solubility ifdia.


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vacancy

1. 2.

3.

4. Diffusionrate:solid


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Calculate the number of atoms in a critically sized nucleus for the

homogeneous nucleation of pure iron. Assume T (undercooling) = 0.2 Tm.

Tutorial3Question1(Nucleation):

*

7 2

10

3

2 2

0.2

2 204 10 J/cm 1808K9.72 10 m

2098J/ cm 0.2 1808K

m m

f f m

T Tr

H T H T

3

3 10 27 34 4vol.ofcriticalsizednucleus * 9.72 10 m 3.85 10 m

3 3r x

For 1criticallysizednucleus,

LatentheatoffusionHf =2098J/cm3, Surfaceenergy=204107J/cm2,

MeltingtemperatureTm =1808K,latticeconstantBCCa=0.28664nm.

r*

3

Tutorial3Question1(Con

3 vol.of1unitcellofFe 0.28664 1a

29 329 3

Averagevolume/atom (includesspace

2.355 1 0 m1.178 1 0 m

2

vol.ofnucleus 3.85

Volume/atom 1.17

For1BCCFeunitcell,

NumberofAtoms

in1nucleus =

Calculate the number of atoms in a c

homogeneous nucleation of pure iron. AssumLatentheatoffusionHf =2098J/cm

3, Surfaceenergy

MeltingtemperatureTm =1808K,latticeconstantBCCa

h

x

y

Calculate the radius of the largest interstitial void in the BCC iron lattice. Theatomic radius of the iron atoms in this lattice is 0.124nm, and the largest

interstitial voids occur at etc., type positions.

Tutorial3Question2(Interstitial):

1 14 2, ,0

2 2

2 25

4 2 16

a ah a

h =RFe +Rvoid

ForBCCstructure,

void Fe 0.160nm 0.124nm 0.036nmR h R

RFe

Rvoid

5

4 0.124nm43 4 0.286nm

3 3

Ra R a

225 5

0.286nm 0.160nm16 16

h a


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[AdaptedfromMST5001LecturenotesChap2]

Interstitialsite inFCCunitcell(Extra)

7

Describe and illustrate the edge and screwt

What types ofstrain fields surround both typ

Tutorial3Question3(Disl

Line imperfection caused by an extra

half plane of atoms between two

normal planes of atoms.

Lin

up

to

sep

EdgeDislocation

Tension

Compression

Shea

Dislocationline

[AdaptedfromMST5001LecturenotesChap4]9

Dislocations:BurgersCircui

Edgedislocation

definesense

of

Dislocation

line

closedcircuitbyRHrule

vectorfromStarttoFinish Burgersvector

T

[AdaptedfromMST5001LecturenotesChap4]


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If there are 400 grains per square inch on a photomicrograph of a ceramic

material at 200x, what is the ASTM grainsize number of the material?

Tutorial3Question4(Grainsize):

[LNChap4p36]

2

2 2100

200400grains/inch 1600grains/inch

100x

xN

x

1 12 1600 2 ln1600 1 ln2 10.64 1 11.64n nN n n

[TBCD:Tutorials Grainsize.htmlorTBExample4.5]

11

a) Calculatetheequilibriumconcentration

purecopperat850oC.Assumethatthee

inpurecopperis1.00eV,constant C=1.

b) Whatisthevacancyfractionat800oC?

Tutorial3Question5(Vac

5

VacancyFraction= 1 exp8.6

2.02 10

vn

N(b)

a) Calculatetheequilibriumconcentrationofvacanciespercubicmeterin

purecopperat850oC.Assumethattheenergyofformationofavacancy

inpurecopperis1.00eV,constantC=1.

b) Whatisthevacancyfractionat800oC?

Tutorial3Question5(Contd):

(AtomicmassofCu=63.54g/mol;DensityofCu=8.96g/cm3;

Avogadrosconstant=6.02x1023 atoms/mol;Boltzmannsconstant=8.62x105 eV/K)

2328 3

3

3

3

8.96 6.023 10 atoms / mol8.49 10 /

. . 63.54g /mol1001

1

AN g xN x atoms mA M cm

cmm

28 3

5

24 3

1eV8.49 10 atoms/ m exp

8.62 10 e V / K 850 273

2.77 10 vacancies/ m

vnK

exp expv v v

v

n E EC n NC

N kT kT

13

WriteequationforFicks firstlawofdiffusion,

Tutorial3Question6(Fick

2

atomsInSIunitform,

m s

dCJ D

dx

where

J = Flux,netflowofatoms

D

=

Diffusitivity, Diffusion

Coefficien

= ConcentrationgradientdC

dx


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Consider the gas carburizing of a gear of 1018 steel (0.18 wt%) at 927oC.

Calculate the time necessary to increase the carbon content to 0.35 wt% at

0.40 mm below the surface of the gear. Assume the carbon content at the

surface to be 1.15 wt%. D (C in iron) at 927oC = 1.28x1011m2/s.

0 2

S X

S

C C xerf

C C Dt

55.90170.8247 erf

t

Co=0.18wt%

T=927oC

t=?Cx =0.35wt%

x=0.4x103 m

Cs=1.15wt%

D=1.28x1011 m2/s

3

11 2

1.15 0.35 0.4 10 m

1.15 0.18 2 1.28 10 m / s

xerf

x t

Tutorial3Question7(Diffusion):

3400

3400sec 56.67mins60

st

s

55.9017

0.9587t

15

z erf (z)

0.95 0.8209

z 0.8247

1 0.8427

1 0.8427 0.8247

1 0.95 0.8427 0.8209

0.9587

z

z

z erf (z)

0.95 0.8209z 0.8247

1 0.8427

17

The diffusivity of copper atoms in the alumi

600oC and 2.50x1015 m2/s at 400oC. Calcula

case in this temperature range. Given R = 8.3

0 exp

QD D

RT

13 2

15 2

7.5 10 /exp

8.314 / 2.5 10 /

m s Q

J mol K m s

3139.3 10 J/Q x

0

1

2

0

exp

exp

DD

D D

Tutorial3Question8(Diff

Thankyou!


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Chap.6Mechanical

cos cosr

0

F

A

i 0

0

l l

l

Ti

F

A 0i

0

ln lnTi

Al

l A

E

(12): 1

(3): 00

FCC

HCP

y ok

d

(Coldwork(loadr.t.)createmoredefects(dislocations,atomsmove

dislocationmovementhinderedby GB&otherdislocations,Anneal

(deform2000%,>0.5Tm+slowstrain+5m+strainsensitive(resistivity),GBsoften,grainsslideandelongated)

(


20/40

ExplanatoryNotesforQn


21/40

Engineering stressstraindatawereobtainedatthe

beginningofatensiletest0.2%Cplaincarbonsteel.

(a) Plottheengineeringstressstraincurve

(b) Determinethe0.2percentoffsetyieldstress

(c) Determinethetensileelasticmodulus

Tutorial4Question1( ): (k psi) (in./in.)0 0

15 0.0005

30 0.001

40 0.0015

50 0.0020

60 0.0035

66 0.004

70 0.006

72 0.008

0

10

20

30

40

50

60

70

80

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009

Engineering

Stress

(1000

psi)

EngineeringStrain(in./in.)

(a)

0.2%( =0.002)yieldstress

=66x103 psi

3

7

30 10 psi

0.001

3.0 10 psi

xE

(b)

(c) [LNChap6p89]

2

A 0.505in.diameter aluminum alloy test

diameter of the bar is 0.490 in. at this load,

(a) the engineering stress and strain and (b)

22

2

0

2

2

0.505

0.200

i2 4

0.4900.189in

4i

d

A r

A

Assumenovolumechange, 0 0 i iA l A l

32

0

25000lb125 10 psi

0.200in

fFx

A

32

25000lb133 10 psi

0.189in

f

T

i

Fx

A

(a)

(b)

Question2(Tensile):

Units: psi:poundforcepersquareinch

a) Why do pure FCC metals like Ag and Cu have low values of critical resolved

shear stress c?

b) What is believed to be responsible for the high values ofc for HCP titanium?

For FCC metals like Ag and Cu, slip takes place on the closepacked {111}

o ctahedral pla nes an d i n the close p ac ked dire ctions .

Lower shear stress is required for slip to occur in densely packed planes.

The high c values associated with HCP titanium is attributed to the mixedcovalent and metallic bonding with the atomic lattice structure.

[TB Chap6 p237]

Question3(CriticalResolvedShearStress):

4

110

[TBChap6p236]

Slipsystem:Thesystem(plane+direction)i

HighestPlanardensity,HighestLineardensi


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111

111

111

111

110

101

011

1 1 2 2 3 3cos

, cos90 0o

u v u v u v u v u v

If perpendi cul ar u v u v

110

101

011

110

101

011

1 2 3u u u u

1 2 3v v v v

110

101

011

7

2 2 21 2 3u u u u

A stress of 75 MPa is applied in the [001] d

(a) the resolved shear stress acting on the

shear stress acting on the sli p s(111)[110]

Question4(Schmids Law

(a)

isbtw &SlipDirtF

isbtw &NormalofSlipPlaneF

cos 54.73

a

a

45Facediagonal

cos cos 75cos45 cos54.7 30.6MPar

Question4(Contd):

9

A stress of 75 MPa is applied in the [001] direction on an FCC single crystal. Calculate

(a) the resolved shear stress acting on the slip system and (b) the resolved

shear stress acting on the sli p system.

(111)[101]

(111)[110]

(b)

cos cos 75cosr

Question4(Contd):A stress of 75 MPa is applied in the [001] d

(a) the resolved shear stress acting on the

shear stress acting on the sli p s(111)[110]

asi


23/40

A 70% Cu 30% Zn brass wire is cold drawn 20 percent to a diameter of 2.80 mm. The

wire is then further colddrawn to a diameter of 2.45 mm. Definition of cold reduction is

given as:

changeincrosssectionalarea%coldreduction 100% 100%

originalcrosssectionalarea

o i

o

A Ax

A

a) Calculate the total percent cold work that the wire undergoes.

b) Estimate the wires tensile, yield strengths and elongation from figure (a) in the

appendix

2 2

1

2

1

2 2 21 1

1

2.80mm

2 220% 100%

2

0.20 7.84mm

3.13 mm

d

d

d d

d

2 2

2

Total%coldwork

3.13mm 2.45mm

2 2100%

3.13mm

2

38.75%

(a)

Question5(ColdWork):

11Figure (a): % Cold work v

work is expressedas a p

For Cold Work 39%,

TensileStrength 76kpsi

YieldStrength 64kpsi

ksi:1000psi

Elongation 7%

Question5(ColdWork):

A 70% Cu 30% Zn brass wire is cold draw

wire is then further colddrawn to a diamet

given as:

a) Calculate the total percent cold work t

b) Estimate the wires tensile, yield stren

appendix

Derivetheleverrulefortheamountinweightpercentofeachphaseintwophase

regionsof abinaryphasediagram.Useaphasediagraminwhichtwoelementsare

completelysolubleineachother. Isomorphous

1L SX X

0 L

S

S L

w w LOX

w w LS

S o

L

S L

w w OSX

w w LS

X:weightfraction

Total B = B in Liquid + B in Solid

Similarly,

0

0

0

0

1

L L S S

S L S S

L S L S S

L S S L

w X w X w

w X w X w

w w X w X w

w w X w w

Question6(LeverRule):

13

0 53 45

58 45

LS

S L

w wX

w w

58 53

58 45

S oL

S L

w wX

w w

0

5 845 58

13 13

17. 3 3 5.7 53 %

L L S Sw X w X w

wt N i

ExtraSlideonLeverRule

0

1 30 0 ,

53 %58 %

45 %

o

S

L

At T C

w wt Ni w wt Ni

w wt Ni

p/s: Tielineisusedin2phaseregiononly


24/40

Ag=88wt%

T=1000oC

Consider the binary eutectic coppersilver phase diagram. Make phase analyses of an

88 wt% Ag 12 wt% Cu alloy at the temperatures (a) 1000oC, (b) 800oC, (c) 780oC + T,(d) 780oC T. T is assumed to be less than 1oC. In the phase analyses, include:(i) The phases present; (ii) The chemical compositions of the phases; (iii) The amounts

of each phase; (iv) Sketch the microstructure by using 2cmdiameter circular fields.

wt% cm

10 1.25

8 1

Question7(PhaseDiagram):

15

(a)T =1000oC

(i)Phasespresent: Liqu

(ii)Compositionofphase: 88wt

(iii)Amountofeachphase:

T=1000oC

wt%ofliquidphase 100%

T=800oC

(b)T = 800oC

(i)Phasespresent: Liquid Beta(ii)Compositionofphase: 78wt%Ag 93wt% Ag

(iii)Amountofeachphase: (iv)Micro

structure

Ag=88wt%

93 88wt%ofliquidphase 100% 33.3%

93 78

88 78

wt%ofbetaphase 100% 66.7%93 78

17

(b)

9378

L

(c)T = 780oC+T

(i)Phasespresent: Liqu

(ii)Compositionofphase: 71.9w

(iii)Amountofeachphase:

A

91.2 88wt%ofliquidphase 10

91.2 71.9

88 71.9wt%ofbetaphase 100

91.2 71.9


25/40

(d)T = 780oC T

(i)Phasespresent: Alpha Beta

(ii)Compositionofphase: 7.9wt%Ag 91.2wt% Ag

(iii)Amountofeachphase: (iv)Microstructure

Ag=88wt%

T=780oC T, T


26/40

Os=30wt%

T=2665oC +T

(c)T = 2665oC+T

(i)Phasespresent: Liquid Beta

(ii)Compositionofphase: 23wt%Os 61.5wt% Os

(iii)Amountofeachphase: (iv)Microstructure

61.5 30wt%ofliquidphase 100% 81.8%

61.5 23

30 23wt%ofbetaphase 100% 18.2%

61.5 23

23Ni=

T=1517oC T

Consider an Fe 4.2 wt% Ni alloy that is s

the amount of phase and phase, respectT to less than 1oC.

Question9(Peritectic):

Ni=4.2wt%

T=1517oC T

4.2 4wt%of phase 100% 66.7%

4.3 4

4.3 4.2wt%of phase 100% 33.3%4.3 4

Peritectic:L+

RH

side:

L

+

LHside:L+L

25

L[LNChap7p15]

Eutectic L +Eutectoid +

Peritectic +L

Peritectoid +

Monotectic L1 +L2


27/40

Martensite:

Metastable phaseconsisting

ofsupersaturatedsolid

solutionofCinBCCorBCC

tetragonaliron.

Causedbyrapidcoolingof

austeniticsteelintoroom

temperature(quenching).


28/40

Pearlite(layered +Fe3C)

Bainite(nonlayered

Pearlite)

Ms

MfMartensite (bct)

250oC

550o

C

+Fe3C

+Fe3C

723oC

IsothermalTransformationDiagramofEutectoid pla

(0.8wt%C)

Time (s

ForTutorial5Question2


29/40

Chap.8MetallicAlloys1. Eutectoid(0.8wt%C,ifnothyper/hypo pro ) AusteniteFCC (at723o

FerriteBCC0.02wt%C+ Cementite/IronCarbideFe3C6.67wt%C

2. Martensite:

Quench to


30/40

Chap.9Polymers1. Polymerization:Chain:freeradical,Stepwise:Condensationbyproduct,N

2.

3. Vinyl,Vinylidene,zigzag,entangled,secondarybondsbtwchains,550n

Copolymers(random,alternating,block,graft)

4. Thermoplastics:PE:CH2=CH2,HD:littlebranching,chainsclosepack,incc

PVC: CH2=CHCl,strongstrengthstrongdipole,brittleCl steric hindrancer

PP:CH2=CHCH3 ,CH3restrictrotationincstrengthhardnessstabilitydec f

PS:CH2=CH ,bulky,rigid,inflexible,brittle,dimensionstable

PTFE:

CF2=CF2,

small

regular

F

dense,

strong

dipole

chemical

resistance

hi

5. Thermosetting:covalentbondnetworke.g.Phenolics rigidstablecreepc

6. Elastomers (Rubbers): Vulcanizationcrosslinking isoprenegreatdimensio

7. SolidificationofThermoplastics:

8. DeformationofThermoplastics:

9. Effectsof

Temperature:

10. StrengtheningofThermoplastics:incmolecularwt.(notmuchincbeyond

crystallinity (effectivepacking),pendant(hinderchainslippage),polar(m

MolecularmassofpolymerDP=

Massof

amer

i i

m

i

f MM

f


31/40

Ferrite

Cementite

orIron

carbide

Eutectoid

Pearlite

Hypo Eutectoid Hyper

0.8wt%C

+Fe3C

Proeutectoid Fe3C+Pearlite

Proeutectoid +Pearlite

Pearlite

A hypoeutectiod plaincarbon steel was slow

carbon concentration (in wt%) in the carbon

723C T? (Hint: see microstructure)

Ferrite

Auste

x

5.9wt%

Tutorial5Question1:

Draw timetemperature cooling paths for a 1080 steel on an isothermal transformation

diagram that will produce the following microstructures. Start with the steels in the

austenitic condition at time = 0 and 850C.

(f)100

%

lower

bainite

1080Steel 0.8wt%C Eutectoidcomposition

P

B

Ms

Mf

(a) 100%martensite

(c)100%finepearlite +Fe

3C

(e)100%upperbainite

Time (s)

Question2:

+Fe3C

Ms

Mf

(b)50%mar

+50%coars

(d) 50%martensite

+50%upperbainite

+Fe3C

+Fe3C

Question2(Contd):

1% 1


32/40

CentralLibraryRBR (2hrs)

Callnumber: TA403Smi 2009

Whatisthemicrostructureproducedafterau

Doesanaustempered steelneedtobetempe

After austempering, a eutectoid plaincarb

Subsequent tempering is not necessary sidistortion and impact energy values comp

with a marquenched and tempered steel.

A

[TBFig.9.34]

Question3:

reheatMartensite atmartensite

Martempering:Tempering:

QuenchAustensite to>Ms,

Holdtill

Tuniform,

then

slow

cool,

finalTempering

Temperedmartensite

Impactenergy>tempered

(Waterquench+Temper) (Marquenching +temper)

[TBFig.9.29] [TBFig.9.33]

Describethefourdecompositionstructurest

whenasupersaturatedsolidsolutionofanA

Supersaturated solid solution (FCGP1 zones: consist of segregated regions in th

few atoms thick (0.4 to 0.6nm) and

in diameter.

They form on the {100} cubic plane

straining the lattice tetragonally

with the matrix.

GP2 zones ( phase): approximately 1 to 4nm thick an

diameter.

These zones are also coherent wit

and have a tetragonal structure.

[TBp.407409]

Question4:


33/40

Supersaturated solid solution (FCC)GP1 zonesGP2 zones ( phase) phase: nucleates heterogeneously and is incoherent with the

matrix.

This phase has a tetragonal structure with a thicknessof 10 to 150nm.

phase (CuAl2): an incoherent equilibrium phase having the

composition CuAl2.

This phase, which forms from or directly from thematrix, has a BCT structure.

[PhaseTransformationsinMetalsandAlloysbyD.A.PorterandK.E.Eastering,p297]

Question4(Contd):Describethefourdecompositionstructuresthatcanbedeveloped

whenasupersaturatedsolidsolutionofanAl4%Cualloyisaged.

Extra:

Determine the mole fraction of polyvinyl chlo

having a molecular weight of 11,000 g/mol and

avav

MW (polymer) 11,000g/mMW (mer)

DP 150

av PVC PVC PVAc PVAc PMW (mer) MW MW f f f

PVCMW 3H 2C 1Cl

3 1 2 1 2 1 35.5

62.5g/ mol

PVAcMW

6 1

86.0g

PVC PVAc0.539 0.461f f

Question5:

Howisitpossibleforapolymerchainsuchasapolyethyleneonetokeepgrowing

spontaneouslyduringpolymerization?

The polymer chains in chain polymerization keep growing spontaneously because the

energy of the chemical system is lowered by the chain polymerization process; the

total energy of newly created polymers is lower than the total energy of the

monomers that have reacted to form them.

Monomerschemicallyreactwitheach

othertoproducelinearpolymersand

asmallmoleculeofbyproduct

[LNChap9p5,6,11]

heat

Question7:

1 . ChainPolymerization:

2. Stepwise

Polymerization

(Condensation):

Explainhowdoestheamountofcrystallin

(a)density,(b)tensilestrength

(a) Density increases, due to increase of t

(b) Tensile strength increases, due to str

between the polymer chains that are p

Question8:

Amorphous


34/40

How much sulphur must be added to 70g of butadiene rubber to cross link 3.0 percent of

the mers? (Assume all sulphur is used to crosslink the mers and that only one sulphur

atom is involved in each crosslinking bond.)

polybutadieneMW

6H 4C 54g/mol

NumberofofmolesofButadiene

70g = 1.2963

54g /mol

Vulcanization

3NumberofofmolesofSulphur= NumberofofmolesofButadiene

100

MassofSulphur=0.03 1.2963mol 32g/mol=1.24g

Question6:(lecturehasntcovered)

How does the presence of a methyl group

polymer chain affect the glass transition temp

polyethylene?

Thesubstitutionofhydrogenatomswithmeth

increasesthebondstrengthsbyrestrictingthe

Consequently,polypropylenehasahigherglas

[MaterialsScienceandEngineering:AnIntroduction

Callister 7th edn.p.501 CLRBRTA403Cal2007]

Question9:(lecturehasntcovered)

Polyethylene

[MaterialsScience

andEngineering:An

Introduction,byW.D.

Callister 7th edn.

p.548]

CentralLibraryRBR

Callnumber:

TA403Cal2007

Extra:FactorsthataffectGlassTransitionTemperatureTg :

[MaterialsScienceandEngineering:AnIntrod

For noncrystalline thermoplastic:

The liquid, upon solidification, changes to a superco

liquid that is in the solid state, which is rubbery or f

leathery, at temperature below TmAs the temperature decreases, the polymer shows a

decrease in specific volume until the temperature r

Below Tg, the polymer becomes glassy and brittle.

For partly crystalline polymer:

Its specific volume drop abruptlyat Tm , due to more

Extra:CharacteristicchangesduringGlassTra


35/40

Chap.10CeramicsOx

. , , , ,

2. AXStructure( +chargebalance):

3. FCCInterstitialsites:Octahedral Tetrahedral vs HCP

r

R

0.410.2250.155

CN=4

ZnS

4. CsCl,NaCl,ZnS,Antifluorite,CaF2

FCC:4atoms4octahed 8tetrahed

HCP:6atoms6octahed 12tetrahed

5. Corundum(Al2O3):6O2 formFCCwithAl3+ at4among6octahed

Li2ONa2O UO2NuclearFuel

emptyoctohed 3 2 2a r R

2 2a r R

3

4

ar R

3Graphite:3sp2formcovalentlayer,1pelectronbond btwlayers ,Spinel (

6. Silicates(SiO42):tetrahedralislandchain/ring(2corners)

2,

7. MaterialspreparationFormDryingSinteringVitrification

~w S con_ ox e

. 2 3, 3 4, , 2

9. Toughness: ZrO2 9mol%MgO cubictetragona IC fK Y a

10. Glasses: Structure,Composition, TemperedGlas

* expoQ

RT


36/40

[BookbyCallister CLRBRTA403Cal2007]

2

3a

3

4

a

Tutorial6Questi

Calculatethecriticalradius

Calculatethedensity ingramspercubiccentimetreofCsI,whichhastheCsCl structure.

IonicradiiareCs+ =0.165nmandI =0.220nm.

3 2a r R

Question2:

8

20.165nm 0.220nm

3

0.445n m 4.45 10 cm

a

22ll132.9

g / mol 126.9

g /mol

un tcell 23 23 .

6.02x10 ions/ mol 6.02x10 ion/ mol

223

3 38

4.32 1 0 g4.90g/cm

4.45 10 cm

m mdensity

v a

4

Calculate the planar densitie

(111) and (110) planes for (a)

Ionic radii are Co2+ = 0.082nm

2

(a)CoO

Question3:

2

0.0820.621

0.132

r o

r O

0.414 0.732r

R


37/40

2


38/40

Ce O2 4Ce4+ 8O2

Question4:Calculate the linear density of Ce4+ and O2 in ions per nm2

in the [111] and [110] directions for CeO2,

which has the calcium fluorite (CaF2) structure.

Ionic radii are Ce4+ = 0.102nm and O2 = 0.132nm.

4 2Ce O

3

4

40.102 0.132 0.540nm

3

a r R

a

4 24 2

L

12 Ce 2O

1Ce 2O2111

10

4 23a 3 0.540nm

1.07Ce 2.14O /nm

44

4L

12 1 Ce

2Ce2110 2.62Ce / nm

2a 2 0.540nm

CaF2

4Ca2+ formsFCC

8F occupytetrahedral

Antifluorite

e.g.Na2O,

Li2O

What causes the lack of plasticity in crystalline

ceramics?

Thelackofplasticity incrystallineceramics

is attributed to their Ionic and Covalent

Question5:

chemicalbonds.

InCovalentlybondedceramics,thebondingbetweenatomsis

specific anddirectional.

Specific:betweenthese2atoms,

Directional:must

be

of

certain

angle 1

2

3

4

5

67

1

2

3

4

5

6

7

[TBChap11p611613]

Therefore,theywillundergobrittle

fracturesascovalentbondsare

brokenwithoutsubsequentreformation.(unlikeinmetallicmaterials)

1 2 3 4 5 6 1 2 3 4 5 612

For Ioniccrystals,theslipofoneplaneofions

involvesionsofdifferentchintocontact,andthusattrac

Question5(Cont

repu s ve orcesmay epro

Forexample,intheNaCltyp

Sliponthe{110}familyofpslipplanesremainattracted

Incontrast,sliponthe{100

samechargeswillrepeleac Hence,therearelimitedsli

lackofplasticity inionically

Polycrystalline ioniccerami

cannotcontinueacrossthe


39/40

(110)

(100)

110

(100)

[010]

(110)

impossible

14

possible

Explain the plastic deform

solids such as NaCl and MgO

What is the preferred slip sy

Question6:

Fo r some sin glecry

compressive stresses at

prior to fracture because

other.

These slip planes, wh

110typ ca y pre er t e

How do (a) porosity and (b) grain size affect the tensile strength of ceramic

materials?

a) Porosity: [LNChap10p24]

Question7:

Pores serve as stress concentration points and induce cracks at lower

threshold values.

They decrease the effective crosssectional area for load distribution as

well.

Consequently, as the porosity of a ceramic material increases, the material

tensile strength decreases.

b Grain Size:

For a porosityfree ceramic, the flaw size, and thus the strength, is solely a

function of grain size

The finer the grain size, the smaller the flaws and the greater the ceramic

tensile strength.

16

s

(surfacedefects, void

[LNlastyear]


40/40

mle1101 ay1213 sem2 detailed tutorial solutions

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