Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.
102 Pages
askelandphulenotes-ch03printable
Course: ME 2733, Fall 2010School: LSU
Rating:
Word Count: 4234
Document Preview
TheScienceandEngineeringof Materials,4thed DonaldR.AskelandPradeepP.Phul Chapter3AtomicandIonic Arrangements 1 ObjectivesofChapter3 Tolearnclassificationofmaterialsbased onatomic/ionicarrangements Todescribethearrangementsincrystalline solidsbasedonlattice,basis,andcrystal structure 2 ChapterOutline 3.1ShortRangeOrderversusLongRange Order 3.2AmorphousMaterials:Principlesand...
Unformatted Document Excerpt
Coursehero >> Louisiana >> LSU >> ME 2733Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
TheScienceandEngineeringof
Materials,4thed
DonaldR.AskelandPradeepP.Phul
Chapter3AtomicandIonic
Arrangements
1
ObjectivesofChapter3
Tolearnclassificationofmaterialsbased
onatomic/ionicarrangements
Todescribethearrangementsincrystalline
solidsbasedonlattice,basis,andcrystal
structure
2
ChapterOutline
3.1ShortRangeOrderversusLongRange
Order
3.2AmorphousMaterials:Principlesand
TechnologicalApplications
3.3Lattice,UnitCells,Basis,andCrystal
Structures
3.4AllotropicorPolymorphic
Transformations
3.5Points,Directions,andPlanesinthe
UnitCell
3.6InterstitialSites
3.7CrystalStructuresofIonicMaterials
3.8CovalentStructures
3
Section3.1Short
RangeOrderversusLongRange
Order
ShortrangeorderTheregularandpredictablearrangementof
theatomsoverashortdistanceusuallyoneortwoatom
spacings.
Longrangeorder(LRO)Aregularrepetitivearrangementof
atomsinasolidwhichextendsoveraverylargedistance.
BoseEinsteincondensate(BEC)Anewlyexperimentally
verifiedstateofamatterinwhichagroupofatomsoccupythe
samequantumgroundstate.
4
Figure3.1Levelsof
atomicarrangementsin
materials:(a)Inert
monoatomicgaseshave
noregularorderingof
atoms:(b,c)Some
materials,includingwater
vapor,nitrogengas,
amorphoussiliconand
silicateglasshaveshort
rangeorder.(d)Metals,
alloys,manyceramicsand
somepolymershave
regularorderingof
atoms/ionsthatextends
throughthematerial.
(c) 2003 Brooks/Cole Publishing / Thomson Learning
5
Figure3.2BasicSi0
tetrahedroninsilicate
glass.
(c) 2003 Brooks/Cole Publishing / Thomson Learning
6
Figure3.3Tetrahedral
arrangementofCHbondsin
polyethylene.
(c) 2003 Brooks/Cole Publishing / Thomson Learning
7
Figure3.4(a)Photographofasilicon
singlecrystal.(b)Micrographofa
polycrystallinestainlesssteelshowing
grainsandgrainboundaries(Courtesy
Dr.M.Hua,Dr.I.Garcia,andDr.A.J.
Deardo.)
8
Figure3.5Liquidcrystaldisplay.Thesematerialsareamorphous
inonestateandundergolocalizedcrystallizationinresponseto
anexternalelectricfieldandarewidelyusedinliquidcrystal
displays.(CourtesyofNickKoudis/PhotoDisc/GettyImages.)
9
(c) 2003 Brooks/Cole Publishing / Thomson
Learning
Figure3.7Classificationofmaterialsbasedonthetypeofatomicorder.
10
Section3.2
AmorphousMaterials:Principlesand
TechnologicalApplications
AmorphousmaterialsMaterials,includingglasses,thathaveno
longrangeorder,orcrystalstructure.
GlassesSolid,noncrystallinematerials(typicallyderivedfrom
themoltenstate)thathaveonlyshortrangeatomicorder.
GlassceramicsAfamilyofmaterialstypicallyderivedfrom
molteninorganicglassesandprocessedintocrystallinematerials
withveryfinegrainsizeandimprovedmechanicalproperties.
11
(c) 2003 Brooks/Cole Publishing / Thomson
Learning
Figure3.9(b)Thisfigureshowsaschematicoftheblowstretchprocess
usedforfabricationofastandardtwoliterPET(polyethyleneterephthalate)
bottlefromapreform.Thestressinducedcrystallizationleadsto
formationofsmallcrystalsthathelpreinforcetheremainingamorphous
matrix.
12
(c) 2003 Brooks/Cole Publishing / Thomson
Learning
Figure3.10Atomicarrangementsincrystallinesiliconandamorphous
silicon.(a)Amorphoussilicon.(b)Crystallinesilicon.Notethe
variationintheinteratomicdistanceforamorphoussilicon.
13
Section3.3
Lattice,UnitCells,Basis,andCrystal
Structures
LatticeAcollectionofpointsthatdividespaceintosmaller
equallysizedsegments.
BasisAgroupofatomsassociatedwithalatticepoint.
UnitcellAsubdivisionofthelatticethatstillretainstheoverall
characteristicsoftheentirelattice.
AtomicradiusTheapparentradiusofanatom,typically
calculatedfromthedimensionsoftheunitcell,usingclose
packeddirections(dependsuponcoordinationnumber).
PackingfactorThefractionofspaceinaunitcelloccupiedby
atoms.
14
Figure3.11The
fourteentypesof
Bravaislattices
groupedinseven
crystalsystems.The
actualunitcellfora
hexagonalsystemis
showninFigures3.12
and3.16.
(c) 2003 Brooks/Cole Publishing / Thomson Learning
15
16
Figure3.12
Definitionofthe
latticeparameters
andtheirusein
cubic,
orthorhombic,and
hexagonalcrystal
systems.
(c) 2003 Brooks/Cole Publishing / Thomson Learning
17
Figure3.13(a)
Illustrationshowing
sharingoffaceand
corneratoms.(b)The
modelsforsimple
cubic(SC),body
centeredcubic(BCC),
andfacecentered
cubic(FCC)unitcells,
assumingonlyone
atomperlatticepoint.
(c) 2003 Brooks/Cole Publishing / Thomson Learning
18
Example3.1Determining
theNumberofLatticePointsinCubicCrystal
Systems
Determinethenumberoflatticepointspercellinthecubiccrystal
systems.Ifthereisonlyoneatomlocatedateachlatticepoint,calculate
thenumberofatomsperunitcell.
Example3.1SOLUTION
IntheSCunitcell:latticepoint/unitcell=(8corners)1/8=1
InBCCunitcells:latticepoint/unitcell=(8
corners)1/8+(1center)(1)=2
InFCCunitcells:latticepoint/unitcell=(8
corners)1/8+(6faces)(1/2)=4
Thenumberofatomsperunitcellwouldbe1,2,and4,forthesimple
cubic,bodycenteredcubic,andfacecenteredcubic,unitcells,
respectively.
19
Example3.2
DeterminingtheRelationshipbetweenAtomic
RadiusandLatticeParameters
Determinetherelationshipbetweentheatomicradiusandthe
latticeparameterinSC,BCC,andFCCstructureswhenone
atomislocatedateachlatticepoint.
(c) 2003 Brooks/Cole Publishing / Thomson
Learning
Figure3.14TherelationshipsbetweentheatomicradiusandtheLattice
parameterincubicsystems(forExample3.2).
20
Example3.2SOLUTION
ReferringtoFigure3.14,wefindthatatomstouchalongthe
edgeofthecubeinSCstructures.
a0
= 2r
InBCCstructures,atomstouchalongthebodydiagonal.There
aretwoatomicradiifromthecenteratomandoneatomicradius
fromeachofthecorneratomsonthebodydiagonal,so
a0
4r
=
3
InFCCstructures,atomstouchalongthefacediagonalofthe
cube.Therearefouratomicradiialongthislengthtworadii
fromthefacecenteredatomandoneradiusfromeachcorner,
so:
a0
4r
=
2
21
(c) 2003 Brooks/Cole Publishing / Thomson
Learning
Figure3.15Illustrationofcoordinationsin(a)SCand(b)BCCunitcells.
SixatomstoucheachatominSC,whiletheeightatomstoucheach
atomintheBCCunitcell.
22
Example3.3
CalculatingthePackingFactor
CalculatethepackingfactorfortheFCCcell.
Example3.3SOLUTION
InaFCCcell,therearefourlatticepointspercell;ifthereisoneatom
perlatticepoint,therearealsofouratomspercell.Thevolumeofone
atomis4r3/3andthevolumeoftheunitcellis.
a
3
0
43
(4 atoms/cell)( r )
3
Packing Factor =
a03
Since, for FCC unit cells, a 0 = 4r/
43
(4)( r )
3
Packing Factor =
=
3
( 4r / 2 )
23
2
0.74
18
Example3.4Determining
theDensityofBCCIron
DeterminethedensityofBCCiron,whichhasalatticeparameterof
0.2866nm.
Example3.4SOLUTION
Atoms/cell=2,a0=0.2866nm=2.866 108cm
Atomicmass=55.847g/mol
3
Volumeofunitcell==(2.866
0
a
108cm)3=23.54 1024cm3/cell
AvogadrosnumberNA=6.02 1023atoms/mol
(number of atoms/cell)(atomic mass of iron)
Density =
(volume of unit cell)(Avogadro' s number)
(2)(55.847)
3
=
= 7.882 g / cm
(23.54 10 24 )(6.02 10 23 )
24
(c) 2003 Brooks/Cole Publishing / Thomson
Learning
Figure3.16Thehexagonalclosepacked(HCP)structure(left)andits
unitcell.
25
26
Section3.4
AllotropicorPolymorphic
Transformations
AllotropyThecharacteristicofanelementbeingabletoexistin
morethanonecrystalstructure,dependingontemperatureand
pressure.
PolymorphismCompoundsexhibitingmorethanonetypeof
crystalstructure.
27
Figure3.17Oxygengassensorsusedincarsand
otherapplicationsarebasedonstabilizedzirconia
compositions.(ImagecourtesyofBoschRobert
BoschGmbH.)
28
Example3.5Calculating
VolumeChangesinPolymorphsofZirconia
Calculatethepercentvolumechangeaszirconiatransformsfrom
atetragonaltomonoclinicstructure.[9]Thelatticeconstantsforthe
monoclinicunitcellsare:a=5.156,b=5.191,andc=5.304,
respectively.Theangleforthemonoclinicunitcellis98.9.The
latticeconstantsforthetetragonalunitcellarea=5.094andc=
5.304,respectively.[10]Doesthezirconiaexpandorcontract
duringthistransformation?Whatistheimplicationofthis
transformationonthemechanicalpropertiesofzirconiaceramics?
29
Example3.5SOLUTION
ThevolumeofatetragonalunitcellisgivenbyV=a2c=
(5.094)2(5.304)=134.333.
ThevolumeofamonoclinicunitcellisgivenbyV=abcsin
=(5.156)(5.191)(5.304)sin(98.9)=140.253.
Thus,thereisanexpansionoftheunitcellasZrO2transformsfroma
tetragonaltomonoclinicform.
Thepercentchangeinvolume
=(finalvolumeinitialvolume)/(initialvolume)100
=(140.25134.333)/140.253*100=4.21%.
Mostceramicsareverybrittleandcannotwithstandmorethana
0.1%changeinvolume.TheconclusionhereisthatZrO2ceramicscannot
beusedintheirmonoclinicformsince,whenzirconiadoestransformtothe
tetragonalform,itwillmostlikelyfracture.Therefore,ZrO2isoftenstabilized
inacubicformusingdifferentadditivessuchasCaO,MgO,andY2O3.
30
Example3.6Designing
aSensortoMeasureVolume
Change
Tostudyhowironbehavesatelevatedtemperatures,wewouldliketo
designaninstrumentthatcandetect(witha1%accuracy)thechange
involumeofa1cm3ironcubewhentheironisheatedthroughits
polymorphictransformationtemperature.At911oC,ironisBCC,witha
latticeparameterof0.2863nm.At913oC,ironisFCC,withalattice
parameterof0.3591nm.Determinetheaccuracyrequiredofthe
measuringinstrument.
Example3.6SOLUTION
ThevolumeofaunitcellofBCCironbeforetransformingis:
VBCC==(0.2863nm)3=0.023467nm3
3
a0
31
Example3.6SOLUTION(Continued)
ThevolumeoftheunitcellinFCCironis:
3
VFCC==(0.3591nm)3=0.046307nm3
a
0
Butthisisthevolumeoccupiedbyfourironatoms,as
therearefouratomsperFCCunitcell.Therefore,wemust
comparetwoBCCcells(withavolumeof2(0.023467)=
0.046934nm3)witheachFCCcell.Thepercentvolumechange
duringtransformationis:
(0.046307 - 0.046934)
Volume change =
100 = 1.34%
0.046934
The1cm3cubeofironcontractsto10.0134=0.9866
cm3aftertransforming;therefore,toassure1%accuracy,the
instrumentmustdetectachangeof:
V=(0.01)(0.0134)=0.000134cm3
32
Section3.5Points,
Directions,andPlanesintheUnitCell
MillerindicesAshorthandnotationtodescribecertain
crystallographicdirectionsandplanesinamaterial.Denotedby
[]brackets.Anegativenumberisrepresentedbyabaroverthe
number.
DirectionsofaformCrystallographicdirectionsthatallhavethe
samecharacteristics,althoughtheirsenseisdifferent.Denoted
byhibrackets.
RepeatdistanceThedistancefromonelatticepointtothe
adjacentlatticepointalongadirection.
LineardensityThenumberoflatticepointsperunitlengthalong
adirection.
PackingfractionThefractionofadirection(linearpacking
fraction)oraplane(planarpackingfactor)thatisactually
coveredbyatomsorions.
33
Figure3.18Coordinatesofselectedpointsin
theunitcell.Thenumberreferstothe
distancefromtheoriginintermsoflattice
parameters.
34
Example3.7
DeterminingMillerIndicesofDirections
DeterminetheMillerindicesofdirectionsA,B,andCin
Figure3.19.
Figure3.19Crystallographic
directionsandcoordinates(for
Example3.7).
(c) 2003 Brooks/Cole Publishing /
Thomson Learning
35
Example3.7SOLUTION
DirectionA
1.Twopointsare1,0,0,and0,0,0
2.1,0,0,0,0,0=1,0,0
3.Nofractionstoclearorintegerstoreduce
4.[100]
DirectionB
1.Twopointsare1,1,1and0,0,0
2.1,1,1,0,0,0=1,1,1
3.Nofractionstoclearorintegerstoreduce
4.[111]
DirectionC
1.Twopointsare0,0,1and1/2,1,0
2.0,0,11/2,1,0=1/2,1,1
3.2(1/2,1,1)=1,2,2
4. [ 1 2 2]
36
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.20Equivalencyofcrystallographicdirectionsofaformin
cubicsystems.
37
38
Figure3.21
Determiningthe
repeatdistance,
lineardensity,and
packingfraction
for[110]direction
inFCCcopper.
(c) 2003 Brooks/Cole Publishing / Thomson Learning
39
Example3.8
DeterminingMillerIndicesofPlanes
DeterminetheMillerindicesofplanesA,B,andCinFigure3.22.
Figure3.22
Crystallographicplanes
andintercepts(forExample
3.8)
(c) 2003 Brooks/Cole Publishing /
Thomson Learning
40
Example3.8SOLUTION
PlaneA
1.x=1,y=1,z=1
2.1/x=1,1/y=1,1/z=1
3.Nofractionstoclear
4.(111)
PlaneB
1.Theplaneneverinterceptsthezaxis,sox=1,y=2,andz=
2.1/x
=1,1/y=1/2,1/z=0
3.Clearfractions:
1/x=2,1/y=1,1/z=0
4.(210)
PlaneC
1.Wemustmovetheorigin,sincetheplanepassesthrough0,0,
0.Letsmovetheoriginonelatticeparameterintheydirection.Then,x
=,y=1,andz=
2.1/x=0,1/y=1,1/z=0
3.Nofractionstoclear.
4. ( 0 1 0)
41
42
Example3.9Calculating
thePlanarDensityandPackingFraction
Calculatetheplanardensityandplanarpackingfractionforthe
(010)and(020)planesinsimplecubicpolonium,whichhasa
latticeparameterof0.334nm.
Figure3.23The
planerdensitiesofthe
(010)and(020)planes
inSCunitcellsare
notidentical(for
Example3.9).
(c) 2003 Brooks/Cole Publishing /
Thomson Learning
43
Example3.9SOLUTION
Thetotalatomsoneachfaceisone.Theplanardensityis:
atom per face
1 atom per face
Planar density (010) =
=
2
area of face
(0.334)
= 8.96 atoms/nm 2 = 8.96 1014 atoms/cm 2
Theplanarpackingfractionisgivenby:
area of atoms per face
(1 atom) (r )
Packing fraction (010) =
=
area of face
(a 0) 2
r 2
=
= 0.79
2
( 2r )
However,noatomsarecenteredonthe(020)planes.Therefore,
theplanardensityandtheplanarpackingfractionarebothzero.
The(010)and(020)planesarenotequivalent!
44
2
Example3.10Drawing
DirectionandPlane
[12 1]
[ 2 10]
Draw(a)thedirectionand(b)theplaneinacubicunit
cell.
Figure3.24
Constructionofa
(a)directionand
(b)planewithina
unitcell(for
Example3.10)
(c) 2003 Brooks/Cole Publishing / Thomson
Learning
45
Example3.10SOLUTION
a.Becauseweknowthatwewillneedtomoveinthenegativey
direction,letslocatetheoriginat0,+1,0.Thetailofthe
directionwillbelocatedatthisneworigin.Asecondpointonthe
directioncanbedeterminedbymoving+1inthexdirection,2in
theydirection,and+1inthezdirection[Figure3.24(a)].
b.Todrawintheplane,firsttakereciprocalsoftheindices
[ 2 10]
toobtaintheintercepts,thatis:
x=1/2=1/2y=1/1=1z=1/0=
Sincethexinterceptisinanegativedirection,andwewishto
drawtheplanewithintheunitcell,letsmovetheorigin+1inthe
xdirectionto1,0,0.Thenwecanlocatethexinterceptat1/2
andtheyinterceptat+1.Theplanewillbeparalleltothezaxis
[Figure3.24(b)].
46
Figure3.25MillerBravaisindices
areobtainedforcrystallographic
planesinHCPunitcellsbyusinga
fouraxiscoordinatesystem.The
planeslabeledAandBandthe
directionlabeledCandDarethose
discussedinExample3.11.
(c) 2003 Brooks/Cole Publishing / Thomson
Learning
47
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.26TypicaldirectionsintheHCPunitcell,usingboththreeand
fouraxissystems.Thedashedlinesshowthatthe[1210]directionis
equivalenttoa[010]direction.
48
Example3.11Determining
theMillerBravaisIndicesforPlanesand
Directions
DeterminetheMillerBravaisindicesforplanesAandBand
directionsCandDinFigure3.25.
Figure3.25MillerBravais
indicesareobtainedfor
crystallographicplanesinHCP
unitcellsbyusingafouraxis
coordinatesystem.Theplanes
labeledAandBandthe
directionlabeledCandDare
thosediscussedinExample
3.11.
(c) 2003 Brooks/Cole Publishing /
Thomson Learning
49
Example3.11SOLUTION
PlaneA
1.a1=a2=a3=,c
=1
2.1/a1=1/a2=1/a3=0,1/c=1
3.Nofractionstoclear
4.(0001)
PlaneB
1.a1=1,a2=1,a3=1/2,c=1
2.1/a1=1,1/a2=1,1/a3=2,1/c=1
3.Nofractionstoclear
(11 2 1)
4.
DirectionC
1.Twopointsare0,0,1and1,0,0.
2.0,0,1,1,0,0=1,0,1
3.Nofractionstoclearorintegerstoreduce.
[ 1 01] or [2113]
4.
50
Example3.11SOLUTION(Continued)
DirectionD
1.Twopointsare0,1,0and1,0,0.
2.0,1,0,1,0,0=1,1,0
3.Nofractionstoclearorintegerstoreduce.
4. [ 1 10] or [ 1 100]
51
52
Figure3.27The
ABABABstacking
sequenceofclose
packedplanes
producestheHCP
structure.
(c) 2003 Brooks/Cole Publishing / Thomson
Learning
53
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.28TheABCABCABCstackingsequenceofclose
packedplanesproducestheFCCstructure.
54
Section3.6Interstitial
Sites
InterstitialsitesLocationsbetweenthenormalatomsorions
inacrystalintowhichanotherusuallydifferentatomorionis
placed.Typically,thesizeofthisinterstitiallocationissmaller
thantheatomorionthatistobeintroduced.
CubicsiteAninterstitialpositionthathasacoordinationnumber
ofeight.Anatomorioninthecubicsitetoucheseightother
atomsorions.
OctahedralsiteAninterstitialpositionthathasacoordination
numberofsix.Anatomorionintheoctahedralsitetouchessix
otheratomsorions.
TetrahedralsiteAninterstitialpositionthathasacoordination
numberoffour.Anatomorioninthetetrahedralsitetouches
fourotheratomsorions.
55
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.29Thelocationoftheinterstitialsitesincubicunitcells.Only
representativesitesareshown.
56
Example3.12Calculating
OctahedralSites
Calculatethenumberofoctahedralsitesthatuniquelybelongto
oneFCCunitcell.
Example3.12SOLUTION
Theoctahedralsitesincludethe12edgesoftheunitcell,with
thecoordinates
1
,0,0
2
1
0, ,0
2
1
0,0,
2
1
1
1
,1,0
,0,1
,1,1
2
2
2
1
1
1
1, ,0 1, ,1 0, ,1
2
2
2
1
1
1
1,0,
1,1,
0,1,
2
2
2
plusthecenterposition,1/2,1/2,1/2.
57
Example3.12SOLUTION(Continued)
Eachofthesitesontheedgeoftheunitcellissharedbetween
fourunitcells,soonly1/4ofeachsitebelongsuniquelytoeach
unitcell.
Therefore,thenumberofsitesbelonginguniquelytoeachcellis:
(12edges)(1/4percell)+1centerlocation
=4octahedralsites
58
59
Example3.13Designof
aRadiationAbsorbingWall
Wewishtoproducearadiationabsorbingwallcomposedof10,000
leadballs,each3cmindiameter,inafacecenteredcubic
arrangement.Wedecidethatimprovedabsorptionwilloccurifwe
fillinterstitialsitesbetweenthe3cmballswithsmallerballs.Design
thesizeofthesmallerleadballsanddeterminehowmanyare
needed.
Figure3.30Calculationofan
octahedralinterstitialsite(for
Example3.13).
(c) 2003 Brooks/Cole Publishing /
Thomson Learning
60
Example3.13SOLUTION
First,wecancalculatethediameteroftheoctahedral
siteslocatedbetweenthe3cmdiameterballs.Figure3.30shows
thearrangementoftheballsonaplanecontaininganoctahedral
site.
LengthAB=2R+2r=2R
2
r=RR=(1)R
2
r/R=0.414
2
ThisisconsistentwithTable36.Sincer=R=0.414,theradiusof
thesmallleadballsis
r=0.414*R=(0.414)(3cm/2)=0.621cm.
FromExample312,wefindthattherearefouroctahedralsitesin
theFCCarrangement,whichalsohasfourlatticepoints.
Therefore,weneedthesamenumberofsmallleadballsaslarge
leadballs,or10,000smallballs.
61
Section3.7Crystal
StructuresofIonicMaterials
Factorsneedtobeconsideredinordertounderstandcrystal
structuresofionicallybondedsolids:
IonicRadii
ElectricalNeutrality
ConnectionbetweenAnionPolyhedra
VisualizationofCrystalStructuresUsingComputers
62
(c) 2003 Brooks/Cole Publishing / Thomson
Learning
Figure3.31Connectionbetweenanionpolyhedra.Differentpossible
connectionsincludesharingofcorners,edges,orfaces.Inthisfigure,
examplesofconnectionsbetweentetrahedraareshown.
63
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.32(a)Thecesiumchloridestructure,aSCunitcellwithtwo
ions(Cs+andCI)perlatticepoint.(b)Thesodiumchloridestructure,a
FCCunitcellwithtwoions(Na++CI)perlatticepoint.Note:Ionsizes
nottoscale.
64
Example3.14Radius
RatioforKCl
Forpotassiumchloride(KCl),(a)verifythatthecompoundhasthe
cesiumchloridestructureand(b)calculatethepackingfactorforthe
compound.
Example3.14SOLUTION
a.FromAppendixB,rK+=0.133nmandrCl=0.181nm,so:
rK+/rCl=0.133/0.181=0.735
Since0.732<0.735<1.000,thecoordinationnumberforeachtype
ofioniseightandtheCsClstructureislikely.
65
Example3.14SOLUTION(Continued)
b.Theionstouchalongthebodydiagonaloftheunitcell,so:
a0=2rK++2rCl=2(0.133)+2(0.181)=0628nm
a0=0.363nm
43
43
rK + (1 K ion) + rCl (1 Cl ion)
3
Packing factor = 3
3
a0
4
4
3
(0.133) + (0.181) 3
3
=3
= 0.725
3
(0.363)
66
Example3.15
IllustratingaCrystalStructureandCalculating
Density
ShowthatMgOhasthesodiumchloridecrystalstructureand
calculatethedensityofMgO.
Example3.15SOLUTION
FromAppendixB,rMg+2=0.066nmandrO2=0.132nm,so:
rMg+2/rO2=0.066/0.132=0.50
Since0.414<0.50<0.732,thecoordinationnumberfor
eachionissix,andthesodiumchloridestructureispossible.
67
Example3.15SOLUTION
Theatomicmassesare24.312and16g/molformagnesiumand
oxygen,respectively.Theionstouchalongtheedgeofthecube,
so:
a0 =2rMg+2+2rO2=2(0.066)+2(0.132)
=0.396nm=3.96 108cm
(4Mg +2 )(24.312) + (4O -2 )(16)
=
= 4.31g / cm3
(3.96 10 8 cm)3 (6.02 10 23 )
68
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.33(a)Thezincblendeunitcell,(b)planview.
69
Example3.16
CalculatingtheTheoreticalDensityof
GaAs
Thelatticeconstantofgalliumarsenide(GaAs)is5.65.
ShowthatthetheoreticaldensityofGaAsis5.33g/cm3.
Example3.16SOLUTION
ForthezincblendeGaAsunitcell,therearefourGaand
fourAsatomsperunitcell.
Fromtheperiodictable(Chapter2):
Eachmole(6.023 1023atoms)ofGahasamassof69.7g.
Therefore,themassoffourGaatomswillbe(4*69.7/6.023
1023)g.
70
Example3.16SOLUTION(Continued)
Eachmole(6.023 1023atoms)ofAshasamassof74.9g.
Therefore,themassoffourAsatomswillbe(4*
74.9/6.023 1023)g.Theseatomsoccupyavolumeof(5.65
108)3cm3.
mass
4(69.7 + 74.9) / 6.023 10 23
density =
=
volume
(5.65 10 8 cm)3
Therefore,thetheoreticaldensityofGaAswillbe5.33g/cm3.
71
(c) 2003 Brooks/Cole Publishing / Thomson
Learning
Figure3.34(a)Fluoriteunitcell,(b)planview.
72
(c) 2003 Brooks/Cole Publishing / Thomson
Learning
Figure3.35TheperovskiteunitcellshowingtheAandBsitecationsand
oxygenionsoccupyingthefacecenterpositionsoftheunitcell.Note:
Ionsarenotshowtoscale.
73
Figure3.36CrystalstructureofanewhighTcceramic
superconductorbasedonayttriumbariumcopperoxide.
Thesematerialsareunusualinthattheyareceramics,yetat
lowtemperaturestheirelectricalresistancevanishes.(Source:
ill.fr/dif/3Dcrystals/superconductor.html;M.Hewat1998.)
74
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.37Corundumstructureofalphaalumina(AI203).
75
Section3.8Covalent
Structures
Covalentlybondedmaterialsfrequentlyhavecomplexstructures
inordertosatisfythedirectionalrestraintsimposedbythe
bonding.
Diamondcubic(DC)Aspecialtypeoffacecenteredcubic
crystalstructurefoundincarbon,silicon,andothercovalently
bondedmaterials.
76
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.38(a)Tetrahedronand(b)thediamondcubic(DC)unit
cell.Thisopenstructureisproducedbecauseoftherequirementsof
covalentbonding.
77
Example3.17
DeterminingthePackingFactorforDiamond
CubicSilicon
Determinethepackingfactorfordiamondcubicsilicon.
(c) 2003 Brooks/Cole Publishing /
Thomson Learning
Figure3.39Determining
therelationshipbetween
latticeparameterand
atomicradiusinadiamond
cubiccell(forExample
3.17).
78
Example3.17SOLUTION
Wefindthatatomstouchalongthebodydiagonalofthecell
(Figure3.39).Althoughatomsarenotpresentatalllocations
alongthebodydiagonal,therearevoidsthathavethesame
diameterasatoms.Consequently:
3a 0 = 8r
4
(8 atoms/cell)( r 3 )
3
Packing factor =
3
a0
43
(8)( r )
3
=
(8r / 3 ) 3
= 0.34
Comparedtoclosepackedstructuresthisisarelativelyopenstructure.
79
Example3.18
CalculatingtheRadius,Density,andAtomic
MassofSilicon
ThelatticeconstantofSiis5.43.Whatwillbetheradiusof
asiliconatom?Calculatethetheoreticaldensityofsilicon.
TheatomicmassofSiis28.1gm/mol.
Example3.18SOLUTION
Forthediamondcubicstructure,
3 0 = 8r
Therefore,substitutinga=5.43,the
radiusofsiliconatom=1.176.
a
ThereareeightSiatomsperunitcell.
mass
8(28.1) / 6.023 10 23
density =
=
= 2.33 g / cm3
volume
(5.43 10 8 cm)3
80
(c) 2003 Brooks/Cole Publishing / Thomson
Learning
Figure3.40Thesiliconoxygentetrahedronandtheresultant
cristobaliteformofsilica.
81
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.41Theunitcellofcrystallinepolyethylene.
82
Example3.19
CalculatingtheNumberofCarbonandHydrogen
AtomsinCrystallinePolyethylene
Howmanycarbonandhydrogenatomsareineachunitcellof
crystallinepolyethylene?Therearetwiceasmanyhydrogenatoms
ascarbonatomsinthechain.Thedensityofpolyethyleneisabout
0.9972g/cm3.
Example3.19SOLUTION
Ifweletxbethenumberofcarbonatoms,then2xisthenumberof
hydrogenatoms.FromthelatticeparametersshowninFigure3.41:
( x)(12 g / mol ) + (2 x)(1g / mol )
=
(7.41 10 8 cm)(4.94 10 8 cm)(2.55 10 8 cm)(6.02 10 23 )
x=4carbonatomspercell
2x=8hydrogenatomspercell
83
Section3.9
DiffractionTechniquesforCrystal
StructureAnalysis
DiffractionTheconstructiveinterference,orreinforcement,ofa
beamofxraysorelectronsinteractingwithamaterial.The
diffractedbeamprovidesusefulinformationconcerningthe
structureofthematerial.
BraggslawTherelationshipdescribingtheangleatwhicha
beamofxraysofaparticularwavelengthdiffractsfrom
crystallographicplanesofagiveninterplanarspacing.
Inadiffractometeramovingxraydetectorrecordsthe2yangles
atwhichthebeamisdiffracted,givingacharacteristicdiffraction
pattern
84
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.43(a)
Destructiveand(b)
reinforcinginteractions
betweenxraysandthe
crystallinematerial.
Reinforcementoccursat
anglesthatsatisfy
Braggslaw.
85
Figure3.44PhotographofaXRD
diffractometer.(CourtesyofH&M
AnalyticalServices.)
86
Figure3.45(a)Diagramofa
diffractometer,showingpowder
sample,incidentanddiffracted
beams.(b)Thediffraction
patternobtainedfromasample
ofgoldpowder.
(c) 2003 Brooks/Cole Publishing / Thomson
Learning
87
Example3.20Examining
XrayDiffraction
Theresultsofaxraydiffractionexperimentusingxrays
with=0.7107(aradiationobtainedfrommolybdenum
(Mo)target)showthatdiffractedpeaksoccuratthefollowing
2angles:
Determinethecrystalstructure,theindicesofthe
planeproducingeachpeak,andthelatticeparameterofthe
material.
88
Example3.20SOLUTION
Wecanfirstdeterminethesin2valueforeachpeak,thendivide
throughbythelowestdenominator,0.0308.
89
Example3.20SOLUTION(Continued)
Wecouldthenuse2valuesforanyofthepeakstocalculate
theinterplanarspacingandthusthelatticeparameter.Picking
peak8:
2=59.42or=29.71
0.7107
d 400 =
=
= 0.71699
2 sin
2 sin( 29.71)
a 0 = d 400 h 2 + k 2 + l 2 = (0.71699)(4) = 2.868
Thisisthelatticeparameterforbodycenteredcubiciron.
90
Figure3.46Photographofa
transmissionelectronmicroscope
(TEM)usedforanalysisofthe
microstructureofmaterials.(Courtesy
ofJEOLUSA,Inc.)
91
Figure3.47ATEMmicrographofanaluminumalloy(Al7055)
sample.Thediffractionpatternattherightshowslargebright
spotsthatrepresentdiffractionfromthemainaluminum
matrixgrains.Thesmallerspotsoriginatefromthenano
scalecrystalsofanothercompoundthatispresentinthe
aluminumalloy.(CourtesyofDr.JrgM.K.Wiezorek,
UniversityofPittsburgh.)
92
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.48DirectionsinacubicunitcellforProblem3.51
93
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.49Directions
inacubicunitcellfor
Problem3.52.
94
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.50PlanesinacubicunitcellforProblem3.53.
95
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.51PlanesinacubicunitcellforProblem3.54.
96
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.52Directionsin
ahexagonallatticefor
Problem3.55.
97
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.53Directionsina
hexagonallatticefor
Problem3.56.
98
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.54Planesina
hexagonallatticefor
Problem3.57.
99
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.55Planesina
hexagonallatticefor
Problem3.58.
100
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.56XRDpatternforProblem3.107.
101
(c) 2003 Brooks/Cole Publishing / Thomson Learning
Figure3.57XRDpatternforProblem3.108.
102
Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more.
Course Hero has millions of course specific materials providing students with the best way to expand
their education.
Below is a small sample set of documents:
Below is a small sample set of documents:
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter4ImperfectionsintheAtomicandIonicArrangements11ObjectivesofChapter4Introducethethreebasictypesofimperfections:pointdefects,linedefects(ordislocations),andsurfacedefects
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter5AtomandIonMovementsinMaterials11ObjectivesofChapter5 Examinetheprinciplesandapplicationsofdiffusioninmaterials. Discuss,howdiffusionisusedinthesynthesisandprocessingo
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter6MechanicalPropertiesandBehavior11ObjectivesofChapter6 Introducethebasicconceptsassociatedwithmechanicalpropertiesofmaterials. Evaluatefactorsthataffectthemechanicalpr
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter7StrainHardeningandAnnealing11ObjectivesofChapter7 Tolearnhowthestrengthofmetalsandalloysisinfluencedbymechanicalprocessingandheattreatments. Tolearnhowtoenhancethestr
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter8PrinciplesofSolidification11ObjectivesofChapter8Studytheprinciplesofsolidificationastheyapplytopuremetals.Examinethemechanismsbywhichsolidificationoccurs.Examinehowte
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter9SolidSolutionsandPhaseEquilibrium11ObjectivesofChapter9Thegoalofthischapteristodescribetheunderlyingphysicalconceptsrelatedtothestructureofmatter.Toexaminetherelation
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter10DispersionStrengtheningandEutecticPhaseDiagrams11ObjectivesofChapter10Discussthefundamentalsofdispersionstrengtheningtodeterminethemicrostructure.Examinethetypesofrea
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter11DispersionStrengtheningbyPhaseTransformationsandHeatTreatment11ObjectivesofChapter11Discussdispersionstrengtheningbystudyingavarietyofsolidstatetransformationprocesses
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter12FerrousAlloys1ObjectivesofChapter12Discusshowtousetheeutectoidreactiontocontrolthestructureandpropertiesofsteelsthroughheattreatmentandalloying.Examinetwospecialclasse
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter13NonferrousAlloys11ObjectivesofChapter13ExplorethepropertiesandapplicationsofCu,Al,andTialloysinloadbearingapplications.22ChapterOutline13.1AluminumAlloys13.2Magnes
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter14CeramicMaterials1ObjectivesofChapter14Toexaminethesynthesis,processing,andapplicationsofceramicmaterials.Recapitulateprocessingandapplicationsofinorganicglassesandglas
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter15Polymers11ObjectivesofChapter15DiscusstheclassificationofPolymersLearntwomainwaysofcreatingaPolymerStudytheeffectoftemperatureonThermoplasticsStudymechanicalpropertie
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter16Composites:TeamworkandSynergyinMaterials1ObjectivesofChapter16Studydifferentcategoriesofcomposites:particulate,fiber,andlaminarFocusoncompositesusedinstructuralormech
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter17ConstructionMaterials1ObjectivesofChapter17Topresentasummaryofthematerialpropertiesofwood,concrete,andasphalt.2ChapterOutline17.1TheStructureofWood17.2MoistureConten
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter18ElectronicMaterials1ObjectivesofChapter18 Tostudyelectronicmaterialsinsulators,dielectrics,conductors,semiconductors,andsuperconductors. Tostudyconductivityinelectroni
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter19MagneticMaterials1ObjectivesofChapter19Tostudythefundamentalbasisforresponsesofcertainmaterialstothepresenceofmagneticfields.Toexaminethepropertiesandapplicationsofdi
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter20PhotonicMaterials1ObjectivesofChapter20Topresentasummaryoffundamentalprinciplesthathaveguidedapplicationsofopticalmaterials.Toexploretwoavenuesbywhichwecanusetheoptica
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter21ThermalPropertiesofMaterials1ObjectivesofChapter21Todiscussheatcapacity,thermalexpansionproperties,andthethermalconductivityofmaterials.2ChapterOutline21.1HeatCapac
LSU - ME - 2733
TheScienceandEngineeringofMaterials,4thedDonaldR.AskelandPradeepP.PhulChapter22CorrosionandWear11ObjectivesofChapter22Tointroducetheprinciplesandmechanismsbywhichcorrosionandwearoccurunderdifferentconditions.Thisincludestheaqueouscorrosionofmeta
LSU - EE - 4580
Constant: K 20log[abs(k)] k>0 0 degrees k<0 -180 degrees Pole at Origin (Integrator) -20 dB/decade passing through 0 dB at w=1 -90 degrees Zero at Origin (Differentiator) +20 dB/decade passing through 0 dB at w=1 +90 degrees Real Pole
LSU - EE - 4580
MATLAB Tutorial IRepresentation of transfer function:b0 sm + b1sm,1 + bm:i T s = na0 s + a1 sn,1 + anMATLAB command isnum = b0 b1 bm ;den = a0 a1 an ;s , z1s , z2 s , zmii T s = K:s , p1s , p2 s , pnMATLAB command isK = ;Z = z1 z2 zm ;P =
LSU - EE - 4580
Classical Feedback ControlIn classical control, feedback is the key concept. In fact theconcept of feedback has been applied to many other branches ofscience.Question: Why feedback?Plant uncertainty: parameter variations, nonlinearities etc.Environm
LSU - EE - 4580
Procedure for Skecth of root locus page 260 of text:Example 1: Gs = s + 1=s2= z1 = ,1; p1 = p2 = 0 and n , m = 1.Step 1: Mark the poles and zeros on the s-plane.Step 2: Draw locus on the real axis: left of s = ,1.Step 3: Draw asymptotes: since n , m
LSU - EE - 4580
Zero-degree Loci for Negative KWe considersm + b1sm,1 + + bm :KGs = ,jK j sn + a sn,1 + + a1nBy convention, we have1 + KGs = 0; K = 0 ! ,1:The angle of Gs is 0o + 360o lChange: 180o + 360o is replaced by 0o + 360o l.On real axis: the number of p
LSU - EE - 4580
x6. Frequency-Response Design Methods1. Frequency Response:We recall inverse Laplace transform:23K5K4L,1 6 s , j! + s + j! 7 = 2jK j cos!ot + 6 K :ooConsider sinusoidal input signalut = Au cos!otapplied to a plant Gs. Then the output in s-dom
LSU - EE - 4580
Example Prob. 6.50 on page 456:K:s1 + s=51 + s=20Design a compensator such thatKGs =ess 0:01 for unit ramp input.P.M.= 45o 3o.ess 1=250 for sinusoidal input with !0:2.Noise be attenuated by a factor of 100 for ! 100.Analysis:essto 1=250 for
LSU - EE - 4580
Modeling of Inverted PendulumWe consider line movement for simplication. This example hasapplications to the launch of rocket or missiles. We will demonstrate the use of Lagrange mechanics in modeling.vyvmvzLu(t)My Step 1: Total kinetic energy
LSU - EE - 4580
Iterative Design AlgorithmWe consider an iterative design method: (a) Design zy , z , and Ky > 0, such that the three zeros of(s) = 0 are in the right locations.z5.67zy5.67 (b) Design Kgy and pg such that the roots of (s) = (s + pg )denp(s) + 3.2
LSU - EE - 4580
Course Title:Control System Design.Course Number:EE4580 | Fall 2002.Instructor:Dr. Guoxiang Gu, ECE 329, Tel : 578-5534, Email: ggu@ee.lsu.edu.O ce Hour:7:30 10:00 AM M. W. at ECE 329.Estimated ABET:Engineering Science: 1 credit; Engineering Desi
LSU - EE - 4580
LSU - EE - 4580
LSU - EE - 4580
LSU - EE - 4580
LSU - EE - 4580
Project 2: Digital Control of the Ball-Beam SystemThis project is based on Project 1 using mainly Simulink toolbox to validate yourdesign. The project consists of the following: Program the continuous-time nonlinear ball-beam model using the Simulink t
LSU - EE - 4580
Solution to Midterm Test1. For the lead compensator, we havedB20.2220200degree45.2220200For the lag compensator, we havedB20.2220200220200degree.204512. (a) For the gain plot, we choose L = 0.1, and H = 1000. Hence we have at
LSU - EE - 4580
H0=zpk([],[0,-1,-5,-10],1);H1=zpk([-2],[0,-1,-5,-10],1);H2=zpk([-2,-6],[0,-1,-5,-10],1);H3=zpk([-2,-4],[0,-1,-5,-10],1);H4=tf([1],[1 3 10]);H5=tf([1],[1 3 10 0]);H6=tf([1 2 8],[1 2 10 0]);H7=tf([1 2 12],[1 2 10 0]);H8=tf([1 0 1],[1 0 4 0]);H9=tf(
LSU - EE - 4580
LSU - EE - 4580
LSU - EE - 4580
Matlab Section of Question 3.41For the lower boundary I chose the number zero to indicate that the number was stillnegative, and then I choose a number smaller than zero to prove that the roots arepositive. This helped me show that the poles are in the
LSU - EE - 4580
LSU - EE - 4580
MATLAB REQUIREMENT FOR HW#2PART 1.The code below is what I used to get the transfer function, zero/pole form, and a graph of the step function.> num1=[500 2500];> den1= conv([1 12 100],[1 25]);> gtf1=tf(num1,den1)Transfer function:500 s + 2500-s^3
LSU - EE - 4580
MATLAB COMMANDS TO GET PLOTS ON THE NEXT PAGE FOR 4.4 a,b,c,d> numA=[1];> numB=[1 2];> numC=[1 8 12];> numD=[1 6 8];> denALL=[1 16 65 50 0];> sysA=tf(numA,denALL)Transfer function:1-s^4 + 16 s^3 + 65 s^2 + 50 s> sysB=tf(numB,denALL)Transfer fun
LSU - EE - 4580
HW #3 MATLAB SECTION OF 5.65.6-AMATLAB INSTRUCTIONS> numA=[1];> denA=[1 8 0 0];> sysA=tf(numA,denA)Transfer function:1-s^3 + 8 s^2> rlocus(sysA)RESULTING GRAPHRoot Locus1510Imag Ax is50-5-10-15-20-15-10-5Real Axis055.6-CMATLAB
LSU - EE - 4580
HW #3 MATLAB SECTION OF 5.65.7-BMATLAB INSTRUCTIONS> numB=[1 2];> denB=[1 16 85 250 0 0];> sysB=tf(numB,denB)Transfer function:s+2-s^5 + 16 s^4 + 85 s^3 + 250 s^2> rlocus(sysB)RESULTING GRAPHRoot Locus201510Imag Ax is50-5-10-15-20-20
LSU - EE - 4580
HW #3 MATLAB SECTION OF 5.85.8-BMATLAB INSTRUCTIONS> numB=[1 2];> denB=[1 10 -1 -10];> sysB=tf(numB,denB)Transfer function:s+2-s^3 + 10 s^2 - s - 10> rlocus(sysB)RESULTING GRAPHRoot Locus1510Imag Ax is50-5-10-15-10-8-6-4Real Axis-
LSU - EE - 4580
Correction to 5-11 of HW4 EE4580 Fall041) The closed-loop transfer function is10 K1Y ( s)Gc ( s ) ==(1)R( s ) s ( s + )( s + 10) + s ( s + ) + 2 K1 s + 10 K1You can use direct computation, Masons rule or block diagram simplification methodto get
LSU - EE - 4580
MATLAB PART OF 5.29PART Dsys29d2 = zpk([-1 -0.5],[0 -1 -10 -55.6 -0.04],10)rlocus(sys29d2);grid on;Root Locus1500.620.480.360.240.120.76140120100100800.886050400.97Imaginary Ax is200200.9740-50600.8880-1001001200.760.6
LSU - EE - 4580
6.5a6.5b6.5c6.5d6.5e-GAVE ME A WARNING SIGN6.5f- GAVE ME A WARNING SIGN