Effect of Fiber Geometry and Representative Volume Element on Elastic and Thermal Properties of ...

13 Pages·2015·3.82 MB·English

Hindawi Publishing Corporation
Journal of Composites
Volume 2014, Article ID 629175, 12 pages
http://dx.doi.org/10.1155/2014/629175
Research Article
Effect of Fiber Geometry and Representative
Volume Element on Elastic and Thermal Properties of
Unidirectional Fiber-Reinforced Composites
SivaBhaskaraRaoDevireddyandSandhyaraniBiswas
DepartmentofMechanicalEngineering,NationalInstituteofTechnologyRourkela,Rourkela769008,India
CorrespondenceshouldbeaddressedtoSivaBhaskaraRaoDevireddy;[email protected]
Received30July2014;Revised22September2014;Accepted27October2014;Published18November2014
AcademicEditor:YanqingYang
Copyright©2014S.B.R.DevireddyandS.Biswas. This is an open access article distributed under the Creative Commons
AttributionLicense,whichpermitsunrestricteduse,distribution,andreproductioninanymedium,providedtheoriginalworkis
properlycited.
Theaimofpresentworkisfocusedontheevaluationofelasticandthermalpropertiesofunidirectionalfiber-reinforcedpolymer
compositeswithdifferentvolumefractionsoffiberupto0.7usingmicromechanicalapproach.Twowaysforcalculatingthematerial
properties,thatis,analyticalandnumericalapproaches,werepresented.Innumericalapproach,finiteelementanalysiswasused
toevaluatetheelasticmodulusandthermalconductivityofcompositefromtheconstituentmaterialproperties.Thefiniteelement
modelbasedonthree-dimensionalmicromechanicalrepresentativevolumeelement(RVE)withasquareandhexagonalpacking
geometrywasimplementedbyusingfiniteelementcodeANSYS.Circularcrosssectionoffiberandsquarecrosssectionoffiberwere
consideredtodevelopRVE.TheperiodicboundaryconditionsareappliedtotheRVEtocalculateelasticmodulusofcomposite.
Thesteadystateheattransfersimulationswereperformedinthermalanalysistocalculatethermalconductivityofcomposite.In
analyticalapproach,theelasticmodulusiscalculatedbyruleofmixture,Halpin-Tsaimodel,andperiodicmicrostructure.Thermal
conductivityiscalculatedanalyticallybyusingruleofmixture,theChawlamodel,andtheHashinmodel.Thematerialproperties
obtainedusingfiniteelementtechniqueswerecomparedwithdifferentanalyticalmethodsandgoodagreementwasachieved.The
resultsareaffectedbyanumberofparameterssuchasvolumefractionofthefibers,geometryoffiber,andRVE.
1.Introduction consideredinobtainingtheglobalpropertiesofcomposites:
(a) macromechanical analysis and (b) micromechanical
Therehasbeenaconsiderableincreaseintheuseoffibercom- analysis.Inmacromechanicalanalysisthecompositematerial
posite materials in various industries like aerospace, auto- is considered as a homogeneous orthotropic continuum. In
motive,infrastructures,andsportinggoodsduetotheirspe- micromechanicalanalysisthestudyofcompositematerialis
cific properties like strength, stiffness, toughness, high cor- atthefiberandmatrixlevel.Typicallytheunitcelltechnique
rosionresistance,highwearresistance,highchemicalresis- combined with the known material properties of fiber and
tance,andreducedcost.Thesematerialscantakeadvantage matrix is used to determine the overall behavior of the
of different properties of their constituents, microstructure, composite[1].Anumberofmethodshavebeendevelopedto
and interaction between constituents in order to improve predictandtosimulatethemechanicalandthermalbehav-
the mechanical behavior of parts made from them. The ior of composites. Basic analytical approaches have been
mechanics of fiber-reinforced composites are complex due reported[2–4]topredictthecompositematerialsproperties,
to their anisotropic and heterogeneous characteristics. The for example, strength, stiffness, and thermal conductivity.
evaluationofeffectivemechanicalandthermalpropertiesof Prediction of boron and aluminium composite properties
compositematerialsisofparamountimportanceinengineer- fromarepresentativevolumeelement(RVE)withsquareand
ing design and application. Generally, two approaches are hexagonalgeometryhasbeenreported[5].Micromechanical
2 JournalofComposites
analysis of unidirectional fiber-reinforced composites with Table1:Mechanicalandthermalpropertiesoftheconstituentmate-
square and hexagonal unit cells has been reported [6, 7] to rials[8,9].
evaluatetheeffectivematerialproperties.
Properties Glassfiber Epoxymatrix
Patnaiketal.[8]studiedthemicromechanicalandther-
mal characteristics of glass-fiber-reinforced polymer com- Density(g/cm3) 2.5 1.15
posites. The experimental results were in good agreement Young’smodulus(GPa) 73 3.76
with finite element model based on representative area ele- Poisson’sratio 0.2 0.39
ment approach. Melro et al. [9] predict the inelastic defor-
Shearmodulus(GPa) 30.42 1.28
mation and fracture of randomly distributed unidirectional
Thermalconductivity(W/mk) 1.3 0.363
fiber-reinforcedpolymercompositematerialswithdifferent
RVEs. Several researchers made their significant contribu-
tioninstudyingthermalcharacteristicsoffibercomposites.
Springer and Tsai [10] studied the composite thermal con-
ductivitiesofunidirectionalcompositesandexpressionsare the unidirectional fiber composite where the fibers are
obtainedforpredictingtheseconductivitiesinthedirections arranged in the square and hexagonal array is shown in
alongthefilamentsandnormaltothem.IslamandPramila Figure1.Byvaryingthevolumefractionoffiberfrom0.1to
[11] predict the effective transverse thermal conductivity of 0.7theelasticandthermalpropertiesofcompositematerial
fiber-reinforcedcompositesbyusingfiniteelementmethod. aredetermined.
Squareandcircularcrosssectionfiberswereusedforperfect
bondingatfiber-matrixinterfaceandwithinterfacialbarrier
2.1.ConstitutiveEquationsforFiberCompositeMaterial. The
byusingfourdifferentsetsofthermalboundaryconditions.
most general form of the anisotropic constitutive equations
Al-Sulaimanetal.[12]predictthethermalconductivityofthe
forhomogeneousandelasticcompositematerialsisgivenby
constituents of fiber-reinforced composite laminates using
Hook’slawasshownin(1)[15].Consider
three empirical formulas. Grove [13] computed transverse
thermalconductivityincontinuousunidirectionalfibercom-
𝜎 𝜎
ptLeuocsh[i1tne4i]qmuuesasetedfroibarolsaunurdasinanrggyecfioonflilfiotebcaeetrlieovmnoelsuncmtheeamnfdreafcsoptriaoctniaasllcuusptlaatttiioostn0ic.oa5fl. {{{{{{{{𝜎𝜎12}}}}}}}} {{{{{{{{𝜎𝜎𝑥𝑦𝑦𝑥}}}}}}}}
3 ≡ 𝑧𝑧
tpsrqeaurniaosrvdeeiarcrsreaaryerffaayenscdtoilvofenlogthncegirrcmcuiralcalurclcoayrnlidanundcdetrivssqiwtuyiathroefhce2yx-dlaiginmodneearnslsaiworrniatahyl {{{{{{{{𝜎𝜎45}}}}}}}} {{{{{{{{𝜏𝜏𝑦𝑧𝑥𝑧}}}}}}}}
{𝜎 } {𝜏 }
foracompleterangeoffibervolumefractions. 6 𝑥𝑦
Althoughagreatdealofworkhasalreadybeendoneon
𝐶 𝐶 𝐶 𝐶 𝐶 𝐶 𝜀
fitmTtihoboorenetdrhe-eo-irlsfdesfiiieumnnbsfdeeionrn,r,gctssheifiqoedunnopaiatbroleejlmeyecclmiretcoimevrsoresemcnosoetfemccatthhnpieoaaonnlpsyiirsocteieafssslefiiwRnsbtVhietwrahErwodcwriilktryihtcihbusdeladaieffrensevqcrrerueeloaonpsrptosei-rRnsatVeegncdEda-. =[[[[[[[[𝐶𝐶𝐶𝐶1111112345 𝐶𝐶𝐶𝐶1222222345 𝐶𝐶𝐶𝐶1233333345 𝐶𝐶𝐶𝐶1234444445 𝐶𝐶𝐶𝐶1234555555 𝐶𝐶𝐶𝐶1234566666]]]]]]]]{{{{{{{{{{{{{{{{𝛾𝛾𝜀𝜀𝑥𝑦𝑦𝑧𝑧𝑦𝑥𝑧𝑥𝑧}}}}}}}}}}}}}}}}
hexagonal packing geometry with circular and square fiber [𝐶 𝐶 𝐶 𝐶 𝐶 𝐶 ]{𝛾 }
16 26 36 46 56 66 𝑥𝑦
crosssections.Anumericalhomogenizationtechniquebased
onthefiniteelementanalysiswasusedtoevaluatetheelastic 𝜀
meleomduenlutsraensudlttshaerremcaolmcopnadreudctiwviitthy othfecoamnaploytsiictea.lThmeethfionditse. {{{{{{{{𝜀𝜀12}}}}}}}}
The aim is to demonstrate applicability of homogenization ≡ 3 ,
tcehcahrnaciqteureisbtiycussininagdfivnaintecee.lementmethodtopredictmaterial {{{{{{{{𝜀𝜀45}}}}}}}}
{𝜀 }
6
2.MaterialsandMethods
(1)
Inthispresentinvestigation,unidirectionalglassfiberasrein-
forcementphaseandepoxyasmatrixphaseforthecomposite where𝜎𝑖𝑗and𝜏𝑖𝑗arenormalandshearcomponentsofstress,
materialwereconsidered.Thefiberandmatrixmaterialsare respectively,𝜀𝑖𝑗and𝛾𝑖𝑗arethenormalandshearcomponents
considered as isotropic and homogeneous. The properties of strain, respectively, and 𝐶𝑖𝑗 is the symmetric stiffness
of the constituent materials are as shown in Table1. In a matrixwith 21 independent, elastic constants. According to
realunidirectionalfiber-reinforcedcomposite,thefibersare their behaviour, composites may be characterized as gen-
arrangedrandomlyanditisdifficulttomodelrandomfiber erallyanisotropic,monoclinic,orthotropic,andtransversely
arrangement. isotropic.Inpresentwork,transverselyisotropiccharacteris-
For this analysis, circular and square cross section fiber ticshavebeenconsideredforthefiber-reinforcedcomposite.
compositematerialisconsidered.Theschematicdiagramof A transversely isotropic material is to be a material whose
JournalofComposites 3
Glass fiber
Glass fiber Matrix
Matrix
Hexagonal unit cell
Square unit cell
(a) (b)
Glass fiber Glass fiber
Matrix
Square unit cell Hexagonal unit cell
Matrix
(c) (d)
Figure1:Arrangementoffibersin(a)squarearraywithcircularfibers,(b)hexagonalarraywithcircularfibers,(c)squarearraywithsquare
fibers,and(d)hexagonalarraywithsquarefibers.
effectivepropertiesareisotropicinoneofitsplanesandthe where𝐸1,𝐸2,]12,and𝐺23 arelongitudinalmodulus,trans-
stiffnesstensorisrepresentedin verse modulus in plane Poisson’s ratio, and in plane shear
modulus,respectively.
𝜎
{{{{{{{𝜎𝜎12}}}}}}} 2m.2e.chGaenniecraaltimonodoeflRsVasEs.umFoerasipmerpiloicdiitcyarreraasnognesm,menotstofmfiibcreors-
3
{{{{{{{𝜎𝜎45}}}}}}} ftohrewsahmicehealaRstVicEcoonrsutannittscealnldcafinbebreviosolulamteed.frThacteioRnVaEshthaes
{𝜎6} composite. The periodic fiber sequences commonly used
are the square array and the hexagonal array. For a square
[[[[𝐶𝐶𝐶111122 𝐶𝐶𝐶122223 𝐶𝐶𝐶122232 000 000 000 ]]]]{{{{{{{𝜀𝜀𝜀12}}}}}}} ptshqaeucokarrienetgRicRVaVlElyEtahaceshsdiheiaovmwabnelteienrfiFboifegrfiubvroeerlsu2ims(ace)aflacrnaudclat2ito(ecnd)bitshye78m.5a4x%im.Fuomr
=[[[[[ 000 000 000 12(𝐶2200−𝐶23) 𝐶0066 𝐶00 ]]]]]{{{{{{{{𝜀𝜀𝜀3456}}}}}}}}. 𝑉𝑓 = 𝑎1𝑎(𝜋1𝑎/24𝑎)3𝑑𝑓2, (4)
66
(2) where 𝑉𝑓 is volume fraction of fiber; 𝑎1, 𝑎2, and 𝑎3 are the
lengthofsquareRVE;and𝑑𝑓isthediameteroffiber.Forthe
hexagonal packing RVE as shown in Figures 2(b) and 2(d)
Once the components of the transversely isotropic stiffness
themaximumtheoreticallyachievablefibervolumefraction
tensor 𝐶 are known, the elastic properties of homogenized
is90.69%.Obviously,withahexagonalpackinggeometrya
materialcanbecomputedby(3)[16].Consider
composite can be made more compact than with a square
packinggeometry.ForhexagonalRVEthediameteroffiber
2𝐶2
𝐸 =𝐶 − 12 , iscalculatedby
1 11 (𝐶 +𝐶 )
22 23 2𝑎 (𝜋/4)𝑑2
1 𝑓
𝑉 = , (5)
[𝐶 (𝐶 +𝐶 )−2𝐶2 ](𝐶 −𝐶 ) 𝑓 𝑎 𝑎 𝑎
𝐸 = 11 22 23 12 22 23 , 1 2 3
2 (𝐶11𝐶22−𝐶122) (3) where𝑎3 =𝑎2tan(60∘)and𝑎2 =4𝑎1.
𝐶
] = 12 ,
12 (𝐶 +𝐶 ) 3.FiniteElementModeling
22 23
1 Inordertoevaluatetheeffectivepropertiesofcomposite,the
𝐺 = (𝐶 −𝐶 ),
23 2 22 23 finiteelementsoftwarepackageANSYSisused.Theprogram
4 JournalofComposites
B
B
A
F
A F
E
a3 a3
E
G
D D
3 G
2 1 a2 a1 a2 a
1
H H
(a) (b)
B
B
A
F
A F
E
a3 a3
E
G
D D
G
a2 a1 a2 a
1
H H
(c) (d)
Figure2:RVE(a)squareRVEwithcircularfiber,(b)hexagonalRVEwithcircularfiber,(c)squareRVEwithsquarefiber,and(d)hexagonal
RVEwithsquarefiber.
iswritteninAPDL(ANSYSProgrammingDesignLanguage), elastic properties and is defined by 20 nodes having three
whichisdeliveredbythesoftwareanditmakesthehandling degreesoffreedomateachnode.Theyaretranslationsinthe
muchmorecomfortable.Forsimplification,therearemany nodal1,2,and3directions.Forthermalconductivityathree-
assumptions considered for the present analysis such as dimensional quadratic brick element SOLID90 is used for
fiberswhicharearrangedinaparticularpattern(squareand discretizationoftheconstituentsandisdefinedby20nodes
hexagonal) in a matrix. The composite is free of voids and withasingledegreeoffreedom(temperature)ateachnode.
otherirregularities,allfibersareuniformlydistributedinthe The meshed model of square and hexagonal RVE at 0.4 of
matrix and perfectly aligned, and the interface between the fibervolumefractionisshowninFigure3.
fiber and matrix is perfectly debonded. In the study of the
micromechanicsoffiber-reinforcedmaterials,itisconvenient 3.1.BoundaryConditionsforEvaluationofElasticProperties.
to use an orthogonal coordinate system that has one axis Compositematerialscanberepresentedasaperiodicarrayof
alignedwiththefiberdirection.Theaxis1isalignedwiththe theRVEs.Therefore,theperiodicboundaryconditionsmust
fiber direction, the axis 2 is in the plane of the RVE and is be applied to the RVE models. This implies that each RVE
perpendiculartothefibers,andtheaxis3isperpendicularto inthecompositehasthesamedeformationmodeandthere
theplaneoftheRVEandisalsoperpendiculartothefibersas is no separation or overlap between the neighboring RVEs
showninFigure2.Dimensionsconsideredfortheanalysisare afterdeformation[17,18].Theresumedboundaryconditions
𝑎1 =1.0×10−5m,𝑎2 =1.0×10−5m,and𝑎3 =1.0×10−5mfor applied are given in Table2. Note that 𝑈, 𝑉, and 𝑊 are
squareRVE.ForhexagonalRVE𝑎1 =1.0×10−5m,𝑎2and𝑎3 the displacements along 1, 2, and 3 directions, respectively,
arecalculatedbyusing(5).Theradiusoffiberscorrespondsto applied on the AEDH, BFCG, ABCD, EFGH, DHGC, and
volumefractionsrangingfrom0.1to0.7.Three-dimensional AEFB faces as shown in Figure2. After applying boundary
structural solid element SOLID186 is used to determine conditionsandthedisplacementconstant,thecorresponding
JournalofComposites 5
Table2:Boundaryconditionsalongthe1,2,and3directionsoftheRVE.
1-directionfaces 2-directionfaces 3-directionfaces
Load Constraint
AEDH BFCG ABCD EFGH DHGC AEFB
U 0 𝑎1
Load11 V 0 0
W 0 0
U 0 0
Load22 V 0 𝑎2
W 0 0
U 0 0
Load33 V 0 0
W 0 𝑎3
(a) (b)
(c) (d)
Figure3:Meshedmodelof(a)squareRVEwithcircularfiber,(b)hexagonalRVEwithcircularfiber,(c)squareRVEwithsquarefiber,and
(d)hexagonalRVEwithsquarefiber.
6 JournalofComposites
Element solution EElleemmeenntt ssoolluuttiioonn
Step=3 SStteepp==22
Sub=1 SSuubb==11
Time=3 TTiimmee==22
SZ (NOAVG) SSZZ ((NNOOAAVVGG))
RSYS=0 RRSSYYSS==00
DMX=10.0004 DDMMXX==4400..33331133
SMN=0.002288 SSMMNN==00..000044335544
SMX=0.017111 SSMMXX==00..001111336622
8 5 2 9 6 3 7 7 4 1 4 3 1 9 9 7 6 5 4 2
8 3 8 2 7 2 1 1 6 1 5 3 1 6 6 4 2 0 8 6
2 9 5 2 8 5 2 8 4 1 3 1 9 6 4 2 0 8 5 3
2 3 5 7 8 0 1 3 5 7 4 5 5 0 7 8 9 9 0 1
0 0 0 0 0 1 0 1 1 1 0 0 0 0 0 0 0 0 1 1
0.0 0.0 0.0 0.0 0.0 0.0 0. 0.0 0.0 0.0 0.0 0.0 0.0 0. 0.0 0.0 0.0 0.0 0.0 0.0
(a) (b)
Element solution Element solution
Step=3 Step=3
Sub=1 Sub=1
Time=3 Time=3
SZ (NOAVG) SZ (NOAVG)
RSYS=0 RSYS=0
DMX=10.0005 DMX=69.2944
SMN=0.818E−03 SMN=0.477E−03
SMX=0.015524 SMX=0.014885
18E−03 0.002452 0.004086 0.00572 0.007354 0.008988 0.010622 0.012256 0.01389 0.015524 77E−03 0.002078 0.003679 0.00528 0.006881 0.008482 0.010082 0.011683 0.013284 0.014885
8 4
0. 0.
(c) (d)
Figure4:Counterofstressin(a)squareRVEwithcircularfiber,(b)hexagonalRVEwithcircularfiber,(c)squareRVEwithsquarefiber,and
(d)hexagonalRVEwithsquarefiber.
engineering constants are calculated as follows in terms of 3.2. Boundary Conditions for Evaluation of Thermal Con-
correspondingstressesandstrainsshownin(6): ductivity. Thesteadystateheattransfersimulationsareper-
formed by using finite element analysis to predict thermal
1
𝜎 = ∫ 𝜎 𝑑𝑉, conductivityofcompositealongthelongitudinalandtrans-
𝑖𝑗 𝑉 𝑖𝑗
𝑉 versedirection.Thethermalboundaryconditionsconsidered
(6)
1 in the present analysis are shown in Figure2. One wall
𝜀 = ∫ 𝜀 𝑑𝑉,
𝑖𝑗 𝑉 𝑖𝑗 (BFGC) is kept isothermal at elevated temperature, while
𝑉
the corresponding wall (AEHD) is subjected 100 Kelvin to
where 𝜎 and 𝜀 are the average stresses and average strains maintaintemperaturedifferenceforthecalculationoflongi-
and𝑉isthevolumeoftheRVE.Theelasticpropertiescanbe tudinalthermalconductivity.Allothersurfacesaresubjected
calculatedbyusingtheconstitutiveequationsofthematerial toinsulationboundaryconditions.Onewall(ABCD)iskept
propertiesastheratioofcorrespondingaveragestressesand isothermalatelevatedtemperature,whilethecorresponding
average strains as shown in (3). Figures 4 and 5 show the wall(EFGH)issubjected100Kelvintomaintaintemperature
counterofstressandstraininsquareandhexagonalRVEat differenceforthecalculationoftransversethermalconduc-
0.4ofvolumefraction. tivity.Allothersurfacesaresubjectedtoinsulationboundary
JournalofComposites 7
Element solution Element solution
Step=3 Step=2
Sub=1 Sub=1
Time=3 Time=2
EPELX (NOAVG) EPELX (NOAVG)
RSYS=0 RSYS=0
DMX=10.0004 DMX=40.3313
SMN=−1.82159 SMN=−0.990065
SMX=1.06714 SMX=1.05493
159 062 965 681 712 742 227 197 166 714 065 844 623 401 118 041 262 483 704 934
−1.82 −1.50 −1.17 0.858 0.537 0.216 0.104 0.425 0.746 1.06 0.990 0.762 0.535 0.308 −0.08 0.146 0.373 0.600 0.827 1.05
− − − − − − −
(a) (b)
Element solution Element solution
Step=3 Step=2
Sub=1 Sub=1
Time=3 Time=2
EPELX (NOAVG) EPELX (NOAVG)
RSYS=0 RSYS=0
DMX=10.0005 DMX=40.4003
SMN=−0.916974 SMN=−1.08927
SMX=1.56775 SMX=0.933549
4 4 4 4 7 7 7 9 7 5 7 4 6 8 1 7 5 3 1 9
7 9 1 3 4 2 0 5 6 7 2 1 5 9 4 1 7 3 9 4
9 8 8 7 3 4 5 5 1 7 9 5 7 9 2 5 2 0 7 5
6 0 4 8 7 3 9 1 9 6 8 4 9 4 0 4 9 4 8 3
1 4 6 8 8 6 3 0 2 5 0 6 3 1 9 3 5 8 0 3
0.9 0.6 0.3 0.0 0.1 0.4 0.7 1. 1. 1. −1. 0.8 0.6 0.4 0.1 0.0 0.2 0.4 0.7 0.9
− − − − − − − −
(c) (d)
Figure5:Counterofstrainin(a)squareRVEwithcircularfiber,(b)hexagonalRVEwithcircularfiber,(c)squareRVEwithsquarefiber,and
(d)hexagonalRVEwithsquarefiber.
conditions. Using these temperature gradients the heat flux 4.AnalyticalMethods
wasobtainedfromANSYSsoftware.
Theeffectivethermalconductivityisestablishedfromthe In order to validate the finite element results, three well-
fundamentalheatconductionlaw,foundbyFourier’s,which existinganalyticalmethodssuchasruleofmixture,semiem-
states that the heat flux is proportional to the temperature piricalmodel,andperiodicmicrostructuremodelshavebeen
gradient[19]: usedforthecurrentstudy.
𝑑𝑇
𝑞=−𝐾 , (7) 4.1. Rule of Mixture Model. The well-known models that
𝑑𝑥
have been proposed and used to evaluate the properties of
unidirectional composites are the Voigt and Reuss models
where 𝑑𝑇/𝑑𝑥 is temperature gradient between two isother- which are also known as the rule of mixture model and
mal surfaces, 𝑞 is the heat flux W/m2, and 𝑘 is the thermal the inverse rule of mixture model. The rule of mixture
conductivity. The temperature distribution in square and is the mathematical expressions which give the property
hexagonal RVE along longitudinal and transverse direction of the composite in terms of the properties, quantity, and
at0.4ofvolumefractionoffiberisshowninFigures6and7. arrangementofitsconstituents.Thelongitudinalpropertyof
8 JournalofComposites
Nodal solution Nodal solution
Step=1 Step=1
Sub=1 Sub=1
Time=1 Time=1
Temp (AVG) Temp (AVG)
RSYS=0 RSYS=0
SMN=300 SMN=300
SMX=373 SMX=373
0 1 2 3 4 6 7 8 9 3 0 1 2 3 4 6 7 8 9 3
0 1 2 3 4 5 6 7 8 7 0 1 2 3 4 5 6 7 8 7
3 1 2 3 4 5 6 7 8 3 3 1 2 3 4 5 6 7 8 3
8. 6. 4. 2. 0. 8. 6. 4. 8. 6. 4. 2. 0. 8. 6. 4.
0 1 2 3 4 4 5 6 0 1 2 3 4 4 5 6
3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
(a) (b)
Nodal solution Nodal solution
Step=1 Step=1
Sub=1 Sub=1
Time=1 Time=1
Temp (AVG) Temp (AVG)
RSYS=0 RSYS=0
SMN=300 SMN=300
SMX=373 SMX=373
0 1 2 3 4 6 7 8 9 3 0 1 2 3 4 6 7 8 9 3
0 1 2 3 4 5 6 7 8 7 0 1 2 3 4 5 6 7 8 7
3 1 2 3 4 5 6 7 8 3 3 1 2 3 4 5 6 7 8 3
8. 6. 4. 2. 0. 8. 6. 4. 8. 6. 4. 2. 0. 8. 6. 4.
0 1 2 3 4 4 5 6 0 1 2 3 4 4 5 6
3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
(c) (d)
Figure6:Longitudinaltemperaturedistributionin(a)squareRVEwithcircularfiber,(b)hexagonalRVEwithcircularfiber,(c)squareRVE
withsquarefiber,and(d)hexagonalRVEwithsquarefiber.
thecompositeiscalculatedbyusingruleofmixtureasshown their models as simple equations by curve fitting to results
in thatarebasedonelasticity.Thematerialpropertiescalculated
inlongitudinaldirectionbyusingHalpin-Tsaimodelarethe
𝜙 =𝜙 𝑉 +𝜙 𝑉 ,
1 𝑓 𝑓 𝑚 𝑚 (8) sameasthoseinruleofmixtures.Forthetransversedirection,
where𝜙1 isthepropertyofcompositeinlongitudinaldirec- thematerialpropertyofcompositeiscalculatedby
tion.𝜙𝑓and𝜙𝑚arethematerialpropertiesoffiberandepoxy 𝜙 1+𝜁𝜂𝑉
matrix.𝑉𝑓and𝑉𝑚arethevolumefractionsoffiberandepoxy 𝜙2 = 1−𝜂𝑉𝑓, (10)
matrix.Thetransversemodulusofcompositeiscalculatedby 𝑚 𝑓
usingruleofmixtureasshownin where𝜂=((𝜙𝑓/𝜙𝑚)−1)/((𝜙𝑓/𝜙𝑚)+𝜁).
1 = 𝑉𝑓 + 𝑉𝑚, Halpin and Tsai found that the value 𝜉 = 2 gave
(9)
𝜙 𝜙 𝜙 an excellent fit to the finite difference elasticity solution of
2 𝑓 𝑚
AdamsandDoner[20]forthetransversemodulusofasquare
where𝜙2isthepropertyofcompositeintransversedirection. arrayofcircularfibers.
4.2.Halpin-TsaiModel. Themostusefulsemiempiricalmod- 4.3. Periodic Microstructure. If the composite has periodic
elsarethosebyHalpinandTsai.HalpinandTsaideveloped microstructure, then Fourier series can be used to estimate
JournalofComposites 9
Nodal solution Nodal solution
Step=1 Step=1
Sub=1 Sub=1
Time=1 Time=1
Temp (AVG) Temp (AVG)
RSYS=0 RSYS=0
SMN=300 SMN=300
SMX=373 SMX=373
0 1 2 3 4 6 7 8 9 3 0 1 2 3 4 6 7 8 9 3
0 1 2 3 4 5 6 7 8 7 0 1 2 3 4 5 6 7 8 7
3 1 2 3 4 5 6 7 8 3 3 1 2 3 4 5 6 7 8 3
8. 6. 4. 2. 0. 8. 6. 4. 8. 6. 4. 2. 0. 8. 6. 4.
0 1 2 3 4 4 5 6 0 1 2 3 4 4 5 6
3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
(a) (b)
Nodal solution Nodal solution
Step=1 Step=1
Sub=1 Sub=1
Time=1 Time=1
Temp (AVG) Temp (AVG)
RSYS=0 RSYS=0
SMN=300 SMN=300
SMX=373 SMX=373
0 1 2 3 4 6 7 8 9 3 0 1 2 3 4 6 7 8 9 3
0 1 2 3 4 5 6 7 8 7 0 1 2 3 4 5 6 7 8 7
3 1 2 3 4 5 6 7 8 3 3 1 2 3 4 5 6 7 8 3
8. 6. 4. 2. 0. 8. 6. 4. 8. 6. 4. 2. 0. 8. 6. 4.
0 1 2 3 4 4 5 6 0 1 2 3 4 4 5 6
3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
(c) (d)
Figure7:Transversetemperaturedistributionin(a)squareRVEwithcircularfiber,(b)hexagonalRVEwithcircularfiber,(c)squareRVE
withsquarefiber,and(d)hexagonalRVEwithsquarefiber.
all the components of the stiffness tensor of the composite. strain. Figure8 shows the effect of fiber content on the
Explicitformulasforacompositereinforcedbylongcircular longitudinalmodulusofcompositesusingruleofmixtures,
cylindricalfiberswerediscussedin[21]. Halpin-Tsai model, periodic microstructure, and finite ele-
mentanalysiswithsquareandhexagonalRVEwithcircular
andsquarefibergeometry.Itcanbeobservedfromthefigure
4.4. Thermal Conductivity Models. Thermal conductivity
that the longitudinal modulus increases with the increase
obtained by finite element analysis, comparing with three
in volume fraction of fiber and there is a good agreement
well-existinganalyticalmethodssuchasruleofmixture,the
betweenfiniteelementresultsandanalyticalmethods.Thisis
Chawlamodel,andtheHashinmodel[22],hasbeenusedfor
becausethestiffnessofthecompositeincreaseswithincrease
thecurrentstudy.
involumefractionoffiber.
The transverse modulus is the response of composite
5.ResultsandDiscussion during the application of load perpendicular to the fiber
direction. It can be defined as the ratio of transverse stress
5.1.EffectofVolumeFractiononElasticProperties. Thelon- to the transverse strain. The effect of fiber volume fraction
gitudinal modulus is the response of composite during the on transverse modulus of composites using finite element
application of load parallel to the fiber direction. It can be analysis and three analytical methods is graphically shown
definedastheratiooflongitudinalstresstothelongitudinal in Figure9. As expected, it is clear from the figure that
10 JournalofComposites
60 0.38
0.36
a) 50
P
ulus (G 40 n's ratio 00..3324
d o
gitudinal mo 2300 n plane Poiss 000...223680
n I
o 10
L 0.24
0 0.22
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
Volume fraction of fiber (Vf) Volume fraction of fiber (Vf)
Rule of mixture FEA circular-hexagonal RVE Rule of mixture FEA circular-hexagonal RVE
Halpin-Tsai FEA square-square RVE Halpin-Tsai FEA square-square RVE
Periodic microstructure FEA square-hexagonal RVE Periodic microstructure FEA square-hexagonal RVE
FEA circular-square RVE FEA circular-square RVE
Figure8:Longitudinalmodulusvalidationwithdifferentvolume Figure10:In-planePoisson’sratiovalidationwithdifferentvolume
fractionoffiber. fractionoffiber.
12
28
24 a) 10
a) GP
ulus (GP 1260 odulus ( 8
Transverse mod 1482 In plane shear m 246
0 0
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
Volume fraction of fiber (Vf) Volume fraction of fiber (Vf)
Rule of mixture FEA circular-hexagonal RVE Rule of mixture FEA circular-hexagonal RVE
Halpin-Tsai FEA square-square RVE Halpin-Tsai FEA square-square RVE
Periodic microstructure FEA square-hexagonal RVE Periodic microstructure FEA square-hexagonal RVE
FEA circular-square RVE FEA circular-square RVE
Figure9:Transversemodulusvalidationwithdifferentvolumefrac- Figure11:In-planeshearmodulusvalidationwithdifferentvolume
tionoffiber. fractionoffiber.
thetransversemodulusincreaseswithincreaseinfibervol- shows the effect of fiber volume fraction on the in-plane
ume fraction. The transverse modulus evaluated by finite shear modulus of composite. It is clear from the figure that
element analysis with hexagonal RVE is more close to the the shear modulus increases with increases in fiber volume
Halpin-Tsaimodelandperiodicmicrostructureascompared fraction.Also,itcanbeobservedthatthereisagoodagree-
totheresultsobtainedfromruleofmixtureandfiniteelement ment between results obtained from finite element analysis
analysiswithsquareRVE. and hexagonal RVE with Halpin-Tsai model and periodic
Figure10showstheeffectoffibervolumefractiononthe microstructure as compared to rule of mixture and finite
in-plane Poisson’s ratio of composite. It is evident from the elementanalysiswithsquareRVE.
figurethatthemajorPoisson’sratiodecreaseswithincrease
in the volume fraction of fiber due to increase in material 5.2.EffectofVolumeFractiononThermalConductivity. Lon-
resistance.Thefiniteelementresultsareingoodagreement gitudinalthermalconductivityofcompositeistheproperty
withanalyticalmethods. of a material to conduct heat in parallel to the direction
In-planeshearmodulusofcompositeistheratioofshear of the fibers. Figure12 shows the effect of fiber content on
stresstotheshearstraininlongitudinaldirection.Figure11 thelongitudinalthermalconductivityusingruleofmixture,