Module 3 Constitutive Equations

1 Module 3 Constitutive EquationsLearning Objectives Understand basic stress-strain response of engineering materials. Quantify the linear elastic stress-strain response in terms of tensorial quantities and inparticular the fourth-order elasticity or stiffness tensor describing Hooke s Law. Understand the relation between internal material symmetries and macroscopic anisotropy,as well as the implications on the structure of the stiffness tensor. Quantify the response of anisotropic materials to loadings aligned as well as rotatedwith respect to the material principal axes with emphasis on orthotropic and transversely-isotropic materials. Understand the nature of temperature effects as a source of thermal expansion strains. Quantify the linear elastic stress and strain tensors from experimental strain-gaugemeasurements. Quantify the linear elastic stress and strain tensors resulting from special materialloading Linear elasticity and Hooke s LawReadings: Reddy the stress strain curve =f( ) of a linear elastic material subjected to uni-axialstress loading conditions (Figure ).

2 4546 Module 3. Constitutive Equations 1E =12E 2 Figure : Stress-strain curve for a linear elastic material subject to uni-axial stress (Notethat this is not uni-axial strain due to Poisson effect)In this expression,Eis Young s Energy DensityFor a given value of the strain , thestrain energy density (per unit volume) = ( ), isdefined as the area under the curve. In this case, ( ) =12E 2We note, that according to this definition, = =E In general, for (possibly non-linear) elastic materials: ij= ij( ) = ij( )Generalized Hooke s LawDefines the most general linear relation among all the components of the stress and straintensor ij=Cijkl kl( )In this expression:Cijklare the components of the fourth-orderstiffnesstensor of materialproperties orElastic moduli. The fourth-order stiffness tensor has 81 and 16 components forthree-dimensional and two-dimensional problems, respectively. The strain energy density TRANSFORMATION OF BASIS FOR THE ELASTICITY TENSOR COMPONENTS47this case is a quadratic function of the strain: ( ) =12 Cijkl ij kl( )Concept Question of Hooke s the Hooke s law from quadratic strain energy function Starting from the quadraticstrain energy function and the definition for the stress components given in the notes,1.

3 Derive the Generalized Hooke s law ij=Cijkl :We start bycomputing: ij kl= ik jlHowever, we have lost the symmetry, the lhs is symmetric with respect toijandwith respect tokl, but the rhs is only symmetric with respect toikandjl. But wecan easily recover the same symmetries as follows: ij kl=12( ik jl+ il jk)Now we use this to compute ijas follows: ij( ) = ij= ij(12 Cklmn kl mn)=12 Cklmn( kl ij mn+ kl mn ij)using the first equation:=12 Cklmn( ki lj kl+ kl mi nj)=12(Cijmn mn+Cklij kl)ifCklij=Cijkl, we obtain: ij=Cijkl Transformation of basis for the elasticity tensorcomponentsReadings: BC , Reddy 3. Constitutive EQUATIONSThe stiffness tensor can be written in two different orthonormal basis as:C=Cijklei ej ek el= Cpqrs ep eq er es( )As we ve done for first and second order tensors, in order to transform the components fromtheeito the ejbasis, we take dot products with the basis vectors ejusing repeatedly thefact that (em en) ek= (en ek)emand obtain: Cijkl=Cpqrs(ep ei)(eq ej)(er ek)(es el)( ) Symmetries of the stiffness tensorReadings: BC stiffness tensor has the followingminor symmetrieswhich result from the symmetryof the stress and strain tensors: ij= ji Cjikl=Cijkl( )Proof by (generalizable) example.

4 From Hooke s law we have 21=C21kl kl, 12=C12kl kland from the symmetry of the stress tensor we have 21= 12 HenceC21kl kl=C12kl klAlso, we have(C21kl C12kl) kl= 0 HenceC21kl=C12klThis reduces the number of material constants from 81 = 3 3 3 3 54 = 6 3 a similar fashion we can make use of the symmetry of the strain tensor ij= ji Cijlk=Cijkl( )This further reduces the number of material constants to 36 = 6 6. To further reduce thenumber of material constants consider equation ( ), ( ): ij= ij=Cijkl kl( ) 2 mn ij= mn(Cijkl kl)( )Cijkl km ln= 2 mn ij( )Cijmn= 2 mn ij( ) ENGINEERING OR VOIGT NOTATION49 Assuming equivalence of the mixed partials:Cijkl= 2 kl ij= 2 ij kl=Cklij( )This further reduces the number of material constants to 21. The most general anisotropiclinear elastic material therefore has 21 material constants. We can write the stress-strainrelations for a linear elastic material exploiting these symmetries as follows: 11 22 33 23 13 12 = C1111C1122C1133C1123C1113C1112C2222C2233 C2223C2213C2212C3333C3323C3313C3312C2323 C2313C2312symmC1313C1312C1212 11 22 332 232 132 12 ( ) Engineering or Voigt notationSince the tensor notation is already lost in the matrix notation, we might as well give indicesto all the components that make more sense for matrix operation: 1 2 3 4 5 6 = C11C12C13C14C15C16C22C23C24C25C26C33C34C 35C36C44C45C46symmC55C56C66 1 2 3 4 5 6 ( )We have: 1) combined pairs of indices as follows: ()11 ()1,()22 ()2,()33 ()3,()23 ()4,()13 ()5,()12 ()6,and, 2) defined theengineering shear strainsas the sum ofsymmetric components, 4= 2 23= 23+ 32, the material has symmetries within its structure the number of material constantsis reduced even further.

5 We now turn to a brief discussion of material symmetries Material Symmetries and anisotropyAnisotropyrefers to the directional dependence of material properties (mechanical or other-wise). It plays an important role in Aerospace Materials due to the wide use of different types of material anisotropy are determined by the existence of symmetriesin the internal structure of the material. The more the internal symmetries, the simplerthe structure of the stiffness tensor. Each type of symmetry results in the invariance of thestiffness tensor to a specificsymmetry transformations(rotations about specific axes and50 Module 3. Constitutive Equations reflections about specific planes). These symmetry transformations can be represented byan orthogonal second order tensor, ej,such thatQ 1=QTand:det(Qij) ={+1 rotation 1 reflectionThe invariance of the stiffness tensor under these transformations is expressed as follows:Cijkl=QipQjqQkrQlsCpqrs( )Let s take a brief look at variousclasses of material symmetry, correspondingsym-metry transformations, implications on theanisotropy of the material, and thestruc-ture of the stiffness tensor:xyzcbabbbbbbbb Triclinic: no symmetry planes, fully anisotropic.}

6 , , <90 Number of independent coefficients: 21 Symmetry transformation: NoneC= C1111C1122C1133C1123C1113C1112C2222C2233 C2223C2213C2212C3333C3323C3313C3312C2323 C2313C2312symmC1313C1312C1212 xyzcbabbbbbbbb Monoclinic:onesymmetryplane(xy).a6=b6=c, = = 90, <90 Number of independent coefficients: 13 Symmetry transformation: reflection aboutz-axisQ= 1 000 100 0 1 C= C1111C1122C113300C1112C2222C223300C2212C 333300C3312C2323C23130symmC13130C1212 Concept Question s consider a monoclinic MATERIAL SYMMETRIES AND ANISOTROPY511. Derive the structure of the stiffness tensor for such a material and show that the tensorhas 13 independent :The symmetry transformations can be represented by an orthogonal secondorder tensor, ej,such thatQ 1=QTand:det(Qij) ={+1 rotation 1 reflectionThe invariance of the stiffness tensor under these transformations is expressed as follows:Cijkl=QipQjqQkrQlsCpqrs( )Herein, in the case of a monoclinic material, there is one symmetry plane (xy).}

7 Hence thesecond order tensorQis written as follows:Q= 1 000 100 0 1 .( )Applying the corresponding symmetry transformation to the stiffness tensor:C11=C1111=Q1pQ1qQ1rQ1sCpqrs= 1p 1q 1r 1sCpqrsthe term 1p 1q 1r 1sis'0 if onlyp= 1 andq= 1 andr= 1 ands= 1:C1111=C1111In the similar manner we obtain:C12=C1122=Q1pQ1qQ2rQ2sCpqrs= 1p 1q 2r 2sCpqrs=C1122C13=C1133=Q1pQ1qQ3rQ3sCpqrs = 1p 1q( 3r)( 3s)Cpqrs=C1133C14=C1123=Q1pQ1qQ2rQ3sCpqr s= 1p 1q 2r( 3s)Cpqrs= C1123= 052 Module 3. Constitutive EQUATIONSC15=C1113=Q1pQ1qQ1rQ3sCpqrs= 1p 1q 1r( 3s)Cpqrs= C1113= 0C16=C1112=Q1pQ1qQ1rQ2sCpqrs= 1p 1q 1r 2sCpqrs=C1112C22=C2222=Q2pQ2qQ2rQ2sCpqrs = 2p 2q 2r 2sCpqrs=C2222C23=C2233=Q2pQ2qQ3rQ3sCpqrs = 2p 2q( 3r)( 3s)Cpqrs=C2233C24=C2223=Q2pQ2qQ2rQ3sCpqr s= 2p 2q 2r( 3s)Cpqrs= C2223= 0C25=C2213=Q2pQ2qQ1rQ3sCpqrs= 2p 2q 1r( 3s)Cpqrs= C2213= 0C26=C2212=Q2pQ2qQ1rQ2sCpqrs= 2p 2q 1r 2sCpqrs=C2212C33=C3333=Q3pQ3qQ3rQ3sCpqrs = ( 3p)( 3q)( 3r)( 3s)Cpqrs=C3333C34=C3323=Q3pQ3qQ2rQ3sCpqr s= ( 3p)( 3q) 2r( 3s)Cpqrs= C3323= MATERIAL SYMMETRIES AND ANISOTROPY53C35=C3313=Q3pQ3qQ1rQ3sCpqrs= ( 3p)( 3q) 1r( 3s)Cpqrs= C3313= 0C36=C3312=Q3pQ3qQ1rQ2sCpqrs= ( 3p)( 3q) 1r 2sCpqrs=C3312C44=C2323=Q2pQ3qQ2rQ3sCpqrs = 2p( 3q) 2r( 3s)Cpqrs=C2323C45=C2313=Q2pQ3qQ1rQ3sCpqr s= 2p( 3q) 1r( 3s)Cpqrs=C2313C46=C2312=Q2pQ3qQ1rQ2sCpqr s= 2p( 3q) 1r 2sCpqrs= C2312= 0C55=C1313=Q1pQ3qQ1rQ3sCpqrs= 1p( 3q) 1r( 3s)

8 Cpqrs=C1313C56=C1312=Q1pQ3qQ1rQ2sCpqrs= 1p( 3q) 1r 2sCpqrs= C1312= 0C66=C1212=Q1pQ2qQ1rQ2sCpqrs= 1p 2q 1r 2sCpqrs=C1212 Hence, the elastic tensor has 13 independant components:C= C1111C1122C113300C1112C2222C223300C2212C 333300C3312C2323C23130symmC13130C1212 54 Module 3. Constitutive Equations xyzcbabbbbbbbb Orthotropic: three mutually orthogonal planes ofreflection , = = = 90 Number of independent coefficients: 9 Symmetry transformations: reflections about allthree orthogonal planesQ= 1 0 001 000 1 Q= 1000 1 0001 Q= 1 000 100 0 1 C= C1111C1122C1133000C2222C2233000C3333000C 232300symmC13130C1212 Concept Question elastic an orthotropic linear elastic material where1,2and3are the orthotropic Use the symmetry transformations corresponding to this material shown in the notesto derive the structure of the elastic In particular, show that the elastic tensor has 9 independent :For the reflection about the plane (1,2), the stressafter reflection is expressed as a function of the stress before the reflection and thereflection transformationR: =RT R( )with (1,2)R=RT= 1 000 100 0 1.

9 ( ) = 11 12 13 12 22 23 13 23 33 .( )and = 11 12 13 12 22 23 13 23 33 .( )or you can write: ij=QipQjp pq( ) MATERIAL SYMMETRIES AND ANISOTROPY55and ij=QipQjp pq( )withQ=R, which leads to: 11= 1p 1q pq= 11 22= 2p 2q pq= 22 33= ( 3p)( 3q) pq= 33 12= 1p 2q pq= 12 23= 2p( 3q) pq= 23 13= 1p( 3q) pq= 13and in the similar manner: 11= 11 22= 22 33= 33 12= 12 23= 23 13= 13 Let s write the consitutive equation =C: 11 22 33 12 23 31 = C11C12C13C14C15C16C22C23C24C25C26C33C34C 35C36C44C45C46C55C56symC66 . 11 22 332 122 232 31 .( )56 Module 3. Constitutive Equations 11=C11 11+C12 22+C13 33+ 2C14 12+ 2C15 23+ 2C16 31 22=C12 11+C22 22+C23 33+ 2C24 12+ 2C25 23+ 2C26 31 33=C13 11+C23 22+C33 33+ 2C34 12+ 2C35 23+ 2C36 31 12=C14 11+C24 22+C34 33+ 2C44 12+ 2C45 23+ 2C46 31 23=C15 11+C25 22+C35 33+ 2C45 12+ 2C55 23+ 2C56 31 31=C16 11+C26 22+C36 33+ 2C46 12+ 2C56 23+ 2C66 31( )and 11=C11 11+C12 22+C13 33+ 2C14 12+ 2C15 23+ 2C16 31 22=C12 11+C22 22+C23 33+ 2C24 12+ 2C25 23+ 2C26 31 33=C13 11+C23 22+C33 33+ 2C34 12+ 2C35 23+ 2C36 31 12=C14 11+C24 22+C34 33+ 2C44 12+ 2C45 23+ 2C46 31 23=C15 11+C25 22+C35 33+ 2C45 12+ 2C55 23+ 2C56 31 31=C16 11+C26 22+C36 33+ 2C46 12+ 2C56 23+ 2C66 31( )Expressing the components of the stress as a function of the components of thestrain.

10 11=C11 11+C12 22+C13 33+ 2C14 12 2C15 23 2C16 31 22=C12 11+C22 22+C23 33+ 2C24 12 2C25 23 2C26 31 33=C13 11+C23 22+C33 33+ 2C34 12 2C35 23 2C36 31 12=C14 11+C24 22+C34 33+ 2C44 12 2C45 23 2C46 31 23=C15 11+C25 22+C35 33+ 2C45 12 2C55 23 2C56 31 31=C16 11+C26 22+C36 33+ 2C46 12 2C56 23 2C66 31and expressing the components of the stress as a function of the components of thestress (equation ): 11= 11 22= 22 33= 33 12= 12 23= 23 31= 31to reach such an equality we need to have:C15=C16=C25=C26=C35=C36=C45=C46= 0 hence the elastic tensor MATERIAL SYMMETRIES AND ANISOTROPY57 C11C12C13C1400C22C23C2400C33C3400C4400C5 50symC66 .( )Now, let s consider the reflection about the plane (2,3), the stress after reflection isexpressed as a function of the stress before the reflection and the reflection matrixR: =RT R( )with (2,3)R=RT= 1 0 001 000 1 .( ) = 11 12 13 12 22 23 13 23 33 .( )and = 11 12 13 12 22 23 13 23 33.