Continuum mechanics V. Constitutive equations. 1. Constitutive equation: definition and basic axioms


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1 Continuum mechanics office Math Mars 16, 2011, Université de Fribourg 1. Constitutive equation: definition and basic axioms Constitutive equation: relation between two physical quantities specific to a material, e.g.: Basic axioms Axiom of causality Axiom of determinism Axiom of equipresence Axiom of neighbourhood Axiom of memory Axiom of objectivity τ ij = τ ij u, {e kl }, {F k l }, T ) Axiom of material invariance Axiom of admissibility
2 1. Basic axioms: causality Axiom of causality: Independent variables in the constitutive laws are: Continuum position y i x, t) Temperature T Dependent variables responses) are e.g.: Helmholtz free energy ϕ thermodynamic potential, measure of the useful work obtainable from a closed thermodynamic system) Strain energy density Ψ Stress tensor τ ij Heat flux q i Internal energy ɛ Entropy S 1. Basic axioms: determinism and equipresence Axiom of determinisim Responses of the constitutive functions at a material point x at time t are determined by the history of the motion and history of the temperature of all points of the body. Axiom of equipresence If an independent variable enters in one function of response, it should be present in all constitutive laws until the proof of the contrary)
3 1. Basic axioms: neighbourhood Axiom of neighbourhood Responses at a point x are not much influenced by values of independent variables temperature and displacement) at a distant point x. Hypothesis: functions yx, t) and T x, t) are sufficiently smooth to be expanded into a Taylor series: y i x, t) = y i x, t)+ y i x j x j x j )+ 1 2 y i x,t 2 x j x k x j x j ) x k x k )+... x,t with negligible higherorder terms. Simple thermomechanical material: Taylor expansion terms with the second+higher derivatives are negligible: τx, t) = T [ yx, t ), y,x x, t ), T x, t ), T,x x, t ); x, t t ] This class of material also called gradient continua. 1. Basic axioms: memory Axiom of memory Values of constitutive variables from a distant past do not affect appreciably the values of constitutive laws now. Smooth memory material: constitutive variables can be expanded to Taylor series in time with negligible higher order terms Fading memory: response functionals must smooth possible discontinuities in memory
4 1. Basic axioms: objectivity Axiom of objectivity Invariance of constitutive laws with respect to rigid body motion of the spatial frame of reference spatial coordinates). Simple consequence: constitutive laws depend of the deformation gradient or strain tensor) rather than yx). 1. Basic axioms: material invariance Axiom of material invariance Invariance of constitutive laws with respect to certain symmetries/transformations of the material frame of reference material coordinates). Symmetries in material properties due to crystallographic orientation. Hemitropic continuum: invariant w.r.t. all rotations Isotropic continuum: hemitropic + invariant to reflection Anisotropic continuum: otherwise can have some invariance properties, but not all) Homogeneous continuum: invariance w.r.t. shift of coord system
5 1. Basic axioms: admissibility Axiom of admissibility Consistence with respect to basic conservation laws mass, momentum, energy), and the 2nd law of thermodynamics entropy). This axiom can also help to eliminate dependences on some constitutive variables. 2. Constitutive laws for simple thermomechanical continua Thermoelastic continua: simple thermomechanical continua with no memory. By applying the basic axioms, all material properties depend only on the current values of deformation and temperature: hence also for Helmholtz free energy: ϕ = ϕ{e ij )}, T ) For simplicity, consider only small deformations Cauchy strain tensor e ij ). Are we able to say more about the form of the constitutive laws in this case?
6 2. Constitutive laws for simple thermomechanical continua Axiom of admissibility: we need to be consistent with thermodynamical laws and basic equilibria. Let us derive ϕ with respect to time ϕ = ϕ e ij ė ij + ϕ T Ṫ and substitute it into the dissipation inequality We get ρ ϕ + ρ Ṫ η τ ij j v i + qi T it 0. ρ ϕ e ij ė ij + ρ ϕ T Ṫ + ρ Ṫ η τ ij j v i + qi T it 0. NB. Due to the symmetry of τ ij = τ ji, we have τ ij j v i = 1 2 τ ij j v i + τ ij i v j ) = τ ij ė ij 2. Constitutive laws for simple thermomechanical continua Hence, the dissipation inequality looks now like: ė ij ρ ϕ ) τ ij + e Ṫ ρ η + ϕ ) + qi ij T T it 0. This inequality must hold for any timedependent process, ie. for any ė ij and Ṫ! Hence there must be: ρ ϕ e ij τ ij = 0 τ ij = ρ ϕ e ij, η + ϕ T η = ϕ T, q i T it 0.
7 3. Simple thermomechanical continuum: large deformation Small deformations: Cauchy stress tensor) τ ij = ρ ϕ, η = ϕ e ij T Large deformations: 2nd PiolaKirchhoff) T ij ϕ = ρ 0, η = ϕ ε ij T,, q i T it 0 q i T it 0 Define strain or stored) energy density Ψ = ρ 0 ϕ, then: T ij = Ψ, η = 1 Ψ ε ij ρ 0 T, q i T it 0 NB: T = Ψ ε is a derivative of a scalar function with respect to a tensor, see M2, Section 2. Hyperelastic material: material for which T ij = Ψ ε ij 4. Hooke s law Robert Hooke ) Neglect temperature: Taylor expansion of strain energy density: Ψe) = Ψ 0 + E ij e ij E ijkl e ij e kl +... with material coefficients E ij, E ijkl called elastic tensors. Suppose small deformations: take only the 3 first terms of the expansion: then from τ ij = Ψ e ij we obtain the Hooke s law 1660): τ ij = E ij + E ijkl e kl with the prestress E ij at initial configuration. If no prestress: τ ij = E ijkl e kl
8 4.1. Hooke s law: elastic tensor E ijkl Elastic tensor E ijkl : 3 4 = 81 components depending only on material coordinates Possible reduction of degrees of freedom: Symmetry of τ ij and e kl symmetry of E ijkl within ij and kl: E ijkl = E jikl = E ijlk = E jilk no. of components reduced to 36. Taylor expansion of Ψe) symmetry in pairs ij and kl: E ijkl = E klij no. of components thus reduced to Hooke s law in deviatorform Consider only small deformations. Physical meaning of Cauchy strain e: dv dv 0 Relative volume change: = e1 1 dv + e2 2 + e3 3 = tre) = el l 0 cf. 1. Kinematics, section 4.) Define: Volumic dilatation e: volumechanging deformation component: e = 1 3 tre) = 1 3 el l Strain deviator ẽ: volumepreserving deformation component ẽ ij = e ij e g ij ẽ i j = e i j e δ i j changes only shape, not the volume, trẽ) = 0. Hydrostatic tension s: forces opposed to volume change s = 1 3 trτ) = 1 3 τ j j Stress deviator τ: forces opposed to shape change: τ ij = τ ij s g ij τ i j = τ i j s δ i j
9 4.2. Hooke s law in deviatorform, shear and bulk moduli Hooke s law for isotropic materials in deviator form): τ j i = 2 µ ẽj i volumepreserving deformations s = 3 K e volumechange Material properties characterized only by 2 constants shear modulus µ characterizes genuine shear bulk modulus K characterizes in)compressibility incompressible material for K ) Total strain tensor: τj i = τ j i + s δj i = 2 µẽj i + 3 K e δj i = 2 µ ej i 1 ) 3 el l δi j + K el l δi j = 2 µ ej i + K 2 µ ) el l 3 δi j 4.3. Hooke s law for isotropic materials: elastic tensor E ijkl Total strain tensor: τ ij = 2 µ e ij + K 2 µ ) el l 3 g ij = 2 µ g ik g jl e kl + K 2 µ ) = µ } 3 {{ } = g ij g kl e kl g ik g jl + g il g jk) e kl + g ij g kl e kl Here, and µ are the so called Lamé coefficients, K = 2 µ The corresponding elastic tensor E ijkl is thus τ ij = E ijkl e kl ): E ijkl = µ g ik g jl + g il g jk) + g ij g kl
10 4.4. Hooke s law for isotropic materials: compliance C ijkl The tensorinverse of E ijkl is called compliance C: τ ij = E ijkl e kl e ij = C ijkl τ kl. Let us inverse the Hooke s law ie. express e as a function of τ): Take a trace: τ = 2 µ e + tre) Id trτ) = 2 µ tre) + 3 tre) = 2 µ + 3 ) tre) Plug back tre) into the Hooke s law above to get Hence, e = 1 2µ τ = 2 µe + τ trτ) ) 2µ+3 Id 2 µ + 3 trτ) Id or e ij = 1 ) g ik g jl 2µ 2µ+3 g kl g ij }{{} C ijkl 4.5. Hooke s law: Young s modulus E, Poisson s ratio ν τ kl Material constants in Hooke s law by analogy with linear springs: In 1D, spring stiffness E = force F relative elongation ε ux,y) τ yyu) Compare with a special case in 3D: F ux,y) τ yyu) ε yy u) = du dy Monoaxial loading Suppose τ ij = 0 for all i, j, except τ Complianceform of a 3D Hooke s law gives: e 11 = 1 ) 1 τ 11 = 2µ 2µ + 3 e 22 = e 33 = 2µ 2µ + 3) τ 11 µ + µ 2µ + 3) τ 11 y x F
11 4.5. Hooke s law: Young s modulus E, Poisson s ratio ν Young s modulus defined as apparent 1D spring stiffness in the case of monoaxial loading, ie: E = τ 11 e 11 = µ + µ 2µ + 3) with τ 11 and e 11 from the monoaxial loading in cartesian coordinates. Poisson s ratio measures transversal vs. axial elongation Relative volume change: ν = e 22 e 11 = 2 µ + ) dv dv 0 dv 0 = e 11 + e 22 + e 33 = 1 2 ν) e 11 ie. ν = 0.5 for incompressible materials 4.6. Hooke s law for isotropic materials: summary Isotropic material characterized by two constants: shear modulus µ and bulk modulus K, µ = E ν) K = 1 3 2µ + 3) = 1 3 E 1 2ν Lamé s coefficients µ and, µ = E ν) = E ν 1 + ν)1 2ν) = K 2 µ 3 Young s modulus E and Poisson s ratio ν, E = µ 2µ + 3) µ + ν = 2µ + )
12 4.6. Hooke s law for isotropic materials: summary Corresponding form of Hooke s law: using shear modulus µ and bulk modulus K, in deviator form: τ i j = 2 µ ẽ i j, s = 3 K e using Lamé s coefficients µ and : τ ij = µ g ik g jl + g il g jk) e kl + g ij g kl e kl Or in global form τ = 2 µe + 2 µ + 3 trτ) Id using Young s modulus E and Poisson s ratio ν: ) τ ij E 2ν = 21 + ν) 1 2ν g ij g kl + g ik g jl + g il g jk large deformations: Saint VenantKirchhoff material Ψε) = µ trε 2 ) + [ ] 2 trε), T ij = Ψ 2 ε ij 4.7. Hooke s law: measuring stressstrain curve for steel e kl stress B A elastic uniform plastic necking 0 strain e 11 1 Ultimate Strength 2 Yield Strength elastic limit) 3 Rupture 4 Strain hardening region 5 Necking region A 1st PiolaKirchhoff stress σ = F A 0 B Euler stress τ = F A
13 5. Linear thermoelasticity: DuhamelNeumann s law Consider also temperature: Taylor expansion of strain energy: Ψe, T ) = Ψ 0 T ) + E ij T ) e ij E ijkl T ) e ij e kl +... ] = Ψ 0 T ) + [E ij ij E T 0 ) + T T T 0) +... e ij + 1 ] [E ijkl ijkl E T 0 ) + 2 T T T 0) +... e ij e kl +... Suppose T T 0 << T 0 and small deformations and neglect all mixed) 3rd order terms and higher. DuhamelNeumann s law: from τ ij = Ψ e ij we obtain: τ ij = E ij T 0 + E ijkl e kl β ij T T 0 ) with β ij E ij = T. For isotropic materials βij = β g ij. Usually, we take T 0 with no prestrain, E ij T 0 = Linear thermoelasticity: DuhamelNeumann s law From the Hooke s law, we can write: τ i j = 2 µ e i j + δ i j e l l βi j T T 0) Let us derive the complianceform e = eτ, T ): Indexcontraction of the above gives τ i i = 2µ+3) e l l βk k T T 0) ie. e l l = 1 2µ + 3 [ ] τi i + βk k T T 0) Substitute it back to the DuhamelNeumann s law to obtain ej i = 1 [ ] δk i 2µ δl j 2µ+3 δi j δk l τl k 1 ) 2µ 2µ+3 δi j βm m βj i T T 0 ) }{{}...thermal dilatation coeff α i j
14 6. Constitutive law for heat flux q: Fourier s law Suppose simple thermomechanical continuum, small deformations: q = qt, T, e) Use firstorder Taylor expansion around a deformationfree configuration at T 0 to approximate: q i = k i 0 + k i 1 T T 0 ) + k ij 2 jt + k ijk 3 e jk with some coefficients k0 i, ki 1 and kijk 3. This law must not contradict the 2nd law of thermodynamics in particular there must be cf. Section 2 and 3 above): q i T it = 1 T ) k0 i + k1 i T T 0 ) + k ij 2 jt + k ijk 3 e jk i T 0 for any state of the continuum, ie. T > 0 and e. This is satisfied only if k0 i = k1 i 0, k ijk 3 0 and [k ij 2 ] is sym.positive definite 6. Constitutive law for heat flux q: Fourier s law Hence, we have derived the Fourier s law for heat flux: q i = k ij j T, q = k T with the heat conductivity tensor k, k ij = k ij 2, [kij ] is a symmetric positive definite matrix. For isotropic materials: k = k Id.
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