Constitutive model for ideally elastic material
Stress–strain curves for various hyperelastic material models. Ahyperelastic orGreen elastic material[ 1] is a type ofconstitutive model for ideallyelastic material for which the stress–strain relationship derives from astrain energy density function . The hyperelastic material is a special case of aCauchy elastic material .
For many materials,linear elastic models do not accurately describe the observed material behaviour. The most common example of this kind of material is rubber, whosestress -strain relationship can be defined as non-linearly elastic,isotropic and incompressible. Hyperelasticity provides a means of modeling the stress–strain behavior of such materials.[ 2] The behavior of unfilled,vulcanized elastomers often conforms closely to the hyperelastic ideal. Filled elastomers andbiological tissues [ 3] [ 4] are also often modeled via the hyperelastic idealization. In addition to being used to model physical materials, hyperelastic materials are also used as fictitious media, e.g. in thethird medium contact method .
Ronald Rivlin andMelvin Mooney developed the first hyperelastic models, theNeo-Hookean andMooney–Rivlin solids. Many other hyperelastic models have since been developed. Other widely used hyperelastic material models include theOgden model and theArruda–Boyce model .
Hyperelastic material models [ edit ] Saint Venant–Kirchhoff model[ edit ] The simplest hyperelastic material model is the Saint Venant–Kirchhoff model which is just an extension of the geometrically linear elastic material model to the geometrically nonlinear regime. This model has the general form and the isotropic form respectivelyS = C : E S = λ tr ( E ) I + 2 μ E . {\displaystyle {\begin{aligned}{\boldsymbol {S}}&={\boldsymbol {C}}:{\boldsymbol {E}}\\{\boldsymbol {S}}&=\lambda ~{\text{tr}}({\boldsymbol {E}}){\boldsymbol {\mathit {I}}}+2\mu {\boldsymbol {E}}{\text{.}}\end{aligned}}} where: {\displaystyle \mathbin {:} } is tensor contraction,S {\displaystyle {\boldsymbol {S}}} is the second Piola–Kirchhoff stress,C ∈ R 3 × 3 × 3 × 3 {\displaystyle {\boldsymbol {C}}\in \mathbb {R} ^{3\times 3\times 3\times 3}} is a fourth orderstiffness tensor andE {\displaystyle {\boldsymbol {E}}} is the Lagrangian Green strain given byE = 1 2 [ ( ∇ X u ) T + ∇ X u + ( ∇ X u ) T ⋅ ∇ X u ] {\displaystyle \mathbf {E} ={\frac {1}{2}}\left[(\nabla _{\mathbf {X} }\mathbf {u} )^{\textsf {T}}+\nabla _{\mathbf {X} }\mathbf {u} +(\nabla _{\mathbf {X} }\mathbf {u} )^{\textsf {T}}\cdot \nabla _{\mathbf {X} }\mathbf {u} \right]\,\!} λ {\displaystyle \lambda } andμ {\displaystyle \mu } are theLamé constants , andI {\displaystyle {\boldsymbol {\mathit {I}}}} is the second order unit tensor.
The strain-energy density function for the Saint Venant–Kirchhoff model isW ( E ) = λ 2 [ tr ( E ) ] 2 + μ tr ( E 2 ) {\displaystyle W({\boldsymbol {E}})={\frac {\lambda }{2}}[{\text{tr}}({\boldsymbol {E}})]^{2}+\mu {\text{tr}}{\mathord {\left({\boldsymbol {E}}^{2}\right)}}}
and the second Piola–Kirchhoff stress can be derived from the relationS = ∂ W ∂ E . {\displaystyle {\boldsymbol {S}}={\frac {\partial W}{\partial {\boldsymbol {E}}}}~.}
Classification of hyperelastic material models [ edit ] Hyperelastic material models can be classified as:
phenomenological descriptions of observed behaviormechanistic models deriving from arguments about the underlying structure of the materialhybrids of phenomenological and mechanistic models Generally, a hyperelastic model should satisfy theDrucker stability criterion.Some hyperelastic models satisfy theValanis-Landel hypothesis which states that the strain energy function can be separated into the sum of separate functions of theprincipal stretches ( λ 1 , λ 2 , λ 3 ) {\displaystyle (\lambda _{1},\lambda _{2},\lambda _{3})} :W = f ( λ 1 ) + f ( λ 2 ) + f ( λ 3 ) . {\displaystyle W=f(\lambda _{1})+f(\lambda _{2})+f(\lambda _{3})\,.}
Stress–strain relations[ edit ] Compressible hyperelastic materials [ edit ] First Piola–Kirchhoff stress[ edit ] IfW ( F ) {\displaystyle W({\boldsymbol {F}})} is the strain energy density function, the1st Piola–Kirchhoff stress tensor can be calculated for a hyperelastic material asP = ∂ W ∂ F or P i K = ∂ W ∂ F i K . {\displaystyle {\boldsymbol {P}}={\frac {\partial W}{\partial {\boldsymbol {F}}}}\qquad {\text{or}}\qquad P_{iK}={\frac {\partial W}{\partial F_{iK}}}.} whereF {\displaystyle {\boldsymbol {F}}} is thedeformation gradient . In terms of theLagrangian Green strain (E {\displaystyle {\boldsymbol {E}}} )P = F ⋅ ∂ W ∂ E or P i K = F i L ∂ W ∂ E L K . {\displaystyle {\boldsymbol {P}}={\boldsymbol {F}}\cdot {\frac {\partial W}{\partial {\boldsymbol {E}}}}\qquad {\text{or}}\qquad P_{iK}=F_{iL}~{\frac {\partial W}{\partial E_{LK}}}~.} In terms of theright Cauchy–Green deformation tensor (C {\displaystyle {\boldsymbol {C}}} )P = 2 F ⋅ ∂ W ∂ C or P i K = 2 F i L ∂ W ∂ C L K . {\displaystyle {\boldsymbol {P}}=2~{\boldsymbol {F}}\cdot {\frac {\partial W}{\partial {\boldsymbol {C}}}}\qquad {\text{or}}\qquad P_{iK}=2~F_{iL}~{\frac {\partial W}{\partial C_{LK}}}~.}
Second Piola–Kirchhoff stress[ edit ] IfS {\displaystyle {\boldsymbol {S}}} is thesecond Piola–Kirchhoff stress tensor thenS = F − 1 ⋅ ∂ W ∂ F or S I J = F I k − 1 ∂ W ∂ F k J . {\displaystyle {\boldsymbol {S}}={\boldsymbol {F}}^{-1}\cdot {\frac {\partial W}{\partial {\boldsymbol {F}}}}\qquad {\text{or}}\qquad S_{IJ}=F_{Ik}^{-1}{\frac {\partial W}{\partial F_{kJ}}}~.} In terms of theLagrangian Green strain S = ∂ W ∂ E or S I J = ∂ W ∂ E I J . {\displaystyle {\boldsymbol {S}}={\frac {\partial W}{\partial {\boldsymbol {E}}}}\qquad {\text{or}}\qquad S_{IJ}={\frac {\partial W}{\partial E_{IJ}}}~.} In terms of theright Cauchy–Green deformation tensor S = 2 ∂ W ∂ C or S I J = 2 ∂ W ∂ C I J . {\displaystyle {\boldsymbol {S}}=2~{\frac {\partial W}{\partial {\boldsymbol {C}}}}\qquad {\text{or}}\qquad S_{IJ}=2~{\frac {\partial W}{\partial C_{IJ}}}~.} The above relation is also known as theDoyle-Ericksen formula in the material configuration.
Similarly, theCauchy stress is given byσ = 1 J ∂ W ∂ F ⋅ F T ; J := det F or σ i j = 1 J ∂ W ∂ F i K F j K . {\displaystyle {\boldsymbol {\sigma }}={\frac {1}{J}}~{\frac {\partial W}{\partial {\boldsymbol {F}}}}\cdot {\boldsymbol {F}}^{\textsf {T}}~;~~J:=\det {\boldsymbol {F}}\qquad {\text{or}}\qquad \sigma _{ij}={\frac {1}{J}}~{\frac {\partial W}{\partial F_{iK}}}~F_{jK}~.} In terms of theLagrangian Green strain σ = 1 J F ⋅ ∂ W ∂ E ⋅ F T or σ i j = 1 J F i K ∂ W ∂ E K L F j L . {\displaystyle {\boldsymbol {\sigma }}={\frac {1}{J}}~{\boldsymbol {F}}\cdot {\frac {\partial W}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{\textsf {T}}\qquad {\text{or}}\qquad \sigma _{ij}={\frac {1}{J}}~F_{iK}~{\frac {\partial W}{\partial E_{KL}}}~F_{jL}~.} In terms of theright Cauchy–Green deformation tensor σ = 2 J F ⋅ ∂ W ∂ C ⋅ F T or σ i j = 2 J F i K ∂ W ∂ C K L F j L . {\displaystyle {\boldsymbol {\sigma }}={\frac {2}{J}}~{\boldsymbol {F}}\cdot {\frac {\partial W}{\partial {\boldsymbol {C}}}}\cdot {\boldsymbol {F}}^{\textsf {T}}\qquad {\text{or}}\qquad \sigma _{ij}={\frac {2}{J}}~F_{iK}~{\frac {\partial W}{\partial C_{KL}}}~F_{jL}~.} The above expressions are valid even for anisotropic media (in which case, the potential function is understood to dependimplicitly on reference directional quantities such as initial fiber orientations). In the special case of isotropy, the Cauchy stress can be expressed in terms of theleft Cauchy-Green deformation tensor as follows:[ 7] σ = 2 J ∂ W ∂ B ⋅ B or σ i j = 2 J B i k ∂ W ∂ B k j . {\displaystyle {\boldsymbol {\sigma }}={\frac {2}{J}}{\frac {\partial W}{\partial {\boldsymbol {B}}}}\cdot ~{\boldsymbol {B}}\qquad {\text{or}}\qquad \sigma _{ij}={\frac {2}{J}}~B_{ik}~{\frac {\partial W}{\partial B_{kj}}}~.}
Incompressible hyperelastic materials [ edit ] For an incompressible materialJ := det F = 1 {\displaystyle J:=\det {\boldsymbol {F}}=1} . The incompressibility constraint is thereforeJ − 1 = 0 {\displaystyle J-1=0} . To ensure incompressibility of a hyperelastic material, the strain-energy function can be written in form:W = W ( F ) − p ( J − 1 ) {\displaystyle W=W({\boldsymbol {F}})-p~(J-1)} where the hydrostatic pressurep {\displaystyle p} functions as aLagrangian multiplier to enforce the incompressibility constraint. The 1st Piola–Kirchhoff stress now becomesP = − p J F − T + ∂ W ∂ F = − p F − T + F ⋅ ∂ W ∂ E = − p F − T + 2 F ⋅ ∂ W ∂ C . {\displaystyle {\boldsymbol {P}}=-p~J{\boldsymbol {F}}^{-{\textsf {T}}}+{\frac {\partial W}{\partial {\boldsymbol {F}}}}=-p~{\boldsymbol {F}}^{-{\textsf {T}}}+{\boldsymbol {F}}\cdot {\frac {\partial W}{\partial {\boldsymbol {E}}}}=-p~{\boldsymbol {F}}^{-{\textsf {T}}}+2~{\boldsymbol {F}}\cdot {\frac {\partial W}{\partial {\boldsymbol {C}}}}~.} This stress tensor can subsequently beconverted into any of the other conventional stress tensors, such as theCauchy stress tensor which is given byσ = P ⋅ F T = − p 1 + ∂ W ∂ F ⋅ F T = − p 1 + F ⋅ ∂ W ∂ E ⋅ F T = − p 1 + 2 F ⋅ ∂ W ∂ C ⋅ F T . {\displaystyle {\boldsymbol {\sigma }}={\boldsymbol {P}}\cdot {\boldsymbol {F}}^{\textsf {T}}=-p~{\boldsymbol {\mathit {1}}}+{\frac {\partial W}{\partial {\boldsymbol {F}}}}\cdot {\boldsymbol {F}}^{\textsf {T}}=-p~{\boldsymbol {\mathit {1}}}+{\boldsymbol {F}}\cdot {\frac {\partial W}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{\textsf {T}}=-p~{\boldsymbol {\mathit {1}}}+2~{\boldsymbol {F}}\cdot {\frac {\partial W}{\partial {\boldsymbol {C}}}}\cdot {\boldsymbol {F}}^{\textsf {T}}~.}
Expressions for the Cauchy stress [ edit ] Compressible isotropic hyperelastic materials [ edit ] Forisotropic hyperelastic materials, the Cauchy stress can be expressed in terms of the invariants of theleft Cauchy–Green deformation tensor (orright Cauchy–Green deformation tensor ). If thestrain energy density function isW ( F ) = W ^ ( I 1 , I 2 , I 3 ) = W ¯ ( I ¯ 1 , I ¯ 2 , J ) = W ~ ( λ 1 , λ 2 , λ 3 ) , {\displaystyle W({\boldsymbol {F}})={\hat {W}}(I_{1},I_{2},I_{3})={\bar {W}}({\bar {I}}_{1},{\bar {I}}_{2},J)={\tilde {W}}(\lambda _{1},\lambda _{2},\lambda _{3}),} thenσ = 2 I 3 [ ( ∂ W ^ ∂ I 1 + I 1 ∂ W ^ ∂ I 2 ) B − ∂ W ^ ∂ I 2 B ⋅ B ] + 2 I 3 ∂ W ^ ∂ I 3 1 = 2 J [ 1 J 2 / 3 ( ∂ W ¯ ∂ I ¯ 1 + I ¯ 1 ∂ W ¯ ∂ I ¯ 2 ) B − 1 J 4 / 3 ∂ W ¯ ∂ I ¯ 2 B ⋅ B ] + [ ∂ W ¯ ∂ J − 2 3 J ( I ¯ 1 ∂ W ¯ ∂ I ¯ 1 + 2 I ¯ 2 ∂ W ¯ ∂ I ¯ 2 ) ] 1 = 2 J [ ( ∂ W ¯ ∂ I ¯ 1 + I ¯ 1 ∂ W ¯ ∂ I ¯ 2 ) B ¯ − ∂ W ¯ ∂ I ¯ 2 B ¯ ⋅ B ¯ ] + [ ∂ W ¯ ∂ J − 2 3 J ( I ¯ 1 ∂ W ¯ ∂ I ¯ 1 + 2 I ¯ 2 ∂ W ¯ ∂ I ¯ 2 ) ] 1 = λ 1 λ 1 λ 2 λ 3 ∂ W ~ ∂ λ 1 n 1 ⊗ n 1 + λ 2 λ 1 λ 2 λ 3 ∂ W ~ ∂ λ 2 n 2 ⊗ n 2 + λ 3 λ 1 λ 2 λ 3 ∂ W ~ ∂ λ 3 n 3 ⊗ n 3 {\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}&={\frac {2}{\sqrt {I_{3}}}}\left[\left({\frac {\partial {\hat {W}}}{\partial I_{1}}}+I_{1}~{\frac {\partial {\hat {W}}}{\partial I_{2}}}\right){\boldsymbol {B}}-{\frac {\partial {\hat {W}}}{\partial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+2{\sqrt {I_{3}}}~{\frac {\partial {\hat {W}}}{\partial I_{3}}}~{\boldsymbol {\mathit {1}}}\\[5pt]&={\frac {2}{J}}\left[{\frac {1}{J^{2/3}}}\left({\frac {\partial {\bar {W}}}{\partial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\frac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}\right){\boldsymbol {B}}-{\frac {1}{J^{4/3}}}~{\frac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+\left[{\frac {\partial {\bar {W}}}{\partial J}}-{\frac {2}{3J}}\left({\bar {I}}_{1}~{\frac {\partial {\bar {W}}}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\frac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}\right)\right]~{\boldsymbol {\mathit {1}}}\\[5pt]&={\frac {2}{J}}\left[\left({\frac {\partial {\bar {W}}}{\partial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\frac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}\right){\bar {\boldsymbol {B}}}-{\frac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}~{\bar {\boldsymbol {B}}}\cdot {\bar {\boldsymbol {B}}}\right]+\left[{\frac {\partial {\bar {W}}}{\partial J}}-{\frac {2}{3J}}\left({\bar {I}}_{1}~{\frac {\partial {\bar {W}}}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\frac {\partial {\bar {W}}}{\partial {\bar {I}}_{2}}}\right)\right]~{\boldsymbol {\mathit {1}}}\\[5pt]&={\frac {\lambda _{1}}{\lambda _{1}\lambda _{2}\lambda _{3}}}~{\frac {\partial {\tilde {W}}}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\frac {\lambda _{2}}{\lambda _{1}\lambda _{2}\lambda _{3}}}~{\frac {\partial {\tilde {W}}}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\frac {\lambda _{3}}{\lambda _{1}\lambda _{2}\lambda _{3}}}~{\frac {\partial {\tilde {W}}}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\end{aligned}}} (See the page onthe left Cauchy–Green deformation tensor for the definitions of these symbols).
Proof 1 Thesecond Piola–Kirchhoff stress tensor for a hyperelastic material is given byS = 2 ∂ W ∂ C {\displaystyle {\boldsymbol {S}}=2~{\frac {\partial W}{\partial {\boldsymbol {C}}}}} whereC = F T ⋅ F {\displaystyle {\boldsymbol {C}}={\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}} is theright Cauchy–Green deformation tensor andF {\displaystyle {\boldsymbol {F}}} is thedeformation gradient . TheCauchy stress is given byσ = 1 J F ⋅ S ⋅ F T = 2 J F ⋅ ∂ W ∂ C ⋅ F T {\displaystyle {\boldsymbol {\sigma }}={\frac {1}{J}}~{\boldsymbol {F}}\cdot {\boldsymbol {S}}\cdot {\boldsymbol {F}}^{T}={\frac {2}{J}}~{\boldsymbol {F}}\cdot {\frac {\partial W}{\partial {\boldsymbol {C}}}}\cdot {\boldsymbol {F}}^{T}} whereJ = det F {\displaystyle J=\det {\boldsymbol {F}}} . LetI 1 , I 2 , I 3 {\displaystyle I_{1},I_{2},I_{3}} be the three principal invariants ofC {\displaystyle {\boldsymbol {C}}} . Then∂ W ∂ C = ∂ W ∂ I 1 ∂ I 1 ∂ C + ∂ W ∂ I 2 ∂ I 2 ∂ C + ∂ W ∂ I 3 ∂ I 3 ∂ C . {\displaystyle {\frac {\partial W}{\partial {\boldsymbol {C}}}}={\frac {\partial W}{\partial I_{1}}}~{\frac {\partial I_{1}}{\partial {\boldsymbol {C}}}}+{\frac {\partial W}{\partial I_{2}}}~{\frac {\partial I_{2}}{\partial {\boldsymbol {C}}}}+{\frac {\partial W}{\partial I_{3}}}~{\frac {\partial I_{3}}{\partial {\boldsymbol {C}}}}~.} Thederivatives of the invariants of the symmetric tensorC {\displaystyle {\boldsymbol {C}}} are∂ I 1 ∂ C = 1 ; ∂ I 2 ∂ C = I 1 1 − C ; ∂ I 3 ∂ C = det ( C ) C − 1 {\displaystyle {\frac {\partial I_{1}}{\partial {\boldsymbol {C}}}}={\boldsymbol {\mathit {1}}}~;~~{\frac {\partial I_{2}}{\partial {\boldsymbol {C}}}}=I_{1}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {C}}~;~~{\frac {\partial I_{3}}{\partial {\boldsymbol {C}}}}=\det({\boldsymbol {C}})~{\boldsymbol {C}}^{-1}} Therefore, we can write∂ W ∂ C = ∂ W ∂ I 1 1 + ∂ W ∂ I 2 ( I 1 1 − F T ⋅ F ) + ∂ W ∂ I 3 I 3 F − 1 ⋅ F − T . {\displaystyle {\frac {\partial W}{\partial {\boldsymbol {C}}}}={\frac {\partial W}{\partial I_{1}}}~{\boldsymbol {\mathit {1}}}+{\frac {\partial W}{\partial I_{2}}}~(I_{1}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}})+{\frac {\partial W}{\partial I_{3}}}~I_{3}~{\boldsymbol {F}}^{-1}\cdot {\boldsymbol {F}}^{-T}~.} Plugging into the expression for the Cauchy stress givesσ = 2 J [ ∂ W ∂ I 1 F ⋅ F T + ∂ W ∂ I 2 ( I 1 F ⋅ F T − F ⋅ F T ⋅ F ⋅ F T ) + ∂ W ∂ I 3 I 3 1 ] {\displaystyle {\boldsymbol {\sigma }}={\frac {2}{J}}~\left[{\frac {\partial W}{\partial I_{1}}}~{\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}+{\frac {\partial W}{\partial I_{2}}}~(I_{1}~{\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}-{\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T})+{\frac {\partial W}{\partial I_{3}}}~I_{3}~{\boldsymbol {\mathit {1}}}\right]} Using theleft Cauchy–Green deformation tensor B = F ⋅ F T {\displaystyle {\boldsymbol {B}}={\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}} and noting thatI 3 = J 2 {\displaystyle I_{3}=J^{2}} , we can writeσ = 2 I 3 [ ( ∂ W ∂ I 1 + I 1 ∂ W ∂ I 2 ) B − ∂ W ∂ I 2 B ⋅ B ] + 2 I 3 ∂ W ∂ I 3 1 . {\displaystyle {\boldsymbol {\sigma }}={\frac {2}{\sqrt {I_{3}}}}~\left[\left({\frac {\partial W}{\partial I_{1}}}+I_{1}~{\frac {\partial W}{\partial I_{2}}}\right)~{\boldsymbol {B}}-{\frac {\partial W}{\partial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+2~{\sqrt {I_{3}}}~{\frac {\partial W}{\partial I_{3}}}~{\boldsymbol {\mathit {1}}}~.} For an incompressible materialI 3 = 1 {\displaystyle I_{3}=1} and henceW = W ( I 1 , I 2 ) {\displaystyle W=W(I_{1},I_{2})} .Then∂ W ∂ C = ∂ W ∂ I 1 ∂ I 1 ∂ C + ∂ W ∂ I 2 ∂ I 2 ∂ C = ∂ W ∂ I 1 1 + ∂ W ∂ I 2 ( I 1 1 − F T ⋅ F ) {\displaystyle {\frac {\partial W}{\partial {\boldsymbol {C}}}}={\frac {\partial W}{\partial I_{1}}}~{\frac {\partial I_{1}}{\partial {\boldsymbol {C}}}}+{\frac {\partial W}{\partial I_{2}}}~{\frac {\partial I_{2}}{\partial {\boldsymbol {C}}}}={\frac {\partial W}{\partial I_{1}}}~{\boldsymbol {\mathit {1}}}+{\frac {\partial W}{\partial I_{2}}}~(I_{1}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}})} Therefore, the Cauchy stress is given byσ = 2 [ ( ∂ W ∂ I 1 + I 1 ∂ W ∂ I 2 ) B − ∂ W ∂ I 2 B ⋅ B ] − p 1 . {\displaystyle {\boldsymbol {\sigma }}=2\left[\left({\frac {\partial W}{\partial I_{1}}}+I_{1}~{\frac {\partial W}{\partial I_{2}}}\right)~{\boldsymbol {B}}-{\frac {\partial W}{\partial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]-p~{\boldsymbol {\mathit {1}}}~.} wherep {\displaystyle p} is an undetermined pressure which acts as aLagrange multiplier to enforce the incompressibility constraint.
If, in addition,I 1 = I 2 {\displaystyle I_{1}=I_{2}} , we haveW = W ( I 1 ) {\displaystyle W=W(I_{1})} and hence∂ W ∂ C = ∂ W ∂ I 1 ∂ I 1 ∂ C = ∂ W ∂ I 1 1 {\displaystyle {\frac {\partial W}{\partial {\boldsymbol {C}}}}={\frac {\partial W}{\partial I_{1}}}~{\frac {\partial I_{1}}{\partial {\boldsymbol {C}}}}={\frac {\partial W}{\partial I_{1}}}~{\boldsymbol {\mathit {1}}}} In that case the Cauchy stress can be expressed asσ = 2 ∂ W ∂ I 1 B − p 1 . {\displaystyle {\boldsymbol {\sigma }}=2{\frac {\partial W}{\partial I_{1}}}~{\boldsymbol {B}}-p~{\boldsymbol {\mathit {1}}}~.}
Proof 2 Theisochoric deformation gradient is defined asF ¯ := J − 1 / 3 F {\displaystyle {\bar {\boldsymbol {F}}}:=J^{-1/3}{\boldsymbol {F}}} , resulting in the isochoric deformation gradient having a determinant of 1, in other words it is volume stretch free. Using this one can subsequently define the isochoric left Cauchy–Green deformation tensorB ¯ := F ¯ ⋅ F ¯ T = J − 2 / 3 B {\displaystyle {\bar {\boldsymbol {B}}}:={\bar {\boldsymbol {F}}}\cdot {\bar {\boldsymbol {F}}}^{T}=J^{-2/3}{\boldsymbol {B}}} .The invariants ofB ¯ {\displaystyle {\bar {\boldsymbol {B}}}} areI ¯ 1 = tr ( B ¯ ) = J − 2 / 3 tr ( B ) = J − 2 / 3 I 1 I ¯ 2 = 1 2 ( tr ( B ¯ ) 2 − tr ( B ¯ 2 ) ) = 1 2 ( ( J − 2 / 3 tr ( B ) ) 2 − tr ( J − 4 / 3 B 2 ) ) = J − 4 / 3 I 2 I ¯ 3 = det ( B ¯ ) = J − 6 / 3 det ( B ) = J − 2 I 3 = J − 2 J 2 = 1 {\displaystyle {\begin{aligned}{\bar {I}}_{1}&={\text{tr}}({\bar {\boldsymbol {B}}})=J^{-2/3}{\text{tr}}({\boldsymbol {B}})=J^{-2/3}I_{1}\\{\bar {I}}_{2}&={\frac {1}{2}}\left({\text{tr}}({\bar {\boldsymbol {B}}})^{2}-{\text{tr}}({\bar {\boldsymbol {B}}}^{2})\right)={\frac {1}{2}}\left(\left(J^{-2/3}{\text{tr}}({\boldsymbol {B}})\right)^{2}-{\text{tr}}(J^{-4/3}{\boldsymbol {B}}^{2})\right)=J^{-4/3}I_{2}\\{\bar {I}}_{3}&=\det({\bar {\boldsymbol {B}}})=J^{-6/3}\det({\boldsymbol {B}})=J^{-2}I_{3}=J^{-2}J^{2}=1\end{aligned}}} The set of invariants which are used to define the distortional behavior are the first two invariants of the isochoric left Cauchy–Green deformation tensor tensor, (which are identical to the ones for the right Cauchy Green stretch tensor), and addJ {\displaystyle J} into the fray to describe the volumetric behaviour.
To express the Cauchy stress in terms of the invariantsI ¯ 1 , I ¯ 2 , J {\displaystyle {\bar {I}}_{1},{\bar {I}}_{2},J} recall thatI ¯ 1 = J − 2 / 3 I 1 = I 3 − 1 / 3 I 1 ; I ¯ 2 = J − 4 / 3 I 2 = I 3 − 2 / 3 I 2 ; J = I 3 1 / 2 . {\displaystyle {\bar {I}}_{1}=J^{-2/3}~I_{1}=I_{3}^{-1/3}~I_{1}~;~~{\bar {I}}_{2}=J^{-4/3}~I_{2}=I_{3}^{-2/3}~I_{2}~;~~J=I_{3}^{1/2}~.} The chain rule of differentiation gives us∂ W ∂ I 1 = ∂ W ∂ I ¯ 1 ∂ I ¯ 1 ∂ I 1 + ∂ W ∂ I ¯ 2 ∂ I ¯ 2 ∂ I 1 + ∂ W ∂ J ∂ J ∂ I 1 = I 3 − 1 / 3 ∂ W ∂ I ¯ 1 = J − 2 / 3 ∂ W ∂ I ¯ 1 ∂ W ∂ I 2 = ∂ W ∂ I ¯ 1 ∂ I ¯ 1 ∂ I 2 + ∂ W ∂ I ¯ 2 ∂ I ¯ 2 ∂ I 2 + ∂ W ∂ J ∂ J ∂ I 2 = I 3 − 2 / 3 ∂ W ∂ I ¯ 2 = J − 4 / 3 ∂ W ∂ I ¯ 2 ∂ W ∂ I 3 = ∂ W ∂ I ¯ 1 ∂ I ¯ 1 ∂ I 3 + ∂ W ∂ I ¯ 2 ∂ I ¯ 2 ∂ I 3 + ∂ W ∂ J ∂ J ∂ I 3 = − 1 3 I 3 − 4 / 3 I 1 ∂ W ∂ I ¯ 1 − 2 3 I 3 − 5 / 3 I 2 ∂ W ∂ I ¯ 2 + 1 2 I 3 − 1 / 2 ∂ W ∂ J = − 1 3 J − 8 / 3 J 2 / 3 I ¯ 1 ∂ W ∂ I ¯ 1 − 2 3 J − 10 / 3 J 4 / 3 I ¯ 2 ∂ W ∂ I ¯ 2 + 1 2 J − 1 ∂ W ∂ J = − 1 3 J − 2 ( I ¯ 1 ∂ W ∂ I ¯ 1 + 2 I ¯ 2 ∂ W ∂ I ¯ 2 ) + 1 2 J − 1 ∂ W ∂ J {\displaystyle {\begin{aligned}{\frac {\partial W}{\partial I_{1}}}&={\frac {\partial W}{\partial {\bar {I}}_{1}}}~{\frac {\partial {\bar {I}}_{1}}{\partial I_{1}}}+{\frac {\partial W}{\partial {\bar {I}}_{2}}}~{\frac {\partial {\bar {I}}_{2}}{\partial I_{1}}}+{\frac {\partial W}{\partial J}}~{\frac {\partial J}{\partial I_{1}}}\\&=I_{3}^{-1/3}~{\frac {\partial W}{\partial {\bar {I}}_{1}}}=J^{-2/3}~{\frac {\partial W}{\partial {\bar {I}}_{1}}}\\{\frac {\partial W}{\partial I_{2}}}&={\frac {\partial W}{\partial {\bar {I}}_{1}}}~{\frac {\partial {\bar {I}}_{1}}{\partial I_{2}}}+{\frac {\partial W}{\partial {\bar {I}}_{2}}}~{\frac {\partial {\bar {I}}_{2}}{\partial I_{2}}}+{\frac {\partial W}{\partial J}}~{\frac {\partial J}{\partial I_{2}}}\\&=I_{3}^{-2/3}~{\frac {\partial W}{\partial {\bar {I}}_{2}}}=J^{-4/3}~{\frac {\partial W}{\partial {\bar {I}}_{2}}}\\{\frac {\partial W}{\partial I_{3}}}&={\frac {\partial W}{\partial {\bar {I}}_{1}}}~{\frac {\partial {\bar {I}}_{1}}{\partial I_{3}}}+{\frac {\partial W}{\partial {\bar {I}}_{2}}}~{\frac {\partial {\bar {I}}_{2}}{\partial I_{3}}}+{\frac {\partial W}{\partial J}}~{\frac {\partial J}{\partial I_{3}}}\\&=-{\frac {1}{3}}~I_{3}^{-4/3}~I_{1}~{\frac {\partial W}{\partial {\bar {I}}_{1}}}-{\frac {2}{3}}~I_{3}^{-5/3}~I_{2}~{\frac {\partial W}{\partial {\bar {I}}_{2}}}+{\frac {1}{2}}~I_{3}^{-1/2}~{\frac {\partial W}{\partial J}}\\&=-{\frac {1}{3}}~J^{-8/3}~J^{2/3}~{\bar {I}}_{1}~{\frac {\partial W}{\partial {\bar {I}}_{1}}}-{\frac {2}{3}}~J^{-10/3}~J^{4/3}~{\bar {I}}_{2}~{\frac {\partial W}{\partial {\bar {I}}_{2}}}+{\frac {1}{2}}~J^{-1}~{\frac {\partial W}{\partial J}}\\&=-{\frac {1}{3}}~J^{-2}~\left({\bar {I}}_{1}~{\frac {\partial W}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\frac {\partial W}{\partial {\bar {I}}_{2}}}\right)+{\frac {1}{2}}~J^{-1}~{\frac {\partial W}{\partial J}}\end{aligned}}} Recall that the Cauchy stress is given byσ = 2 I 3 [ ( ∂ W ∂ I 1 + I 1 ∂ W ∂ I 2 ) B − ∂ W ∂ I 2 B ⋅ B ] + 2 I 3 ∂ W ∂ I 3 1 . {\displaystyle {\boldsymbol {\sigma }}={\frac {2}{\sqrt {I_{3}}}}~\left[\left({\frac {\partial W}{\partial I_{1}}}+I_{1}~{\frac {\partial W}{\partial I_{2}}}\right)~{\boldsymbol {B}}-{\frac {\partial W}{\partial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+2~{\sqrt {I_{3}}}~{\frac {\partial W}{\partial I_{3}}}~{\boldsymbol {\mathit {1}}}~.} In terms of the invariantsI ¯ 1 , I ¯ 2 , J {\displaystyle {\bar {I}}_{1},{\bar {I}}_{2},J} we haveσ = 2 J [ ( ∂ W ∂ I 1 + J 2 / 3 I ¯ 1 ∂ W ∂ I 2 ) B − ∂ W ∂ I 2 B ⋅ B ] + 2 J ∂ W ∂ I 3 1 . {\displaystyle {\boldsymbol {\sigma }}={\frac {2}{J}}~\left[\left({\frac {\partial W}{\partial I_{1}}}+J^{2/3}~{\bar {I}}_{1}~{\frac {\partial W}{\partial I_{2}}}\right)~{\boldsymbol {B}}-{\frac {\partial W}{\partial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+2~J~{\frac {\partial W}{\partial I_{3}}}~{\boldsymbol {\mathit {1}}}~.} Plugging in the expressions for the derivatives ofW {\displaystyle W} in terms ofI ¯ 1 , I ¯ 2 , J {\displaystyle {\bar {I}}_{1},{\bar {I}}_{2},J} , we haveσ = 2 J [ ( J − 2 / 3 ∂ W ∂ I ¯ 1 + J − 2 / 3 I ¯ 1 ∂ W ∂ I ¯ 2 ) B − J − 4 / 3 ∂ W ∂ I ¯ 2 B ⋅ B ] + 2 J [ − 1 3 J − 2 ( I ¯ 1 ∂ W ∂ I ¯ 1 + 2 I ¯ 2 ∂ W ∂ I ¯ 2 ) + 1 2 J − 1 ∂ W ∂ J ] 1 {\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}&={\frac {2}{J}}~\left[\left(J^{-2/3}~{\frac {\partial W}{\partial {\bar {I}}_{1}}}+J^{-2/3}~{\bar {I}}_{1}~{\frac {\partial W}{\partial {\bar {I}}_{2}}}\right)~{\boldsymbol {B}}-J^{-4/3}~{\frac {\partial W}{\partial {\bar {I}}_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+\\&\qquad 2~J~\left[-{\frac {1}{3}}~J^{-2}~\left({\bar {I}}_{1}~{\frac {\partial W}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\frac {\partial W}{\partial {\bar {I}}_{2}}}\right)+{\frac {1}{2}}~J^{-1}~{\frac {\partial W}{\partial J}}\right]~{\boldsymbol {\mathit {1}}}\end{aligned}}} or,σ = 2 J [ 1 J 2 / 3 ( ∂ W ∂ I ¯ 1 + I ¯ 1 ∂ W ∂ I ¯ 2 ) B − 1 J 4 / 3 ∂ W ∂ I ¯ 2 B ⋅ B ] + [ ∂ W ∂ J − 2 3 J ( I ¯ 1 ∂ W ∂ I ¯ 1 + 2 I ¯ 2 ∂ W ∂ I ¯ 2 ) ] 1 {\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}&={\frac {2}{J}}~\left[{\frac {1}{J^{2/3}}}~\left({\frac {\partial W}{\partial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\frac {\partial W}{\partial {\bar {I}}_{2}}}\right)~{\boldsymbol {B}}-{\frac {1}{J^{4/3}}}~{\frac {\partial W}{\partial {\bar {I}}_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]\\&\qquad +\left[{\frac {\partial W}{\partial J}}-{\frac {2}{3J}}\left({\bar {I}}_{1}~{\frac {\partial W}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\frac {\partial W}{\partial {\bar {I}}_{2}}}\right)\right]{\boldsymbol {\mathit {1}}}\end{aligned}}} In terms of the deviatoric part ofB {\displaystyle {\boldsymbol {B}}} , we can writeσ = 2 J [ ( ∂ W ∂ I ¯ 1 + I ¯ 1 ∂ W ∂ I ¯ 2 ) B ¯ − ∂ W ∂ I ¯ 2 B ¯ ⋅ B ¯ ] + [ ∂ W ∂ J − 2 3 J ( I ¯ 1 ∂ W ∂ I ¯ 1 + 2 I ¯ 2 ∂ W ∂ I ¯ 2 ) ] 1 {\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}&={\frac {2}{J}}~\left[\left({\frac {\partial W}{\partial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\frac {\partial W}{\partial {\bar {I}}_{2}}}\right)~{\bar {\boldsymbol {B}}}-{\frac {\partial W}{\partial {\bar {I}}_{2}}}~{\bar {\boldsymbol {B}}}\cdot {\bar {\boldsymbol {B}}}\right]\\&\qquad +\left[{\frac {\partial W}{\partial J}}-{\frac {2}{3J}}\left({\bar {I}}_{1}~{\frac {\partial W}{\partial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\frac {\partial W}{\partial {\bar {I}}_{2}}}\right)\right]{\boldsymbol {\mathit {1}}}\end{aligned}}} For an incompressible materialJ = 1 {\displaystyle J=1} and henceW = W ( I ¯ 1 , I ¯ 2 ) {\displaystyle W=W({\bar {I}}_{1},{\bar {I}}_{2})} .Thenthe Cauchy stress is given byσ = 2 [ ( ∂ W ∂ I ¯ 1 + I 1 ∂ W ∂ I ¯ 2 ) B ¯ − ∂ W ∂ I ¯ 2 B ¯ ⋅ B ¯ ] − p 1 . {\displaystyle {\boldsymbol {\sigma }}=2\left[\left({\frac {\partial W}{\partial {\bar {I}}_{1}}}+I_{1}~{\frac {\partial W}{\partial {\bar {I}}_{2}}}\right)~{\bar {\boldsymbol {B}}}-{\frac {\partial W}{\partial {\bar {I}}_{2}}}~{\bar {\boldsymbol {B}}}\cdot {\bar {\boldsymbol {B}}}\right]-p~{\boldsymbol {\mathit {1}}}~.} wherep {\displaystyle p} is an undetermined pressure-like Lagrange multiplier term. In addition, ifI ¯ 1 = I ¯ 2 {\displaystyle {\bar {I}}_{1}={\bar {I}}_{2}} , we haveW = W ( I ¯ 1 ) {\displaystyle W=W({\bar {I}}_{1})} and hencethe Cauchy stress can be expressed asσ = 2 ∂ W ∂ I ¯ 1 B ¯ − p 1 . {\displaystyle {\boldsymbol {\sigma }}=2{\frac {\partial W}{\partial {\bar {I}}_{1}}}~{\bar {\boldsymbol {B}}}-p~{\boldsymbol {\mathit {1}}}~.}
Proof 3 To express the Cauchy stress in terms of thestretches λ 1 , λ 2 , λ 3 {\displaystyle \lambda _{1},\lambda _{2},\lambda _{3}} recall that∂ λ i ∂ C = 1 2 λ i R T ⋅ ( n i ⊗ n i ) ⋅ R ; i = 1 , 2 , 3 . {\displaystyle {\frac {\partial \lambda _{i}}{\partial {\boldsymbol {C}}}}={\frac {1}{2\lambda _{i}}}~{\boldsymbol {R}}^{T}\cdot (\mathbf {n} _{i}\otimes \mathbf {n} _{i})\cdot {\boldsymbol {R}}~;~~i=1,2,3~.} The chain rule gives∂ W ∂ C = ∂ W ∂ λ 1 ∂ λ 1 ∂ C + ∂ W ∂ λ 2 ∂ λ 2 ∂ C + ∂ W ∂ λ 3 ∂ λ 3 ∂ C = R T ⋅ [ 1 2 λ 1 ∂ W ∂ λ 1 n 1 ⊗ n 1 + 1 2 λ 2 ∂ W ∂ λ 2 n 2 ⊗ n 2 + 1 2 λ 3 ∂ W ∂ λ 3 n 3 ⊗ n 3 ] ⋅ R {\displaystyle {\begin{aligned}{\frac {\partial W}{\partial {\boldsymbol {C}}}}&={\frac {\partial W}{\partial \lambda _{1}}}~{\frac {\partial \lambda _{1}}{\partial {\boldsymbol {C}}}}+{\frac {\partial W}{\partial \lambda _{2}}}~{\frac {\partial \lambda _{2}}{\partial {\boldsymbol {C}}}}+{\frac {\partial W}{\partial \lambda _{3}}}~{\frac {\partial \lambda _{3}}{\partial {\boldsymbol {C}}}}\\&={\boldsymbol {R}}^{T}\cdot \left[{\frac {1}{2\lambda _{1}}}~{\frac {\partial W}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\frac {1}{2\lambda _{2}}}~{\frac {\partial W}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\frac {1}{2\lambda _{3}}}~{\frac {\partial W}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\right]\cdot {\boldsymbol {R}}\end{aligned}}} The Cauchy stress is given byσ = 2 J F ⋅ ∂ W ∂ C ⋅ F T = 2 J ( V ⋅ R ) ⋅ ∂ W ∂ C ⋅ ( R T ⋅ V ) {\displaystyle {\boldsymbol {\sigma }}={\frac {2}{J}}~{\boldsymbol {F}}\cdot {\frac {\partial W}{\partial {\boldsymbol {C}}}}\cdot {\boldsymbol {F}}^{T}={\frac {2}{J}}~({\boldsymbol {V}}\cdot {\boldsymbol {R}})\cdot {\frac {\partial W}{\partial {\boldsymbol {C}}}}\cdot ({\boldsymbol {R}}^{T}\cdot {\boldsymbol {V}})} Plugging in the expression for the derivative ofW {\displaystyle W} leads toσ = 2 J V ⋅ [ 1 2 λ 1 ∂ W ∂ λ 1 n 1 ⊗ n 1 + 1 2 λ 2 ∂ W ∂ λ 2 n 2 ⊗ n 2 + 1 2 λ 3 ∂ W ∂ λ 3 n 3 ⊗ n 3 ] ⋅ V {\displaystyle {\boldsymbol {\sigma }}={\frac {2}{J}}~{\boldsymbol {V}}\cdot \left[{\frac {1}{2\lambda _{1}}}~{\frac {\partial W}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\frac {1}{2\lambda _{2}}}~{\frac {\partial W}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\frac {1}{2\lambda _{3}}}~{\frac {\partial W}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\right]\cdot {\boldsymbol {V}}} Using thespectral decomposition ofV {\displaystyle {\boldsymbol {V}}} we haveV ⋅ ( n i ⊗ n i ) ⋅ V = λ i 2 n i ⊗ n i ; i = 1 , 2 , 3. {\displaystyle {\boldsymbol {V}}\cdot (\mathbf {n} _{i}\otimes \mathbf {n} _{i})\cdot {\boldsymbol {V}}=\lambda _{i}^{2}~\mathbf {n} _{i}\otimes \mathbf {n} _{i}~;~~i=1,2,3.} Also note thatJ = det ( F ) = det ( V ) det ( R ) = det ( V ) = λ 1 λ 2 λ 3 . {\displaystyle J=\det({\boldsymbol {F}})=\det({\boldsymbol {V}})\det({\boldsymbol {R}})=\det({\boldsymbol {V}})=\lambda _{1}\lambda _{2}\lambda _{3}~.} Therefore, the expression for the Cauchy stress can be written asσ = 1 λ 1 λ 2 λ 3 [ λ 1 ∂ W ∂ λ 1 n 1 ⊗ n 1 + λ 2 ∂ W ∂ λ 2 n 2 ⊗ n 2 + λ 3 ∂ W ∂ λ 3 n 3 ⊗ n 3 ] {\displaystyle {\boldsymbol {\sigma }}={\frac {1}{\lambda _{1}\lambda _{2}\lambda _{3}}}~\left[\lambda _{1}~{\frac {\partial W}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+\lambda _{2}~{\frac {\partial W}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\lambda _{3}~{\frac {\partial W}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\right]} For an incompressible materialλ 1 λ 2 λ 3 = 1 {\displaystyle \lambda _{1}\lambda _{2}\lambda _{3}=1} and henceW = W ( λ 1 , λ 2 ) {\displaystyle W=W(\lambda _{1},\lambda _{2})} . Following Ogden[ 1] p. 485, we may writeσ = λ 1 ∂ W ∂ λ 1 n 1 ⊗ n 1 + λ 2 ∂ W ∂ λ 2 n 2 ⊗ n 2 + λ 3 ∂ W ∂ λ 3 n 3 ⊗ n 3 − p 1 {\displaystyle {\boldsymbol {\sigma }}=\lambda _{1}~{\frac {\partial W}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+\lambda _{2}~{\frac {\partial W}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\lambda _{3}~{\frac {\partial W}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}-p~{\boldsymbol {\mathit {1}}}~} Some care is required at this stage because, when an eigenvalue is repeated, it is in general onlyGateaux differentiable , but notFréchet differentiable .[ 8] [ 9] A rigoroustensor derivative can only be found by solving another eigenvalue problem.
If we express the stress in terms of differences between components,σ 11 − σ 33 = λ 1 ∂ W ∂ λ 1 − λ 3 ∂ W ∂ λ 3 ; σ 22 − σ 33 = λ 2 ∂ W ∂ λ 2 − λ 3 ∂ W ∂ λ 3 {\displaystyle \sigma _{11}-\sigma _{33}=\lambda _{1}~{\frac {\partial W}{\partial \lambda _{1}}}-\lambda _{3}~{\frac {\partial W}{\partial \lambda _{3}}}~;~~\sigma _{22}-\sigma _{33}=\lambda _{2}~{\frac {\partial W}{\partial \lambda _{2}}}-\lambda _{3}~{\frac {\partial W}{\partial \lambda _{3}}}} If in addition to incompressibility we haveλ 1 = λ 2 {\displaystyle \lambda _{1}=\lambda _{2}} then a possible solution to the problemrequiresσ 11 = σ 22 {\displaystyle \sigma _{11}=\sigma _{22}} and we can write the stress differences asσ 11 − σ 33 = σ 22 − σ 33 = λ 1 ∂ W ∂ λ 1 − λ 3 ∂ W ∂ λ 3 {\displaystyle \sigma _{11}-\sigma _{33}=\sigma _{22}-\sigma _{33}=\lambda _{1}~{\frac {\partial W}{\partial \lambda _{1}}}-\lambda _{3}~{\frac {\partial W}{\partial \lambda _{3}}}}
Incompressible isotropic hyperelastic materials [ edit ] For incompressibleisotropic hyperelastic materials, thestrain energy density function isW ( F ) = W ^ ( I 1 , I 2 ) {\displaystyle W({\boldsymbol {F}})={\hat {W}}(I_{1},I_{2})} . The Cauchy stress is then given byσ = − p 1 + 2 [ ( ∂ W ^ ∂ I 1 + I 1 ∂ W ^ ∂ I 2 ) B − ∂ W ^ ∂ I 2 B ⋅ B ] = − p 1 + 2 [ ( ∂ W ∂ I ¯ 1 + I 1 ∂ W ∂ I ¯ 2 ) B ¯ − ∂ W ∂ I ¯ 2 B ¯ ⋅ B ¯ ] = − p 1 + λ 1 ∂ W ∂ λ 1 n 1 ⊗ n 1 + λ 2 ∂ W ∂ λ 2 n 2 ⊗ n 2 + λ 3 ∂ W ∂ λ 3 n 3 ⊗ n 3 {\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}&=-p~{\boldsymbol {\mathit {1}}}+2\left[\left({\frac {\partial {\hat {W}}}{\partial I_{1}}}+I_{1}~{\frac {\partial {\hat {W}}}{\partial I_{2}}}\right){\boldsymbol {B}}-{\frac {\partial {\hat {W}}}{\partial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]\\&=-p~{\boldsymbol {\mathit {1}}}+2\left[\left({\frac {\partial W}{\partial {\bar {I}}_{1}}}+I_{1}~{\frac {\partial W}{\partial {\bar {I}}_{2}}}\right)~{\bar {\boldsymbol {B}}}-{\frac {\partial W}{\partial {\bar {I}}_{2}}}~{\bar {\boldsymbol {B}}}\cdot {\bar {\boldsymbol {B}}}\right]\\&=-p~{\boldsymbol {\mathit {1}}}+\lambda _{1}~{\frac {\partial W}{\partial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+\lambda _{2}~{\frac {\partial W}{\partial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\lambda _{3}~{\frac {\partial W}{\partial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\end{aligned}}} wherep {\displaystyle p} is an undetermined pressure. In terms of stress differencesσ 11 − σ 33 = λ 1 ∂ W ∂ λ 1 − λ 3 ∂ W ∂ λ 3 ; σ 22 − σ 33 = λ 2 ∂ W ∂ λ 2 − λ 3 ∂ W ∂ λ 3 {\displaystyle \sigma _{11}-\sigma _{33}=\lambda _{1}~{\frac {\partial W}{\partial \lambda _{1}}}-\lambda _{3}~{\frac {\partial W}{\partial \lambda _{3}}}~;~~\sigma _{22}-\sigma _{33}=\lambda _{2}~{\frac {\partial W}{\partial \lambda _{2}}}-\lambda _{3}~{\frac {\partial W}{\partial \lambda _{3}}}} If in additionI 1 = I 2 {\displaystyle I_{1}=I_{2}} , thenσ = 2 ∂ W ∂ I 1 B − p 1 . {\displaystyle {\boldsymbol {\sigma }}=2{\frac {\partial W}{\partial I_{1}}}~{\boldsymbol {B}}-p~{\boldsymbol {\mathit {1}}}~.} Ifλ 1 = λ 2 {\displaystyle \lambda _{1}=\lambda _{2}} , thenσ 11 − σ 33 = σ 22 − σ 33 = λ 1 ∂ W ∂ λ 1 − λ 3 ∂ W ∂ λ 3 {\displaystyle \sigma _{11}-\sigma _{33}=\sigma _{22}-\sigma _{33}=\lambda _{1}~{\frac {\partial W}{\partial \lambda _{1}}}-\lambda _{3}~{\frac {\partial W}{\partial \lambda _{3}}}}
Consistency with linear elasticity [ edit ] Consistency with linear elasticity is often used to determine some of the parameters of hyperelastic material models. These consistency conditions can be found by comparingHooke's law with linearized hyperelasticity at small strains.
Consistency conditions for isotropic hyperelastic models [ edit ] For isotropic hyperelastic materials to be consistent with isotropiclinear elasticity , the stress–strain relation should have the following form in theinfinitesimal strain limit:σ = λ t r ( ε ) 1 + 2 μ ε {\displaystyle {\boldsymbol {\sigma }}=\lambda ~\mathrm {tr} ({\boldsymbol {\varepsilon }})~{\boldsymbol {\mathit {1}}}+2\mu {\boldsymbol {\varepsilon }}} whereλ , μ {\displaystyle \lambda ,\mu } are theLamé constants . The strain energy density function that corresponds to the above relation is[ 1] W = 1 2 λ [ t r ( ε ) ] 2 + μ t r ( ε 2 ) {\displaystyle W={\tfrac {1}{2}}\lambda ~[\mathrm {tr} ({\boldsymbol {\varepsilon }})]^{2}+\mu ~\mathrm {tr} {\mathord {\left({\boldsymbol {\varepsilon }}^{2}\right)}}} For an incompressible materialt r ( ε ) = 0 {\displaystyle \mathrm {tr} ({\boldsymbol {\varepsilon }})=0} and we haveW = μ t r ( ε 2 ) {\displaystyle W=\mu ~\mathrm {tr} {\mathord {\left({\boldsymbol {\varepsilon }}^{2}\right)}}} For any strain energy density functionW ( λ 1 , λ 2 , λ 3 ) {\displaystyle W(\lambda _{1},\lambda _{2},\lambda _{3})} to reduce to the above forms for small strains the following conditions have to be met[ 1] W ( 1 , 1 , 1 ) = 0 ; ∂ W ∂ λ i ( 1 , 1 , 1 ) = 0 ∂ 2 W ∂ λ i ∂ λ j ( 1 , 1 , 1 ) = λ + 2 μ δ i j {\displaystyle {\begin{aligned}&W(1,1,1)=0~;~~{\frac {\partial W}{\partial \lambda _{i}}}(1,1,1)=0\\&{\frac {\partial ^{2}W}{\partial \lambda _{i}\partial \lambda _{j}}}(1,1,1)=\lambda +2\mu \delta _{ij}\end{aligned}}}
If the material isincompressible, then the above conditions may be expressed in the following form.W ( 1 , 1 , 1 ) = 0 ∂ W ∂ λ i ( 1 , 1 , 1 ) = ∂ W ∂ λ j ( 1 , 1 , 1 ) ; ∂ 2 W ∂ λ i 2 ( 1 , 1 , 1 ) = ∂ 2 W ∂ λ j 2 ( 1 , 1 , 1 ) ∂ 2 W ∂ λ i ∂ λ j ( 1 , 1 , 1 ) = i n d e p e n d e n t o f i , j ≠ i ∂ 2 W ∂ λ i 2 ( 1 , 1 , 1 ) − ∂ 2 W ∂ λ i ∂ λ j ( 1 , 1 , 1 ) + ∂ W ∂ λ i ( 1 , 1 , 1 ) = 2 μ ( i ≠ j ) {\displaystyle {\begin{aligned}&W(1,1,1)=0\\&{\frac {\partial W}{\partial \lambda _{i}}}(1,1,1)={\frac {\partial W}{\partial \lambda _{j}}}(1,1,1)~;~~{\frac {\partial ^{2}W}{\partial \lambda _{i}^{2}}}(1,1,1)={\frac {\partial ^{2}W}{\partial \lambda _{j}^{2}}}(1,1,1)\\&{\frac {\partial ^{2}W}{\partial \lambda _{i}\partial \lambda _{j}}}(1,1,1)=\mathrm {independentof} ~i,j\neq i\\&{\frac {\partial ^{2}W}{\partial \lambda _{i}^{2}}}(1,1,1)-{\frac {\partial ^{2}W}{\partial \lambda _{i}\partial \lambda _{j}}}(1,1,1)+{\frac {\partial W}{\partial \lambda _{i}}}(1,1,1)=2\mu ~~(i\neq j)\end{aligned}}} These conditions can be used to find relations between the parameters of a given hyperelastic model and shear and bulk moduli.
Consistency conditions for incompressibleI 1 based rubber materials [ edit ] Many elastomers are modeled adequately by a strain energy density function that depends only onI 1 {\displaystyle I_{1}} . For such materials we haveW = W ( I 1 ) {\displaystyle W=W(I_{1})} .The consistency conditions for incompressible materials forI 1 = 3 , λ i = λ j = 1 {\displaystyle I_{1}=3,\lambda _{i}=\lambda _{j}=1} may then be expressed asW ( I 1 ) | I 1 = 3 = 0 and ∂ W ∂ I 1 | I 1 = 3 = μ 2 . {\displaystyle \left.W(I_{1})\right|_{I_{1}=3}=0\quad {\text{and}}\quad \left.{\frac {\partial W}{\partial I_{1}}}\right|_{I_{1}=3}={\frac {\mu }{2}}\,.} The second consistency condition above can be derived by noting that∂ W ∂ λ i = ∂ W ∂ I 1 ∂ I 1 ∂ λ i = 2 λ i ∂ W ∂ I 1 and ∂ 2 W ∂ λ i ∂ λ j = 2 δ i j ∂ W ∂ I 1 + 4 λ i λ j ∂ 2 W ∂ I 1 2 . {\displaystyle {\frac {\partial W}{\partial \lambda _{i}}}={\frac {\partial W}{\partial I_{1}}}{\frac {\partial I_{1}}{\partial \lambda _{i}}}=2\lambda _{i}{\frac {\partial W}{\partial I_{1}}}\quad {\text{and}}\quad {\frac {\partial ^{2}W}{\partial \lambda _{i}\partial \lambda _{j}}}=2\delta _{ij}{\frac {\partial W}{\partial I_{1}}}+4\lambda _{i}\lambda _{j}{\frac {\partial ^{2}W}{\partial I_{1}^{2}}}\,.} These relations can then be substituted into the consistency condition for isotropic incompressible hyperelastic materials.
^a b c d e R.W. Ogden, 1984,Non-Linear Elastic Deformations ,ISBN 0-486-69648-0 , Dover. ^ Muhr, A. H. (2005). "Modeling the stress–strain behavior of rubber".Rubber Chemistry and Technology .78 (3):391– 425.doi :10.5254/1.3547890 . ^ Gao, H; Ma, X; Qi, N; Berry, C; Griffith, BE; Luo, X (2014)."A finite strain nonlinear human mitral valve model with fluid-structure interaction" .Int J Numer Methods Biomed Eng .30 (12):1597– 613.doi :10.1002/cnm.2691 .PMC 4278556 .PMID 25319496 . ^ Jia, F; Ben Amar, M; Billoud, B; Charrier, B (2017)."Morphoelasticity in the development of brown algaEctocarpus siliculosus : from cell rounding to branching" .J R Soc Interface .14 (127) 20160596.doi :10.1098/rsif.2016.0596 .PMC 5332559 .PMID 28228537 . ^ Arruda, E.M.; Boyce, M.C. (1993)."A three-dimensional model for the large stretch behavior of rubber elastic materials" (PDF) .J. Mech. Phys. Solids .41 :389– 412.doi :10.1016/0022-5096(93)90013-6 .S2CID 136924401 . ^ Buche, M.R.; Silberstein, M.N. (2020). "Statistical mechanical constitutive theory of polymer networks: The inextricable links between distribution, behavior, and ensemble".Phys. Rev. E .102 (1) 012501.arXiv :2004.07874 .Bibcode :2020PhRvE.102a2501B .doi :10.1103/PhysRevE.102.012501 .PMID 32794915 .S2CID 215814600 . ^ Y. Basar, 2000, Nonlinear continuum mechanics of solids, Springer, p. 157. ^ Fox & Kapoor,Rates of change of eigenvalues and eigenvectors ,AIAA Journal , 6 (12) 2426–2429 (1968) ^ Friswell MI.The derivatives of repeated eigenvalues and their associated eigenvectors. Journal of Vibration and Acoustics (ASME) 1996; 118:390–397.