Modelo constitutivo de un material idealmente elástico
Curvas de tensión-deformación para varios modelos de materiales hiperelásticos. Un material hiperelástico o elástico de Green [1] es un tipo de modelo constitutivo de material idealmente elástico para el cual la relación tensión-deformación se deriva de una función de densidad de energía de deformación . El material hiperelástico es un caso especial de un material elástico de Cauchy .
Para muchos materiales, los modelos elásticos lineales no describen con precisión el comportamiento observado del material. El ejemplo más común de este tipo de material es el caucho, cuya relación tensión - deformación puede definirse como no linealmente elástica, isotrópica e incompresible . La hiperelasticidad proporciona un medio para modelar el comportamiento tensión-deformación de tales materiales. [2] El comportamiento de los elastómeros vulcanizados sin relleno a menudo se ajusta estrechamente al ideal hiperelástico. Los elastómeros rellenos y los tejidos biológicos [3] [4] también se modelan a menudo mediante la idealización hiperelástica. Además de usarse para modelar materiales físicos, los materiales hiperelásticos también se usan como medios ficticios, por ejemplo, en el tercer método de contacto del medio .
Ronald Rivlin y Melvin Mooney desarrollaron los primeros modelos hiperelásticos, los sólidos neo-hookeanos y de Mooney-Rivlin . Desde entonces se han desarrollado muchos otros modelos hiperelásticos. Otros modelos de materiales hiperelásticos ampliamente utilizados incluyen el modelo de Ogden y el modelo de Arruda-Boyce .
Modelos de materiales hiperelásticos
Modelo de Saint Venant-Kirchhoff El modelo de material hiperelástico más simple es el modelo de Saint Venant–Kirchhoff, que es simplemente una extensión del modelo de material elástico geométricamente lineal al régimen geométricamente no lineal. Este modelo tiene la forma general y la forma isotrópica respectivamente
donde es la contracción del tensor, es la segunda tensión de Piola–Kirchhoff, es un tensor de rigidez de cuarto orden y es la deformación de Green lagrangiana dada por y son las constantes de Lamé , y es el tensor unitario de segundo orden. S = 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}}} : {\displaystyle \mathbin {:} } S {\displaystyle {\boldsymbol {S}}} C : R 3 × 3 → R 3 × 3 {\displaystyle {\boldsymbol {C}}:\mathbb {R} ^{3\times 3}\to \mathbb {R} ^{3\times 3}} E {\displaystyle {\boldsymbol {E}}} E = 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 } μ {\displaystyle \mu } I {\displaystyle {\boldsymbol {\mathit {I}}}}
La función de densidad de energía de tensión para el modelo de Saint Venant-Kirchhoff es W ( 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)}}}
y la segunda tensión de Piola-Kirchhoff se puede derivar de la relación S = ∂ W ∂ E . {\displaystyle {\boldsymbol {S}}={\frac {\partial W}{\partial {\boldsymbol {E}}}}~.}
Clasificación de modelos de materiales hiperelásticos Los modelos de materiales hiperelásticos se pueden clasificar como:
descripciones fenomenológicas del comportamiento observadomodelos mecanicistas derivados de argumentos sobre la estructura subyacente del materialHíbridos de modelos fenomenológicos y mecanicistas En general, un modelo hiperelástico debería satisfacer el criterio de estabilidad de Drucker . Algunos modelos hiperelásticos satisfacen la hipótesis de Valanis-Landel, que establece que la función de energía de deformación puede descomponerse en la suma de funciones independientes de los estiramientos principales : ( 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})\,.}
Relaciones tensión-deformación
Materiales hiperelásticos compresibles
Si es la función de densidad de energía de deformación, el primer tensor de tensión de Piola-Kirchhoff se puede calcular para un material hiperelástico como
donde es el gradiente de deformación . En términos de la deformación de Green lagrangiana ( )
En términos del tensor de deformación de Cauchy-Green derecho ( ) W ( F ) {\displaystyle W({\boldsymbol {F}})} P = ∂ 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}}}.} F {\displaystyle {\boldsymbol {F}}} 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}}}~.} 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}}}~.}
Si es el segundo tensor de tensión de Piola-Kirchhoff entonces
En términos de la deformación de Green lagrangiana
En términos del tensor de deformación de Cauchy-Green derecho
La relación anterior también se conoce como fórmula de Doyle-Ericksen en la configuración del material. S {\displaystyle {\boldsymbol {S}}} S = 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}}}~.} 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}}}~.} 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}}}~.}
Estrés de Cauchy De manera similar, la tensión de Cauchy se da por
En términos de la deformación de Green lagrangiana
En términos del tensor de deformación de Cauchy-Green derecho
Las expresiones anteriores son válidas incluso para medios anisotrópicos (en cuyo caso, se entiende que la función potencial depende implícitamente de cantidades direccionales de referencia como las orientaciones iniciales de las fibras). En el caso especial de isotropía, la tensión de Cauchy se puede expresar en términos del tensor de deformación de Cauchy-Green izquierdo de la siguiente manera: [7] σ = 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}~.} 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}~.} 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}~.} 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}}}~.}
Materiales hiperelásticos incompresibles Para un material incompresible . La restricción de incompresibilidad es, por lo tanto , . Para garantizar la incompresibilidad de un material hiperelástico, la función de deformación-energía se puede escribir en la forma:
donde la presión hidrostática funciona como un multiplicador de Lagrange para hacer cumplir la restricción de incompresibilidad. La primera tensión de Piola-Kirchhoff ahora se convierte en Este tensor de tensión se puede convertir
posteriormente en cualquiera de los otros tensores de tensión convencionales, como el tensor de tensión de Cauchy que se da por J := det F = 1 {\displaystyle J:=\det {\boldsymbol {F}}=1} J − 1 = 0 {\displaystyle J-1=0} W = W ( F ) − p ( J − 1 ) {\displaystyle W=W({\boldsymbol {F}})-p~(J-1)} p {\displaystyle p} P = − 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}}}}~.} σ = 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}}~.}
Expresiones para el estrés de Cauchy
Materiales hiperelásticos isotrópicos compresibles Para materiales hiperelásticos isotrópicos , la tensión de Cauchy se puede expresar en términos de los invariantes del tensor de deformación de Cauchy-Green izquierdo (o tensor de deformación de Cauchy-Green derecho ). Si la función de densidad de energía de deformación es entonces
(Consulte la página sobre el tensor de deformación de Cauchy-Green izquierdo para conocer las definiciones de estos símbolos). W ( 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}),} σ = 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}}}
Prueba 1 El segundo tensor de tensión de Piola-Kirchhoff para un material hiperelástico viene dado por
donde es el tensor de deformación de Cauchy-Green derecho y es el gradiente de deformación . La tensión de Cauchy viene dada por
donde . Sean los tres invariantes principales de . Entonces
Las derivadas de los invariantes del tensor simétrico son
Por lo tanto, podemos escribir
Sustituyendo en la expresión para la tensión de Cauchy se obtiene
Usando el tensor de deformación de Cauchy-Green izquierdo y notando que , podemos escribir
Para un material incompresible y por lo tanto .
Entonces Por lo tanto, la tensión de Cauchy viene dada por
donde es una presión indeterminada que actúa como un multiplicador de Lagrange para imponer la restricción de incompresibilidad. S = 2 ∂ W ∂ C {\displaystyle {\boldsymbol {S}}=2~{\frac {\partial W}{\partial {\boldsymbol {C}}}}} C = F T ⋅ F {\displaystyle {\boldsymbol {C}}={\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}} F {\displaystyle {\boldsymbol {F}}} σ = 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}} J = det F {\displaystyle J=\det {\boldsymbol {F}}} I 1 , I 2 , I 3 {\displaystyle I_{1},I_{2},I_{3}} C {\displaystyle {\boldsymbol {C}}} ∂ 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}}}}~.} C {\displaystyle {\boldsymbol {C}}} ∂ 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}} ∂ 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}~.} σ = 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]} B = F ⋅ F T {\displaystyle {\boldsymbol {B}}={\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}} I 3 = J 2 {\displaystyle I_{3}=J^{2}} σ = 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}}}~.} I 3 = 1 {\displaystyle I_{3}=1} W = W ( I 1 , I 2 ) {\displaystyle W=W(I_{1},I_{2})} ∂ 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}})} σ = 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}}}~.} p {\displaystyle p}
Si, además , tenemos y por lo tanto
En ese caso la tensión de Cauchy se puede expresar como I 1 = I 2 {\displaystyle I_{1}=I_{2}} W = W ( I 1 ) {\displaystyle W=W(I_{1})} ∂ 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}}}} σ = 2 ∂ W ∂ I 1 B − p 1 . {\displaystyle {\boldsymbol {\sigma }}=2{\frac {\partial W}{\partial I_{1}}}~{\boldsymbol {B}}-p~{\boldsymbol {\mathit {1}}}~.}
Prueba 2 El gradiente de deformación isocórico se define como , lo que da como resultado que el gradiente de deformación isocórico tenga un determinante de 1, en otras palabras, no tenga estiramiento de volumen. Usando esto, se puede definir posteriormente el tensor de deformación de Cauchy-Green izquierdo isocórico . Los invariantes de son
El conjunto de invariantes que se usan para definir el comportamiento distorsionante son los dos primeros invariantes del tensor de deformación de Cauchy-Green izquierdo isocórico (que son idénticos a los del tensor de estiramiento de Cauchy-Green derecho), y se suman al conjunto para describir el comportamiento volumétrico. F ¯ := J − 1 / 3 F {\displaystyle {\bar {\boldsymbol {F}}}:=J^{-1/3}{\boldsymbol {F}}} B ¯ := F ¯ ⋅ F ¯ T = J − 2 / 3 B {\displaystyle {\bar {\boldsymbol {B}}}:={\bar {\boldsymbol {F}}}\cdot {\bar {\boldsymbol {F}}}^{T}=J^{-2/3}{\boldsymbol {B}}} B ¯ {\displaystyle {\bar {\boldsymbol {B}}}} I ¯ 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}}} J {\displaystyle J}
Para expresar la tensión de Cauchy en términos de los invariantes, recordemos que
La regla de la cadena de diferenciación nos da
Recordemos que la tensión de Cauchy está dada por
En términos de los invariantes tenemos
Introduciendo las expresiones para las derivadas de en términos de , tenemos
o,
En términos de la parte desviatoria de , podemos escribir
Para un material incompresible y por lo tanto . Entonces la tensión de Cauchy está dada por
donde es un término multiplicador de Lagrange indeterminado similar a la presión. Además, si , tenemos y por lo tanto la tensión de Cauchy se puede expresar como I ¯ 1 , I ¯ 2 , J {\displaystyle {\bar {I}}_{1},{\bar {I}}_{2},J} I ¯ 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}~.} ∂ 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}}} σ = 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}}}~.} I ¯ 1 , I ¯ 2 , J {\displaystyle {\bar {I}}_{1},{\bar {I}}_{2},J} σ = 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}}}~.} W {\displaystyle W} I ¯ 1 , I ¯ 2 , J {\displaystyle {\bar {I}}_{1},{\bar {I}}_{2},J} σ = 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}}} 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}}} B {\displaystyle {\boldsymbol {B}}} σ = 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}}} J = 1 {\displaystyle J=1} W = W ( I ¯ 1 , I ¯ 2 ) {\displaystyle W=W({\bar {I}}_{1},{\bar {I}}_{2})} σ = 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}}}~.} p {\displaystyle p} I ¯ 1 = I ¯ 2 {\displaystyle {\bar {I}}_{1}={\bar {I}}_{2}} W = W ( I ¯ 1 ) {\displaystyle W=W({\bar {I}}_{1})} σ = 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}}}~.}
Prueba 3 Para expresar la tensión de Cauchy en términos de los estiramientos, recuerde que
La regla de la cadena da
La tensión de Cauchy viene dada por
Sustituir la expresión para la derivada de conduce a
Utilizando la descomposición espectral de tenemos
Observe también que
Por lo tanto, la expresión para la tensión de Cauchy se puede escribir como
Para un material incompresible y, por tanto , . Siguiendo a Ogden [1] p. 485, podemos escribir
Se requiere cierto cuidado en esta etapa porque, cuando se repite un valor propio, en general solo es diferenciable de Gateaux , pero no de Fréchet . [8] [9] Una derivada tensorial rigurosa solo se puede encontrar resolviendo otro problema de valor propio. λ 1 , λ 2 , λ 3 {\displaystyle \lambda _{1},\lambda _{2},\lambda _{3}} ∂ 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~.} ∂ 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}}} σ = 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}})} W {\displaystyle W} σ = 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}}} V {\displaystyle {\boldsymbol {V}}} V ⋅ ( 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.} J = 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}~.} σ = 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]} λ 1 λ 2 λ 3 = 1 {\displaystyle \lambda _{1}\lambda _{2}\lambda _{3}=1} W = W ( λ 1 , λ 2 ) {\displaystyle W=W(\lambda _{1},\lambda _{2})} σ = 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}}}~}
Si expresamos la tensión en términos de diferencias entre componentes,
si además de la incompresibilidad tenemos entonces una posible solución al problema requiere y podemos escribir las diferencias de tensión como 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}}}} λ 1 = λ 2 {\displaystyle \lambda _{1}=\lambda _{2}} σ 11 = σ 22 {\displaystyle \sigma _{11}=\sigma _{22}} σ 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}}}}
Materiales hiperelásticos isotrópicos incompresibles Para materiales hiperelásticos isotrópicos incompresibles , la función de densidad de energía de deformación es . La tensión de Cauchy se obtiene entonces mediante
donde es una presión indeterminada. En términos de diferencias de tensión
Si además , entonces
Si , entonces W ( F ) = W ^ ( I 1 , I 2 ) {\displaystyle W({\boldsymbol {F}})={\hat {W}}(I_{1},I_{2})} σ = − 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}}} p {\displaystyle p} 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}}}} I 1 = I 2 {\displaystyle I_{1}=I_{2}} σ = 2 ∂ W ∂ I 1 B − p 1 . {\displaystyle {\boldsymbol {\sigma }}=2{\frac {\partial W}{\partial I_{1}}}~{\boldsymbol {B}}-p~{\boldsymbol {\mathit {1}}}~.} λ 1 = λ 2 {\displaystyle \lambda _{1}=\lambda _{2}} σ 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}}}}
Consistencia con elasticidad lineal La consistencia con elasticidad lineal se utiliza a menudo para determinar algunos de los parámetros de los modelos de materiales hiperelásticos. Estas condiciones de consistencia se pueden determinar comparando la ley de Hooke con la hiperelasticidad linealizada a pequeñas deformaciones.
Condiciones de consistencia para modelos hiperelásticos isotrópicos Para que los materiales hiperelásticos isotrópicos sean consistentes con la elasticidad lineal isotrópica , la relación tensión-deformación debe tener la siguiente forma en el límite de deformación infinitesimal :
donde son las constantes de Lamé . La función de densidad de energía de deformación que corresponde a la relación anterior es [1]
Para un material incompresible y tenemos
Para que cualquier función de densidad de energía de deformación se reduzca a las formas anteriores para deformaciones pequeñas, se deben cumplir las siguientes condiciones [1] σ = λ t r ( ε ) 1 + 2 μ ε {\displaystyle {\boldsymbol {\sigma }}=\lambda ~\mathrm {tr} ({\boldsymbol {\varepsilon }})~{\boldsymbol {\mathit {1}}}+2\mu {\boldsymbol {\varepsilon }}} λ , μ {\displaystyle \lambda ,\mu } 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)}}} t r ( ε ) = 0 {\displaystyle \mathrm {tr} ({\boldsymbol {\varepsilon }})=0} W = μ t r ( ε 2 ) {\displaystyle W=\mu ~\mathrm {tr} {\mathord {\left({\boldsymbol {\varepsilon }}^{2}\right)}}} W ( λ 1 , λ 2 , λ 3 ) {\displaystyle W(\lambda _{1},\lambda _{2},\lambda _{3})} 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}}}
Si el material es incompresible, las condiciones anteriores pueden expresarse de la siguiente forma.
Estas condiciones pueden utilizarse para encontrar relaciones entre los parámetros de un modelo hiperelástico dado y los módulos de esfuerzo cortante y de volumen. 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}}}
Condiciones de consistencia para materiales incompresiblesYo 1materiales a base de caucho Muchos elastómeros se modelan adecuadamente mediante una función de densidad de energía de deformación que depende solo de . Para tales materiales tenemos . Las condiciones de consistencia para materiales incompresibles para pueden entonces expresarse como
La segunda condición de consistencia anterior se puede derivar observando que
Estas relaciones pueden entonces sustituirse en la condición de consistencia para materiales hiperelásticos incompresibles isotrópicos. I 1 {\displaystyle I_{1}} W = W ( I 1 ) {\displaystyle W=W(I_{1})} I 1 = 3 , λ i = λ j = 1 {\displaystyle I_{1}=3,\lambda _{i}=\lambda _{j}=1} W ( 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}}\,.} ∂ 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}}}\,.}
Referencias ^ abcde RW Ogden, 1984, Deformaciones elásticas no lineales , ISBN 0-486-69648-0 , Dover. ^ Muhr, AH (2005). "Modelado del comportamiento de tensión-deformación del caucho". Química y tecnología del caucho . 78 (3): 391–425. doi :10.5254/1.3547890. ^ Gao, H; Ma, X; Qi, N; Berry, C; Griffith, BE; Luo, X (2014). "Un modelo de válvula mitral humana no lineal de deformación finita con interacción fluido-estructura". 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). "Morfoelasticidad en el desarrollo del alga parda Ectocarpus siliculosus: desde el redondeo celular hasta la ramificación". JR Soc Interface . 14 (127): 20160596. doi :10.1098/rsif.2016.0596. PMC 5332559 . PMID 28228537. ^ Arruda, EM; Boyce, MC (1993). "Un modelo tridimensional para el comportamiento de gran estiramiento de materiales elásticos de caucho" (PDF) . J. Mech. Phys. Solids . 41 : 389–412. doi :10.1016/0022-5096(93)90013-6. S2CID 136924401. ^ Buche, MR; Silberstein, MN (2020). "Teoría constitutiva mecánica estadística de redes de polímeros: los vínculos inextricables entre distribución, comportamiento y conjunto". 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, Mecánica continua no lineal de sólidos, Springer, pág. 157. ^ Fox y Kapoor, Tasas de cambio de valores propios y vectores propios , AIAA Journal , 6 (12) 2426–2429 (1968) ^ Friswell MI. Las derivadas de valores propios repetidos y sus vectores propios asociados. Revista de vibración y acústica (ASME) 1996; 118:390–397.
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![{\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]\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c8b59fed4fc1f45e08b5f6ae13fe5e5f2586951)
![{\displaystyle W({\boldsymbol {E}})={\frac {\lambda }{2}}[{\text{tr}}({\boldsymbol {E}})]^{2}+\mu {\text{tr}}{\mathord {\left({\boldsymbol {E}}^{2}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/de41de18a428e0920a40050db04bdf7653ef0c5f)
![{\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 {\parcial {\bar {W}}}{\parcial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\frac {\parcial {\bar {W}}}{\parcial {\bar {I}}_{2}}}\right){\boldsymbol {B}}-{\frac {1}{J^{4/3}}}~{\frac {\parcial {\bar {W}}}{\parcial {\bar {I}}_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+\left[{\frac {\parcial {\bar {W}}}{\parcial J}}-{\frac {2}{3J}}\left({\bar {I}}_{1}~{\frac {\parcial {\bar {W}}}{\parcial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\frac {\parcial {\bar {W}}}{\parcial {\bar {I}}_{2}}}\right)\right]~{\boldsymbol {\mathit {1}}}\\[5pt]&={\frac {2}{J}}\left[\left({\frac {\parcial {\bar {W}}}{\parcial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\frac {\parcial {\bar {W}}}{\parcial {\bar {I}}_{2}}}\right){\bar {\boldsymbol {B}}}-{\frac {\parcial {\bar {W}}}{\parcial {\bar {I}}_{2}}}~{\bar {\boldsymbol {B}}}\cdot {\bar {\boldsymbol {B}}}\derecha]+\izquierda[{\frac {\parcial {\bar {W}}}{\parcial J}}-{\frac {2}{3J}}\izquierda({\bar {I}}_{1}~{\frac {\parcial {\bar {W}}}{\parcial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\frac {\parcial {\bar {W}}}{\parcial {\bar {I}}_{2}}}\derecha)\derecha]~{\boldsymbol {\mathit {1}}}\\[5pt]&={\frac {\lambda _{1}}{\lambda _{1}\lambda _{2}\lambda _{3}}}~{\frac {\parcial {\tilde {W}}}{\parcial \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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc63e121bafc2f5ebc99e2cc102ff55e68618452)
![{\displaystyle {\boldsymbol {\sigma }}={\frac {2}{J}}~\left[{\frac {\parcial W}{\parcial I_{1}}}~{\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}+{\frac {\parcial W}{\parcial I_{2}}}~(I_{1}~{\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}-{\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T})+{\frac {\parcial W}{\parcial I_{3}}}~I_{3}~{\boldsymbol {\mathit {1}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2fceb8e2cf5d2f77071ed86209e41e3b391829c7)
![{\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}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c5e6b445b86b2c353a1f18494fc0b3d6cee61325)
![{\displaystyle {\boldsymbol {\sigma }}=2\left[\left({\frac {\parcial W}{\parcial I_{1}}}+I_{1}~{\frac {\parcial W}{\parcial I_{2}}}\right)~{\boldsymbol {B}}-{\frac {\parcial W}{\parcial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]-p~{\boldsymbol {\mathit {1}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/29f1bd839ed56c70acb2202948b7027a97467764)
![{\displaystyle {\boldsymbol {\sigma }}={\frac {2}{J}}~\left[\left({\frac {\parcial W}{\parcial I_{1}}}+J^{2/3}~{\bar {I}}_{1}~{\frac {\parcial W}{\parcial I_{2}}}\right)~{\boldsymbol {B}}-{\frac {\parcial W}{\parcial I_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]+2~J~{\frac {\parcial W}{\parcial I_{3}}}~{\boldsymbol {\mathit {1}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4976e9a59607a67563a2168f67cac160453c8a59)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d92fc8cc0a376769e1a55384831822bfe7278570)
![{\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}&={\frac {2}{J}}~\left[{\frac {1}{J^{2/3}}}~\left({\frac {\parcial W}{\parcial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\frac {\parcial W}{\parcial {\bar {I}}_{2}}}\right)~{\boldsymbol {B}}-{\frac {1}{J^{4/3}}}~{\frac {\parcial W}{\parcial {\bar {I}}_{2}}}~{\boldsymbol {B}}\cdot {\boldsymbol {B}}\right]\\&\qquad +\left[{\frac {\parcial W}{\parcial J}}-{\frac {2}{3J}}\left({\bar {I}}_{1}~{\frac {\parcial W}{\parcial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\frac {\parcial W}{\parcial {\bar {I}}_{2}}}\right)\right]{\boldsymbol {\mathit {1}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/33fb2175d1cf8867bc4aeb7f3991b4d237cf1543)
![{\displaystyle {\begin{aligned}{\boldsymbol {\sigma }}&={\frac {2}{J}}~\left[\left({\frac {\parcial W}{\parcial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\frac {\parcial W}{\parcial {\bar {I}}_{2}}}\right)~{\bar {\boldsymbol {B}}}-{\frac {\parcial W}{\parcial {\bar {I}}_{2}}}~{\bar {\boldsymbol {B}}}\cdot {\bar {\boldsymbol {B}}}\right]\\&\qquad +\left[{\frac {\parcial W}{\parcial J}}-{\frac {2}{3J}}\left({\bar {I}}_{1}~{\frac {\parcial W}{\parcial {\bar {I}}_{1}}}+2~{\bar {I}}_{2}~{\frac {\partial W}{\partial {\bar {I}}_{2}}}\right)\right]{\boldsymbol {\mathit {1}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6bab10de5208332a92b44c2e19af8c19a8dca207)
![{\displaystyle {\boldsymbol {\sigma }}=2\left[\left({\frac {\parcial W}{\parcial {\bar {I}}_{1}}}+I_{1}~{\frac {\parcial W}{\parcial {\bar {I}}_{2}}}\right)~{\bar {\boldsymbol {B}}}-{\frac {\parcial W}{\parcial {\bar {I}}_{2}}}~{\bar {\boldsymbol {B}}}\cdot {\bar {\boldsymbol {B}}}\right]-p~{\boldsymbol {\mathit {1}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e5f43a4b8e83bf20b8c88b401395570654aaa89)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/376db081f7668da0294fb408efd68f30e246fba8)
![{\displaystyle {\boldsymbol {\sigma }}={\frac {2}{J}}~{\boldsymbol {V}}\cdot \left[{\frac {1}{2\lambda _{1}}}~{\frac {\parcial W}{\parcial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+{\frac {1}{2\lambda _{2}}}~{\frac {\parcial W}{\parcial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+{\frac {1}{2\lambda _{3}}}~{\frac {\parcial W}{\parcial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\derecha]\cdot {\boldsymbol {V}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c1cbf72e429aed44eb8dfa08ee2108ef209c94b)
![{\displaystyle {\boldsymbol {\sigma }}={\frac {1}{\lambda _{1}\lambda _{2}\lambda _{3}}}~\left[\lambda _{1}~{\frac {\parcial W}{\parcial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+\lambda _{2}~{\frac {\parcial W}{\parcial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\lambda _{3}~{\frac {\parcial W}{\parcial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32b4bfca4342f7dc22085470c19916ece1ea702b)
![{\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 {\parcial W}{\parcial {\bar {I}}_{2}}}~{\bar {\boldsymbol {B}}}\cdot {\bar {\boldsymbol {B}}}\right]\\&=-p~{\boldsymbol {\mathit {1}}}+\lambda _{1}~{\frac {\parcial W}{\parcial \lambda _{1}}}~\mathbf {n} _{1}\otimes \mathbf {n} _{1}+\lambda _{2}~{\frac {\parcial W}{\parcial \lambda _{2}}}~\mathbf {n} _{2}\otimes \mathbf {n} _{2}+\lambda _{3}~{\frac {\parcial W}{\parcial \lambda _{3}}}~\mathbf {n} _{3}\otimes \mathbf {n} _{3}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ec076b6ccffceae7e2ba84414d011247098baac)
![{\displaystyle W={\tfrac {1}{2}}\lambda ~[\mathrm {tr} ({\boldsymbol {\varepsilon }})]^{2}+\mu ~\mathrm {tr} {\mathord {\left({\boldsymbol {\varepsilon }}^{2}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6bcd60974de818734aa8e47cea766e49c221fc7e)
