Modelo constitutivo de material idealmente elástico.
Curvas tensión-deformación para varios modelos de materiales hiperelásticos. Un material hiperelástico o elástico verde [1] es un tipo de modelo constitutivo para 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 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 se puede definir como no linealmente elástica, isotrópica e incompresible . La hiperelasticidad proporciona un medio para modelar el comportamiento tensión-deformación de dichos 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 suelen modelarse mediante la idealización hiperelástica. Además de utilizarse para modelar materiales físicos, los materiales hiperelásticos también se utilizan como medios ficticios, por ejemplo, en el método de contacto del tercer medio .
Ronald Rivlin y Melvin Mooney desarrollaron los primeros modelos hiperelásticos, los sólidos Neo-Hookean y 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 solo 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 Lagrangiano de Green dada por y son las constantes de Lamé , y es la unidad de segundo orden. tensor. 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 deformación 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 el segundo estrés 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 material.híbridos de modelos fenomenológicos y mecanicistas Generalmente, 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 se puede separar en la suma de funciones separadas de los tramos 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 está el gradiente de deformación . En términos de la deformación de Lagrange Green ( )
En términos del tensor de deformación derecho de Cauchy-Green ( ) 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 lagrangiana de Green
En términos del tensor de deformación derecho de Cauchy-Green
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 está dada por
En términos de la deformación de Lagrange Green
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 la tensión direccional de referencia cantidades como las orientaciones iniciales de las fibras). En el caso especial de la 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 . Por tanto, la restricción de incompresibilidad es . Para garantizar la incompresibilidad de un material hiperelástico, la función de energía de deformación se puede escribir en la forma:
donde la presión hidrostática funciona como un multiplicador lagrangiano para imponer 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 viene dado 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 las invariantes del tensor de deformación de Cauchy-Green izquierdo (o del tensor de deformación de Cauchy-Green derecho ). Si la función de densidad de energía de deformación es entonces
(consulte la página de la izquierda Tensor de deformación de Cauchy-Green 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
dónde está el tensor de deformación de Cauchy-Green derecho y es el gradiente de deformación . La tensión de Cauchy viene dada por
dónde . Sean los tres invariantes principales de . Entonces ,
las derivadas de las invariantes del tensor simétrico son
Por lo tanto, podemos escribir
Sustituyendo la expresión para la tensión de Cauchy se obtiene
Usando el tensor de deformación de Cauchy-Green izquierdo y observando que , podemos escribir
Para un material incompresible y por lo tanto .
Entonces Por lo tanto, la tensión de Cauchy viene dada por
donde hay 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 el estrés 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 isocórica se define como , lo que da como resultado que el gradiente de deformación isocórica tenga un determinante de 1; en otras palabras, no tiene estiramiento de volumen. Usando esto se puede definir posteriormente el tensor de deformación isocórico izquierdo de Cauchy-Green . Los invariantes de son
El conjunto de invariantes que se utilizan para definir el comportamiento distorcional son los dos primeros invariantes del tensor isocórico de deformación de Cauchy-Green izquierdo (que son idénticos a los del tensor de estiramiento derecho de Cauchy Green) y suman en la refriega 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 invariantes, recuerde que
La regla de la cadena de diferenciación nos da
Recuerde que la tensión de Cauchy está dada por
En términos de las invariantes tenemos
Al reemplazar 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 viene dada por
donde es un término multiplicador de Lagrange similar a una presión indeterminada. Además, si , tenemos y por tanto el estrés 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 al
Sustituir la expresión para la derivada de conduce a
Usando la descomposición espectral de tenemos
También tenga en cuenta 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 diferenciable de Fréchet . [8] [9] Una derivada tensorial rigurosa sólo se puede encontrar resolviendo otro problema de valores propios. 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 , 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 viene dada entonces por
donde hay una presión indeterminada. En términos de diferencias de estrés
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 la elasticidad lineal. La coherencia con la elasticidad lineal se utiliza a menudo para determinar algunos de los parámetros de los modelos de materiales hiperelásticos. Estas condiciones de consistencia se pueden encontrar comparando la ley de Hooke con la hiperelasticidad linealizada en deformaciones pequeñas.
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 :
¿dónde están 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, entonces las condiciones anteriores pueden expresarse de la siguiente forma.
Estas condiciones se pueden utilizar para encontrar relaciones entre los parámetros de un modelo hiperelástico dado y los módulos de corte y 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 incompresibles.yo 1materiales a base de caucho Muchos elastómeros se modelan adecuadamente mediante una función de densidad de energía de deformación que depende únicamente de . Para tales materiales tenemos . Las condiciones de consistencia para materiales incompresibles se pueden expresar como
La segunda condición de consistencia anterior se puede derivar observando que
estas relaciones se pueden sustituir en la condición de consistencia para materiales hiperelásticos isotrópicos incompresibles. 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 tensión-deformación del caucho". Química y Tecnología del Caucho . 78 (3): 391–425. doi : 10.5254/1.3547890. ^ Gao, H; Mamá, X; Qi, N; Baya, C; Griffith, BE; Luo, X (2014). "Un modelo de válvula mitral humana no lineal de tensión finita con interacción fluido-estructura". Int J Métodos numéricos 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: del redondeo celular a la ramificación". Interfaz JR Soc . 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. Física. Sólidos . 41 : 389–412. doi :10.1016/0022-5096(93)90013-6. S2CID 136924401. ^ Buche, señor; Silberstein, Minnesota (2020). "Teoría constitutiva mecánica estadística de redes de polímeros: los vínculos inextricables entre distribución, comportamiento y conjunto". Física. Rev. E. 102 (1): 012501. arXiv : 2004.07874 . Código Bib : 2020PhRvE.102a2501B. doi : 10.1103/PhysRevE.102.012501. PMID 32794915. S2CID 215814600. ^ Y. Basar, 2000, Mecánica continua no lineal de sólidos, Springer, p. 157. ^ Fox & 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.
Ver también
![{\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 {\ sombrero {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 {\ barra {I}}_{2}}}\right)\right]~{\boldsymbol {\mathit {1}}}\\[5pt]&={\frac {2}{J}}\left[\ izquierda({\frac {\partial {\bar {W}}}{\partial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\frac {\ parcial {\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]+\ izquierda[{\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 {\parcial {\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 {\partial W}{\partial I_{1}}}~{\boldsymbol {F} }\cdot {\boldsymbol {F}}^{T}+{\frac {\partial W}{\partial I_{2}}}~(I_{1}~{\boldsymbol {F}}\cdot {\ negritasymbol {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]}](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 {\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 {\ símbolo en negrita {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 {\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}}}~.}](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 {\ W parcial}{\parcial {\bar {I}}_{1}}}+J^{-2/3}~{\bar {I}}_{1}~{\frac {\W parcial}{ \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 {\partial W}{\partial {\bar {I}}_{1}}}+{\bar {I}}_{1}~{\frac {\partial W}{\ parcial {\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}}}\ derecha)\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 {\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}}\ izquierda({\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}} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6bab10de5208332a92b44c2e19af8c19a8dca207)
![{\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}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e5f43a4b8e83bf20b8c88b401395570654aaa89)
![{\displaystyle {\begin{alineado}{\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 {\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}}}](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 {\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]}](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 {\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}}}](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)
