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.
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.
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}}} tensor de rigidez : {\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 } constantes de Lamé 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. Primera tensión de Piola-Kirchhoff 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 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}}}.} gradiente de deformación cepa Verde Lagrangiana 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}}}~.} tensor de deformación derecho de Cauchy-Green 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}}}~.} Segundo estrés de Piola-Kirchhoff Si es el segundo tensor de tensiones de Piola-Kirchhoff, entonces 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}}}~.} cepa Lagrangian Green 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}}}~.} tensor de deformación derecho de Cauchy-Green 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}}}~.} fórmula de Doyle-Ericksen estrés de cauchy De manera similar, el estrés de Cauchy viene dado por
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}~.} cepa Lagrangian Green 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}~.} tensor de deformación derecho de Cauchy-Green 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}~.} implícitamente izquierdo [7] σ = 2 J ∂ W ∂ B ⋅ B or σ i j = 2 J B i k ∂ W ∂ B k j . {\displaystyle {\boldsymbol {\sigma }}={\frac {2}{J}}{\frac {\partial W}{\partial {\boldsymbol {B}}}}\cdot ~{\boldsymbol {B}}\qquad {\text{or}}\qquad \sigma _{ij}={\frac {2}{J}}~B_{ik}~{\frac {\partial W}{\partial B_{kj}}}~.} Materiales hiperelásticos incompresibles. Para un material incompresible . Por tanto, la restricción de incompresibilidad es . Para asegurar la incompresibilidad de un material hiperelástico, la función de energía de deformación se puede escribir en la forma: 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)} multiplicador lagrangiano 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}}}}~.} convertir tensor de tensión de Cauchy 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
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}}} la izquierda Tensor de deformación de Cauchy-Green Prueba 1 El segundo tensor de tensión de Piola-Kirchhoff para un material hiperelástico viene dado por
S = 2 ∂ W ∂ C {\displaystyle {\boldsymbol {S}}=2~{\frac {\partial W}{\partial {\boldsymbol {C}}}}} donde es el
tensor de deformación de Cauchy-Green derecho y es el
gradiente de deformación . La
tensión de Cauchy está dada por
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}} dónde . Sean los tres invariantes principales de . Entonces
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}}}}~.} Las
derivadas de las invariantes del tensor simétrico son
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}} Por lo tanto, podemos escribir
∂ 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}~.} Al conectar la expresión para el estrés de Cauchy se obtiene
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]} Usando el
tensor de deformación de Cauchy-Green izquierdo y observando que , podemos escribir
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}}}~.} Para un material
incompresible y por tanto . Entonces
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}})} Por lo tanto, el estrés de Cauchy viene dado por
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}}}~.} donde es una presión indeterminada que actúa como
multiplicador de Lagrange para imponer la restricción de incompresibilidad.
p {\displaystyle p} Si, además, tenemos y por tanto 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}}}} En ese caso, la tensión de Cauchy se puede expresar como
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 . Las invariantes de son 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}}} 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 se suman a la refriega para describir el comportamiento volumétrico.
J {\displaystyle J} Para expresar el estrés de Cauchy en términos de invariantes, recuerde que 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}~.} La regla de la cadena de diferenciación nos da
∂ 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}}} Recuerde que el estrés de Cauchy está dado por
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}}}~.} En términos de las invariantes tenemos
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}}}~.} Sustituyendo las expresiones para las derivadas de en términos de , tenemos
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}}} o,
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}}} En términos de la parte desviatoria de , podemos escribir
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}}} Para un material
incompresible y por tanto . Entonces la tensión de Cauchy viene dada por
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}}}~.} donde es un término multiplicador de Lagrange similar a la presión indeterminada. Además, si , tenemos y por tanto el estrés de Cauchy se puede expresar como
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 estiramientos , recuerde que 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~.} La regla de la cadena da
∂ 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}}} La tensión de Cauchy está dada por
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}})} Al introducir la expresión para la derivada de se obtiene
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}}} Usando la
descomposición espectral de tenemos
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.} También tenga en cuenta que
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}~.} Por lo tanto, la expresión para el estrés de Cauchy se puede escribir como
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]} Para un material
incompresible y por tanto . Siguiendo a Ogden
[1] p. 485, podemos escribir
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}}}~} 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.
Si expresamos la tensión en términos de diferencias entre componentes,
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}}}} Si además de la incompresibilidad tenemos entonces una posible solución al problema requiere y podemos escribir las diferencias de tensión como
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 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 :
σ = λ t r ( ε ) 1 + 2 μ ε {\displaystyle {\boldsymbol {\sigma }}=\lambda ~\mathrm {tr} ({\boldsymbol {\varepsilon }})~{\boldsymbol {\mathit {1}}}+2\mu {\boldsymbol {\varepsilon }}} constantes de Lamé [1] 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)}}} [1] 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.
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 de caucho incompresibles a base de I 1 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 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". Método numérico Int J 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)
