Ogden类超弹性本构推导

应变能定义为:
W = μ 2 ( I 1 − 3 − 2 ln ⁡ J ) + λ 2 ( ln ⁡ J ) 2 W=\frac{\mu}{2}(I_{1}-3-2\ln J)+\frac{\lambda}{2}(\ln J)^{2} W=2μ(I132lnJ)+2λ(lnJ)2
其中 I 1 = t r ( C ) I_{1}=\mathrm{tr}(\mathbf{C}) I1=tr(C)
在推导之前需要知道的张量运算(如果之前从来没见过,最好自己推一遍,这里不给出证明)
∂ t r ( C ) ∂ C = I ∂ J ∂ C = J 2 C − 1 ∂ C − 1 ∂ C = − C ⊙ C = − 1 2 ( C i k − 1 C j l − 1 + C i l − 1 C j k − 1 ) \frac{\partial\mathrm{tr}(\mathbf{C})}{\partial\mathbf{C}}=\mathbf{I}\quad \frac{\partial J}{\partial\mathbf{C}}=\frac{J}{2}\mathbf{C}^{-1}\quad \frac{\partial\mathbf{C}^{-1}}{\partial\mathbf{C}}=-\mathbf{C}\odot\mathbf{C}=-\frac{1}{2}(C_{ik}^{-1}C_{jl}^{-1}+C_{il}^{-1}C_{jk}^{-1}) Ctr(C)=ICJ=2JC1CC1=CC=21(Cik1Cjl1+Cil1Cjk1)

考虑right Cauchy-Green变形张量 C = F T F \mathbf{C}=\mathbf{F}^{T}\mathbf{F} C=FTF 和对应的Green-Lagrange应变张量 E = 1 2 ( F T F − I ) = 1 2 ( C − I ) \mathbf{E}=\frac{1}{2}(\mathbf{F}^{T}\mathbf{F}-\mathbf{I})=\frac{1}{2}(\mathbf{C}-\mathbf{I}) E=21(FTFI)=21(CI).
按照定义,可以写出second Piola-Kirchhoff stress:
S = δ W δ E = 2 ∂ W ∂ C = 2 { μ 2 ( ∂ I 1 ∂ C − 2 J ∂ J ∂ C ) + λ 2 ( 2 ln ⁡ J ) 1 J ∂ J ∂ C } = 2 { μ 2 ( I − 2 J ⋅ J 2 C − 1 ) + λ ln ⁡ J ⋅ 1 J ⋅ J 2 C − 1 } = μ ( I − C − 1 ) + λ ln ⁡ J C − 1 \begin{aligned} \mathbf{S}&=\frac{\delta W}{\delta\mathbf{E}}=\frac{2\partial W}{\partial\mathbf{C}}\\ &=2\{\frac{\mu}{2}(\frac{\partial I_{1}}{\partial\mathbf{C}}-\frac{2}{J}\frac{\partial J}{\partial\mathbf{C}})+\frac{\lambda}{2}(2\ln J)\frac{1}{J}\frac{\partial J}{\partial\mathbf{C}}\}\\ &=2\{\frac{\mu}{2}(\mathbf{I}-\frac{2}{J}\cdot\frac{J}{2}\mathbf{C}^{-1}) +\lambda\ln J\cdot\frac{1}{J}\cdot\frac{J}{2}\mathbf{C}^{-1} \}\\ &=\mu(\mathbf{I}-\mathbf{C}^{-1})+\lambda\ln J\mathbf{C}^{-1} \end{aligned} S=δEδW=C2W=2{2μ(CI1J2CJ)+2λ(2lnJ)J1CJ}=2{2μ(IJ22JC1)+λlnJJ12JC1}=μ(IC1)+λlnJC1
如果使用参考构型,请务必将其转换为1st Piola-Kirchhoff stress:
P = ∂ W ∂ F = F S \mathbf{P}=\frac{\partial W}{\partial\mathbf{F}}=\mathbf{F}\mathbf{S} P=FW=FS
如果是使用当前构型,需要将其转换为Cauchy stress: σ = J − 1 F S F T \mathbf{\sigma}=J^{-1}\mathbf{F}\mathbf{S}\mathbf{F}^{T} σ=J1FSFT

接下来在上诉公式的基础上进行本构推导:
J a c = δ S δ E = 2 ∂ S ∂ C = 2 μ ⋅ ( − ∂ C − 1 ∂ C ) + 2 λ ⋅ 1 J ∂ J ∂ C ⊗ C − 1 + 2 λ ln ⁡ J ⋅ ∂ C − 1 ∂ C = 2 μ C − 1 ⊙ C − 1 + 2 λ 1 J ⋅ J 2 C − 1 ⊗ C − 1 + 2 λ ln ⁡ J ( − C − 1 ⊙ C ) = 2 μ C − 1 ⊙ C − 1 + λ C − 1 ⊗ C − 1 − 2 λ ln ⁡ J C − 1 ⊙ C − 1 \begin{aligned} \mathrm{Jac}&=\frac{\delta\mathbf{S}}{\delta\mathbf{E}}=\frac{2\partial\mathbf{S}}{\partial\mathbf{C}}\\ &=2\mu\cdot(-\frac{\partial\mathbf{C}^{-1}}{\partial\mathbf{C}})+ 2\lambda\cdot\frac{1}{J}\frac{\partial J}{\partial\mathbf{C}}\otimes\mathbf{C}^{-1} +2\lambda\ln J\cdot\frac{\partial\mathbf{C}^{-1}}{\partial\mathbf{C}}\\ &=2\mu\mathbf{C}^{-1}\odot\mathbf{C}^{-1}+2\lambda\frac{1}{J}\cdot\frac{J}{2}\mathbf{C}^{-1}\otimes\mathbf{C}^{-1}+2\lambda\ln J(-\mathbf{C}^{-1}\odot\mathbf{C})\\ &=2\mu\mathbf{C}^{-1}\odot\mathbf{C}^{-1}+\lambda\mathbf{C}^{-1}\otimes\mathbf{C}^{-1} -2\lambda\ln J\mathbf{C}^{-1}\odot\mathbf{C}^{-1} \end{aligned} Jac=δEδS=C2S=2μ(CC1)+2λJ1CJC1+2λlnJCC1=2μC1C1+2λJ12JC1C1+2λlnJ(C1C)=2μC1C1+λC1C12λlnJC1C1

你可能感兴趣的:(固体力学,固体力学,超弹性材料,本构关系推导,大变形,张量推导)