1.2.1. Hooke


Features

Linear elasticity


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Stress-strain relationship

The connection between strain and stress is based on the Green-Lagrange strain $\tnsr E$ and its work-conjugate, the second Piola-Kirchhoff stress $\tnsr S$. \begin{align} \label{eq:Hooke} \tnsr S &= \tnsrfour C : \tnsr E = \frac{1}{2} \tnsrfour C : \left({\tnsr F_\text e}^\text T \,\tnsr F_\text e - \tnsr I\right) \end{align}

Tangent $\partial\tnsr S / \partial \tnsr F_\text e$

The derivative of the second Piola-Kirchhoff stress with respect to the elastic deformation gradient is required in the implicit stress calculation. For simpler analysis we conceptually split the fourth order stiffness tensor $\tnsrfour C$ into a product of two second order tensors $\tnsr A \otimes \tnsr B$ (cf. tensor notation scheme). \begin{align*} \tnsr S,{\scriptscriptstyle\tnsr F_\text e} &= \left[\frac{1}{2} (\tnsr A \otimes \tnsr B) : \left({\tnsr F_\text e}^\text T \,\tnsr F_\text e - \tnsr I\right)\right],_{\scriptscriptstyle\tnsr F_\text e} \\ &= \frac{1}{2} \left[(\tnsr A \otimes \tnsr B) : \left({\tnsr F_\text e}^\text T \,\tnsr F_\text e\right)\right],_{\scriptscriptstyle\tnsr F_\text e} \\ &= \frac{1}{2} \left[\tnsr A \left(\tnsr B : {\tnsr F_\text e}^\text T \,\tnsr F_\text e\right)\right],_{\scriptscriptstyle\tnsr F_\text e} \\ &= \frac{1}{2} \tnsr A \odot \left[ \tnsr B : \left({\tnsr F_\text e}^\text T \,\tnsr F_\text e\right)\right],_{\scriptscriptstyle\tnsr F_\text e} \\ &= \frac{1}{2} \tnsr A \odot \left( \tnsr B \dblContInOut \left[{\tnsr F_\text e}^\text T \,\tnsr F_\text e\right],_{\scriptscriptstyle\tnsr F_\text e}\right) \\ &= \frac{1}{2} \tnsr A \odot \left( \tnsr B \dblContInOut \left(\tnsr I\boxtimes\tnsr F_\text e + {\tnsr F_\text e}^\text T\otimes\tnsr I \right)\right) \\ &= \frac{1}{2} \tnsr A \odot \left( \tnsr F_\text e \tnsr B^\text T + \tnsr F_\text e \tnsr B\right) \\ &= \frac{1}{2} \tnsr A \odot \left( \tnsr F_\text e \left(\tnsr B^\text T + \tnsr B\right) \right) \end{align*} Due to the symmetry of the stiffness tensor, there is $\tnsr B^\text T=\tnsr B$ and one obtains \begin{align} \tnsr S,{\scriptscriptstyle\tnsr F_\text e} &= \tnsr A \odot \left( \tnsr F_\text e \tnsr B^\text T\right) \end{align} It is useful to rewrite this equation in index notation. \begin{align} \frac{\partial S_{ij}}{\partial {F_\text e}_{kl}} \vctr g^i \otimes \vctr g_k \otimes \vctr g_l \otimes \vctr g^j &= A_{ij} {F_e}^{km} B^{l}_{\cdot m} \, \vctr g^i \otimes \vctr g_k \otimes \vctr g_l \otimes \vctr g^j \\ &= C_{ij\cdot\cdot}^{\phantom{ij}lm} {F_e}_{\cdot m}^{k} \, \vctr g^i \otimes \vctr g_k \otimes \vctr g_l \otimes \vctr g^j \end{align}
Topic revision: r7 - 22 Jan 2014, ChristophKords


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