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Technically, no force was labeled, it was only stresses, $T,\,\sigma$, and unit vectors, $\mathbf e_i$. But note that there totally are 3 sets of $\sigma$ arrows on each face! See: They're not actually labeled, but they're definitely there (just hidden). Since there are only three orthogonal directions here, $\hat{\mathbf{x}},\,\hat{\mathbf ...


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For each surface on a unit cube (see below), the stress on that surface can point in each of the three directions. (source) Since it is not necessarily the case that $\sigma_{11}=\sigma_{31}=\sigma_{21}$ (all pointing the in the same $\mathbf{e}_1$ direction)--or any of the other $\sigma_{ij}$ combinations, we need to have 9 components describing it, ...


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Stress is a tensor1 because it describes things happening in two directions simultaneously. You can have an $x$-directed force pushing along an interface of constant $y$; this would be $\sigma_{xy}$. If we assemble all such combinations $\sigma_{ij}$, the collection of them is the stress tensor. Pressure is part of the stress tensor. The diagonal elements ...


1

As $W$ is a potential for the stresses in $F_{ij}$, a Legendre-Fenchel-transformation to change the independent variable from strain to stress gives you the dual potential $W^*(P_{ij})$, where $ P_{ij}$ are the first Piola Kirchhoff stresses. It is given by $$ W^\ast(\mathbf{P})=\sup[{\mathbf{P}\cdot\mathbf{F}-W(\mathbf{F})}] \text{ over } \mathbf{F}, $$ ...


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If you write your equation in component form, you get \begin{equation} P_{ij} = \frac{\partial W}{\partial F_{ij}}, \end{equation} which is equivalent to \begin{equation} W = \int P_{ij}dF_{ij} = \int \tilde{P} \colon d\tilde{F}, \end{equation} where $\tilde{P}$ and $\tilde{F}$ are the stress and strain tensors in Gibbs notation. You can immediately see the ...



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