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I have some trouble to understand the relationship between internal stress tensor and the external traction force (in this example, the traction force is a volumic centripetal force). Consider a cube made of a very high elastic modulus homogeneous material (which can be approximated to a quasi-rigid body), it is rotating about axis-z in a constant spin speed $\Omega$. For sake of simplicity, the three axes are attached to the cube. enter image description here

In its stationnary configuration, with respect to the attached rotating reference, the centrifugal force on any point in this cube should be equilibrated by an internal elastic force, in order to keep the cube from falling apart. Given the coordinate of a generic point $P$ in this cube as $\boldsymbol{x} = [x\;y\;z]^\text{T}$, the centripetal force density can be given by $$\boldsymbol{f} = \rho [\boldsymbol{\omega}]^2 \boldsymbol{x}$$ where $$[\boldsymbol{\omega}] = \begin{bmatrix} 0 & -\Omega & 0 \\ \Omega & 0 & 0 \\ 0& 0 & 0 \end{bmatrix}$$ My question is how to express the corresponding true/cauchy stress tensor on this point for equilibrating the centripetal force density as $$\boldsymbol{\sigma}_{\boldsymbol{f}} = \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz}& \sigma_{yz} & \sigma_{zz} \end{bmatrix}$$

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