Determine the tensor of contraint and deformation of a cube under compression

Question

We have a cube under compression with dimension l1*l2*l3, is put between 2 rigid plates in the axis 1 (two plates block the deformation of the cube in thí axis), the cube is also put on a rigid plate, and it has a rigid plate above it; and also this cube is under the force F from above (axis 3). The material of the cube is isotropic, linear elastic. The deformation of the cube is ∆l2 and ∆l3 (since the deformation of the axis 1 is block)

The question is: How to write/determine the tensor of constrain and tensor of deformation? Is this below one is correct?

$\left( \begin{matrix} {{\sigma }_{11}} & {{\sigma }_{12}} & {{\sigma }_{13}} \\ {{\sigma }_{21}} & {{\sigma }_{22}} & {{\sigma }_{23}} \\ {{\sigma }_{31}} & {{\sigma }_{32}} & {{\sigma }_{33}} \\ \end{matrix} \right)=\left( \begin{matrix} {{\sigma }_{11}} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & {{\sigma }_{33}} \\ \end{matrix} \right)$

and

$\left( \begin{matrix} 0 & 0 & 0 \\ 0 & {{\varepsilon }_{22}}={}^{\Delta {{l}_{2}}}\!\!\diagup\!\!{}_{{{l}_{2}}}\; & 0 \\ 0 & 0 & {{\varepsilon }_{33}}=-{}^{\Delta {{l}_{3}}}\!\!\diagup\!\!{}_{{{l}_{3}}}\; \\ \end{matrix} \right)$

Answer 1

Your answer for the strain tensor is correct, but for not the stress (on utilise "stress" en anglais, pas "contraint"). Since you know $\epsilon_{ij}$ you can immediately calculate the stress tensor from Hooke's law:

$$\sigma_{ij}=2\mu\epsilon_{ij} +\lambda \operatorname{tr}(\epsilon) \delta_{ij}$$

to get $$\sigma=\left( \begin{array}{ccc} \lambda \left(\frac{\Delta l_2}{l_2}+\frac{\Delta l_3}{l_3}\right) & 0 & 0 \\ 0 & \frac{(\lambda +2 \mu ) \Delta l_2}{l_2}+\frac{\lambda \Delta l_3}{l_3} & 0 \\ 0 & 0 & \frac{\lambda \Delta l_2}{l_2}+\frac{(\lambda +2 \mu ) \Delta l_3}{l_3} \end{array} \right)$$