Taylor series expansion for stresses on an element vs equal stress values on opposite faces I only recently realized this "confusion" despite having cleared all my exams long ago. Consider these 2 diagrams:


*

*
and 2. 
Clearly they yield different equations. I have seen both in different sources, but cant reconcile them since in both pictures the element has sides that tend to 0. Anyone sees the problem?
 A: In situation 2, for simplicity, we’re considering a region of uniform stress. It doesn’t need to be plane stress; it just needs to not change much locally. This model is useful for establishing the stress tensor and defining the types of stress. (For instance, we can review that normal stresses act perpendicular to faces, whereas shear stresses act parallel to faces; furthermore, $\tau_{xy}$ must equal $\tau_{yx}$ to satisfy static equilibrium.) 
In the specific plane stress configuration shown in situation 2, for example, the stress tensor is
$$\boldsymbol{\sigma}=\begin{pmatrix}\sigma_{x} & \tau_{xy} & 0\\&\sigma_{y}&0\\&&0\\\end{pmatrix}=\begin{pmatrix}\sigma_{xx} & \sigma_{xy} & 0\\&\sigma_{yy}&0\\&&0\\\end{pmatrix}.$$
Situation 1 is more complex but essential if we’re going to move beyond  just defining stress and start considering how to establish constitutive laws of how the stress changes throughout a finite 3D body in response to applied loads and inertia, for example. 
Consider, for instance, the addition of a rightward volumetric body force $B$ and a density $\rho$. Then, I can sum the forces in the $x$-direction and divide by $\Delta x\,\Delta y\,\Delta z$ to obtain $$\frac{\partial\sigma_x}{\partial x}+\frac{\partial\tau_{xy}}{\partial y}+B=\rho\frac{d^2 \mathbf{x}}{dt^2}$$
where $\mathbf{x}$ is the $x$-coordinate of the element. Now I can start solving for how the body deforms and translates.
