Stress tensor and equality of normal stresses on opposite faces

Question

Consider a body arbitrarily loaded as shown,

At a particular point in the body, I take an element and show all the stresses acting on its faces.

To specify a plane I will be using the the axis which is perpendicular to it. For instance, the front face is the +z face and the face opposite to the +z face is the -z face.

All the sources that I follow, state that the normal stress acting on the +z and -z face will be equal. Similarly, the normal stress on the faces +x and -x as well as +y and -y face will be equal. However, I feel that might not necessarily be the case.

The normal stress on the +z and -z face, can be different, but could be such that theses normal stresses along with the shear stresses acting along z direction, on the +y,-y and +x, -x faces vectorially sum to zero, so that equilibrium is maintained along the z direction.

$$\sigma_z - \sigma_z' + \tau_{xz} - \tau_{xz}' + \tau_{yz} - \tau_{yz}' = 0$$

Same arguments can hold true for equilibrium along x and y directions.

So, it might not be necessary that the normal stresses on opposite faces are equal, then why in the general state of stress at a point they are shown equal?

A similar question was asked here

Answer 1

The point is that the claimed identity is valid in the limit of vanishing size $2\delta$ of the considered cubic element. If the element is cetered on the origin and you consider two opposite faces, e.g., normal to the axis $z$, you have, assuming that the stress tensor is differentiable at the origin, $$\sigma(0,0,\pm\delta) \cdot {\bf e}_z= \sigma(0,0,0)\cdot {\bf e}_z + O_\pm(\delta)\:.$$ At this order of approximation, the stresses on the two opposite faces are equal in norm and have opposite directions (since the normal outward vectors are opposite: $\pm {\bf e}_z$).

Since what I wrote above concerns the stresses as vectors, not only the normal components of the stresses are equal in norm, but also the tangential components are: $$\sigma_{az}(0,0,z)= \sigma_{az}(0,0,-z)+ O(\delta)$$ for $a=x,y,z$. Since this argument applies to each pair of opposite faces, in your equation $$(\sigma_z - \sigma_z') + (\tau_{xz} - \tau_{xz}') + (\tau_{yz} - \tau_{yz}') \simeq 0$$ the three differences vanish separately with the said approximation.