Cauchy's formula and equilibrium equations

Cauchy's formula

Take a tetrahedron at a point with three faces on the coordinate planes and a fourth inclined face of area $dA$ and unit normal $\mathbf{n} = (l, m, n)$ (direction cosines). The coordinate-plane faces then have areas

$$dA_x = l\,dA,\quad dA_y = m\,dA,\quad dA_z = n\,dA$$

Let $\mathbf{t} = (t_x, t_y, t_z)$ be the traction (stress vector) on the inclined face. The faces on the coordinate planes have outward normals along $-x, -y, -z$, so their stresses act opposite to the positive components.

        z
        |  C
        | /|\
        |/ | \
        A--+--B  (inclined face ABC)
       /  O
      y       x

Force equilibrium in the x-direction (body force and inertia are of higher order, since volume $\propto h^3$ while area $\propto h^2$ as the tetrahedron shrinks):

$$\begin{aligned} t_x\, dA - \sigma_x (l\,dA) - \tau_{yx}(m\,dA) - \tau_{zx}(n\,dA) &= 0 \\ t_x &= \sigma_x l + \tau_{yx} m + \tau_{zx} n \end{aligned}$$

Similarly for $y$ and $z$. In matrix form (symmetric tensor):

$$\begin{Bmatrix} t_x \\ t_y \\ t_z \end{Bmatrix} = \begin{bmatrix} \sigma_x & \tau_{xy} & \tau_{xz} \\ \tau_{xy} & \sigma_y & \tau_{yz} \\ \tau_{xz} & \tau_{yz} & \sigma_z \end{bmatrix} \begin{Bmatrix} l \\ m \\ n \end{Bmatrix}$$

or $t_i = \sigma_{ij} n_j$. The normal and shear stress on the plane are then

$$\sigma_n = \mathbf{t}\cdot\mathbf{n},\qquad \tau = \sqrt{|\mathbf{t}|^2 - \sigma_n^2}$$

So the six components of $\sigma_{ij}$ determine the traction on every plane through the point.

Equilibrium equations

For a body force per unit volume $(X, Y, Z)$:

$$\begin{aligned} \frac{\partial \sigma_x}{\partial x} + \frac{\partial \tau_{xy}}{\partial y} + \frac{\partial \tau_{xz}}{\partial z} + X &= 0 \\ \frac{\partial \tau_{xy}}{\partial x} + \frac{\partial \sigma_y}{\partial y} + \frac{\partial \tau_{yz}}{\partial z} + Y &= 0 \\ \frac{\partial \tau_{xz}}{\partial x} + \frac{\partial \tau_{yz}}{\partial y} + \frac{\partial \sigma_z}{\partial z} + Z &= 0 \end{aligned}$$

Moment equilibrium gives the symmetry $\tau_{ij}=\tau_{ji}$. At the boundary, the traction must satisfy $t_i = \sigma_{ij}n_j$ equal to the applied surface load.