If there was an object and we wanted to determine the components of stress along any plane in the body, we could use what is called a traction vector. Think of the traction vector as a way to measure the intensity of forces on an object's plane at a certain point. We can denote the traction vector as
[ 1 ]
$$ \begin{Bmatrix} t \end{Bmatrix} = \begin{Bmatrix} t_x \\ t_y \\ t_z \end{Bmatrix} $$
A way to visualize the traction vector can be found in Figure 1.
Fig. 1 - Visualization of the traction vector on a tetrahedron.
One way to relate the traction vector back to our stress field is through Cauchy's equation. In order to set up Cauchy's equation, all that is needed is to make a vector of the stresses on each face of the object in Figure 1, for example. We can start by looking at the y-z face of the object. We can vectorize the stresses on that face and since our normal stress is in the x direction, we would associate this with the x direction of the traction vector. So it would look something like this:
[ 2 ]
$$ \begin{Bmatrix} t \end{Bmatrix}^x = -\sigma_{xx} \hat{i} -\tau_{xy} \hat{j} -\tau_{xz} \hat{k} $$
And for the x-z face:
[ 3 ]
$$ \begin{Bmatrix} t \end{Bmatrix}^y = -\tau_{yx} \hat{i} -\sigma_{yy} \hat{j} -\tau_{yz} \hat{k} $$
For the x-y face:
[ 4 ]
$$ \begin{Bmatrix} t \end{Bmatrix}^z = -\tau_{zx} \hat{i} -\tau_{zy} \hat{j} -\sigma_{zz} \hat{k} $$
And finally for the slanted face:
[ 5 ]
$$ \begin{Bmatrix} t \end{Bmatrix}^n = t_x \hat{i} + t_y \hat{j} + t_z \hat{k} $$
Once the traction vectors are defined, we can add them all up and set them equal to zero because we are in equilibrium, so $ \Sigma F = 0 $. Then we have
[ 6 ]
$$ \begin{Bmatrix} t \end{Bmatrix}^x dA_x + \begin{Bmatrix} t \end{Bmatrix}^y dA_y + \begin{Bmatrix} t \end{Bmatrix}^z dA_z + \begin{Bmatrix} t \end{Bmatrix}^n dA_n = 0 $$
However, we can relate our $ dA_x $, $ dA_y $, and $ dA_z $ with the normal vector to each face because
[ 7 ]
$$ dA_x = dA_n (\hat{n} • \hat{i}) = dA_n • n_x $$ $$ dA_y = dA_n (\hat{n} • \hat{j}) = dA_n • n_y $$ $$ dA_z = dA_n (\hat{n} • \hat{k}) = dA_n • n_z $$
We can then substitute the $ dA $'s from Eq. (7) into Eq. (6) and divide by $ dA_n $. One last step before we get our full Cauchy's Equation is we need to define the normal vector $ \hat{n} $ as
[ 8 ]
$$ \begin{Bmatrix} n \end{Bmatrix} = n_x \hat{i} + n_y \hat{j} + n_z \hat{k} $$
We can then rearrange the substituted equation Eq. (6) and write it in matrix form.
[ 9 ]
$$ \begin{Bmatrix} t \end{Bmatrix} = \begin{Bmatrix} t_x \\ t_y \\ t_z \end{Bmatrix} = \begin{bmatrix} \sigma_x & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \sigma_y & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \sigma_z \end{bmatrix} \begin{Bmatrix} n_x \\ n_y \\ n_z \end{Bmatrix} $$
We have just derived Cauchy's Equation! This is used often for boundary element problems for validating assumptions and determining how intense forces are in each direction at a point on a surface.