Taylor's series + Stress Equilibrium Equation (Cauchy's equilibrium equation) I am now studying theory of plasticity and elasticity and I stumbled upon the derivation of the equation of stress equilibrium (or Cauchy's equilibrium equation) that I would like to have more clarifications on.
You can see the image below, while I understand that, in order to formulate a more accurate opposite force, it will be infintismally different from the applied force (hence the inclusion of the differential equations), but almost all text books just include the first two terms of a Taylor's series expansion as explained. Why? and how do they get the two terms from the taylor's series?
thanks!

 A: To take the first 2 terms of Taylor expansion is equivalent to say that for small enough interval, the function can be considered linear.
In the case, considering a cubic small element $\Delta x \Delta y \Delta z$ of a body at equilibrium, by taking the components of the net force on it:
$$F_{x+\Delta x} - F_x = 0$$
$$F_{y+\Delta y} - F_y = 0$$
$$F_{z+\Delta z} - F_z = 0$$
$$(\sigma_{xx(x + \Delta x)} - \sigma_{xx(x)})\Delta y \Delta z + (\sigma_{yx(y + \Delta y)} - \sigma_{yx(y)})\Delta x \Delta y + (\sigma_{zx(z + \Delta z)} - \sigma_{zx(z)})\Delta x \Delta z = 0$$
Dividing by $\Delta x \Delta y \Delta z$ and taking the limit when deltas go to zero:
$$\frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \sigma_{yx}}{\partial y}  + \frac{\partial \sigma_{zx}}{\partial z} =  0$$  The same  for the $y$ and $z$.
A: Let us consider the force balance in the $x$ direction around the point $(x, y, z)$, and taking as side lengths $\Delta x$, $\Delta y$, and $\Delta z$.
We have the following
$$0 = \left(\sigma_{xx} + \frac{\partial \sigma_{xx}}{\partial x}(x, y, z)\Delta x \right)\Delta y \Delta z - \sigma_{xx}(x, y, z)\Delta y \Delta z + \left(\sigma_{yx}(x, y, z) + \frac{\sigma_{yx}}{\partial y}(x, y, z)\Delta y\right)\Delta x \Delta z - \sigma_{yx}(x, y, z)\Delta x \Delta z + \left(\sigma_{zx}(x, y, z) + \frac{\partial\sigma_{zx}}{z}(x, y, z)\Delta z\right)\Delta x \Delta y - \sigma_{zx}(x, y, z)\Delta x \Delta y + b_x(x, y, z) \Delta x \Delta y \Delta z\, .$$
If we expand this, we get
$$0 = \left(\frac{\partial\sigma_{xx}}{\partial x}
+\frac{\partial\sigma_{yx}}{\partial y} 
+\frac{\partial\sigma_{zx}}{\partial z} + b_x\right)\Delta x \Delta y \Delta z\, ,$$
and, after dividing by $\Delta x\Delta y\Delta z$,
$$0 = \frac{\partial\sigma_{xx}}{\partial x}
+\frac{\partial\sigma_{yx}}{\partial y} 
+\frac{\partial\sigma_{zx}}{\partial z} + b_x\, .$$

I would suggest that it is better if you look at it globally, that is, look at the balance of forces in the whole body. This corresponds with the sum of surface attractions ($\mathbf{t}$) and body forces ($\mathbf{b}$), that is,
$$\int\limits_A \mathbf{t} \mathrm{d}A
+ \int\limits_V \mathbf{f} \mathrm{d}V = \mathbf{0}\, ,$$
where we can replace the definition of traction
$$\mathbf{t} = \hat{\mathbf{n}}\cdot\mathbf{\sigma}\, ,$$
to get
$$\int\limits_A \hat{\mathbf{n}}\cdot\mathbf{\sigma} \mathrm{d}A
+ \int\limits_V \mathbf{f} \mathrm{d}V = \mathbf{0}\, .$$
If we use the divergence theorem, we get
$$\int\limits_V \left(\nabla\cdot\sigma + \mathbf{f}\right) \mathrm{d}V = \mathbf{0}\, .$$
Considering that this balance should hold for any volume $V$ we get that the integrated should be zero
$$\nabla\cdot\sigma + \mathbf{f} = \mathbf{0}\, .$$
A: 
this might give you some insightstrong text
