Equations of Equilibrium

Consider a small cube of sides $\Delta x,\ \Delta y$ and $\Delta z$ with its centroid at point Q. Since different faces of the cube do not pass through the same point, the state of stress on them will be different. We assume that each component of stress is a continuously differentiable function of $x,\ y$ and z. For a continuously differentiable function f of x,

$$f(x + \Delta x) \simeq f(x) + \frac{\partial f}{\partial x}\Delta x.\tag{3.1}$$ (3.1)

The components of stress on the six faces of the cube are shown in Fig. 3.1

Let the gravitational force g with components $g_x,\ g_y$ and g_z act at every point of the cube; g is measured as force/mass. The equilibrium of forces in the x-direction gives
$$\begin{align}&\left(\sigma_{xx} + \frac{\partial\sigma_{xx}}{\partial x}\Delta x... ..._{zx}\Delta x\Delta y + \rho g_x\Delta x\Delta y\Delta z=0,\tag{3.2} \end{align}$$
or

$$\frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial\sigm... ...frac{\partial\sigma_{zx}}{\partial z} + \rho g_x = 0.\tag{3.3}$$ (3.3)

Similarly $$\sum F_y = 0$$, and $$\sum F_z = 0$$ give
$$\begin{gather}\begin{split} &\frac{\partial \sigma_{xy}}{\partial x} + \frac{\pa... ...partial\sigma_{zz}}{\partial z} + \rho g_z = 0.\end{split}\tag{3.5} \end{gather}$$
Here $\rho$ is the mass density per unit volume of the body. Now take moments about the z-axis passing through the centroid of the cube. $$\sum M_z = 0$$ gives
$$\begin{align}&\left(\sigma_{xy} + \frac{\partial\sigma_{xy}}{\partial x} \Delta ... ... y}{2} - \sigma_{yx}\Delta x\Delta z\frac{\Delta y}{2} = 0.\tag{3.6} \end{align}$$
Divide by $(\Delta x\Delta y\Delta z)/2$ to obtain

$$\sigma_{xy} - \sigma_{yx} + \frac{\partial\sigma_{xy}}{\parti... ... \frac{\partial\sigma_{yx}}{\partial y} \Delta y = 0.\tag{3.7}$$ (3.7)

Let $\Delta x \rightarrow 0,\ \Delta y\rightarrow 0$ and $\Delta z \rightarrow 0$, and assume that $\partial\sigma_{xy}/\partial x$ and $\partial\sigma_{yx}/\partial y$ are bounded. Equation (3.7) reduces to

$$\sigma_{xy} - \sigma_{yx} = 0. \tag{3.8}$$ (3.8)

Similarly
$$\begin{align}&\sigma_{yz} - \sigma_{zy} = 0,\tag{3.9}\\ &\sigma_{zx} - \sigma_{xz} = 0.\tag{3.10} \end{align}$$
Thus the balance of moments implies the symmetry of the stress components.

Results: Equations of Equilibrium:
$$\begin{align}&\frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \sigma_{y... ...},\ \sigma_{yz} = \sigma_{zy},\ \sigma_{zx} = \sigma_{xz}.\tag{3.12} \end{align}$$
In condensed notation, we write

$$\{\sigma\}^T = [\sigma_{xx}\ \ \sigma_{yy}\ \ \sigma_{zz}\ \ \sigma_{yz}\ \ \sigma_{zx}\ \ \sigma_{xy}],\tag{3.13}$$ (3.13)

and in matrix notation

$$[{\mbox{\boldmath {$\sigma$ }}} ]= \left[\begin{array}{ccc} \... ..._{zx} & \sigma_{zy} & \sigma_{zz}\end{array}\right].\tag{3.14}$$ (3.14)