Equilibrium equations

A solid body is in static equilibrium when the resultant force and the resultant moment about every axis are zero. Applied to every part of the body, these conditions lead to the equilibrium equations. In this article we first derive the equilibrium equations from the equilibrium of an infinitesimal element and then, more elegantly, from the equilibrium of an arbitrary volume with the divergence theorem. Both approaches also show that the stress tensor is symmetric.

In this article tensile stresses are taken as positive: a positive stress component acts in the positive coordinate direction on a face whose outward normal points in a positive coordinate direction. In geomechanics the opposite convention, with compressive stresses positive, is often used (see Mohr’s circle); all stress terms then change sign and the equilibrium equations become $\sigma_{ji,j}=f_{i}$ instead of Eq. (5).

Equilibrium of forces

Surface and body forces act on a solid body in static equilibrium. We cut an infinitesimal parallelepiped (element) inside the body and analyze the forces that act on it, as shown in Fig. 1. We assume that the stress field is continuous and differentiable inside the whole body.

The stress components on each face of the element are functions of the position, since the stress field is non-uniform but continuous. For example, on face $4$ the normal stress is $\sigma_{11}(x_{1},x_{2},x_{3})=\sigma_{11}$. On the opposite face $2$ the normal stress is $\sigma_{11}(x_{1}+\mathrm{d}x_{1},x_{2},x_{3})$. Using a first-order Taylor expansion we may write:

$$\sigma_{11}(x_{1}+\mathrm{d}x_{1},x_{2},x_{3})=\sigma_{11}+\dfrac{\partial\sigma_{11}}{\partial x_{1}}\mathrm{d}x_{1} \tag{1}$$

where the higher-order terms have been omitted because they vanish in the limit $\mathrm{d}x_{1}\to 0$. We follow the same procedure for all the components, as shown in Fig. 1.

Equilibrium of the element requires that the resultant force is zero. By summing up the forces parallel to the axis $x_{1}$ we get:

where $\mathrm{d}x_{1}$, $\mathrm{d}x_{2}$ and $\mathrm{d}x_{3}$ are the dimensions of the element and $f_{1}$ is the component of the body force parallel to $x_{1}$. The body force $f_{i}$ is a force per unit volume; for example, the weight of a material with density $\rho$ is $f_{i}=(0,0,-\rho g)$ when the axis $x_{3}$ points upwards. By dividing by $\mathrm{d}x_{1}\mathrm{d}x_{2}\mathrm{d}x_{3}$ we get:

$$\dfrac{\partial\sigma_{11}}{\partial x_{1}}+\dfrac{\partial\sigma_{21}}{\partial x_{2}}+\dfrac{\partial\sigma_{31}}{\partial x_{3}}+f_{1}=0 \tag{3}$$

Similarly, we can obtain the equations for the other two directions. The final set of equilibrium equations is:

$$\begin{array}{l}\dfrac{\partial \sigma_{11}}{\partial x_{1}}+\dfrac{\partial \sigma_{21}}{\partial x_{2}}+\dfrac{\partial \sigma_{31}}{\partial x_{3}}+f_{1}=0\\ \dfrac{\partial \sigma_{12}}{\partial x_{1}}+\dfrac{\partial \sigma_{22}}{\partial x_{2}}+\dfrac{\partial \sigma_{32}}{\partial x_{3}}+f_{2}=0\\ \dfrac{\partial \sigma_{13}}{\partial x_{1}}+\dfrac{\partial \sigma_{23}}{\partial x_{2}}+\dfrac{\partial \sigma_{33}}{\partial x_{3}}+f_{3}=0\end{array} \tag{4}$$

By using index notation we may write the three equilibrium equations in compact form:

$$\sigma_{ji,j}+f_{i}=0 \tag{5}$$

Symmetry of the stress tensor

The resultant moment about each axis must also vanish. We take moments about an axis parallel to $x_{3}$ through the corner of the element where faces $3$, $4$ and $6$ meet, i.e. through the point $(x_{1},x_{2},x_{3})$, with counterclockwise moments (seen from the positive $x_{3}$ axis) taken as positive. By taking into account all the forces that contribute to the moment about this axis we may write:

By dividing by $\mathrm{d}x_{1}\mathrm{d}x_{2}\mathrm{d}x_{3}$ and collecting terms we get:

and by taking the limit $\mathrm{d}x_{1}\to 0$, $\mathrm{d}x_{2}\to 0$ and $\mathrm{d}x_{3}\to 0$ only the first two terms remain:

$$\sigma_{12}=\sigma_{21} \tag{8}$$

Following the same procedure for the other two axes leads to the conclusion that the stress tensor is symmetric:

$$\sigma_{ij}=\sigma_{ji} \tag{9}$$

Because of the symmetry, the equilibrium equations (5) may also be written as $\sigma_{ij,j}+f_{i}=0$ or, in vector form, $\nabla\cdot\boldsymbol{\sigma}+\mathbf{f}=\mathbf{0}$.

It should be noted that the above symmetry holds only if there are no distributed body couples (moments per unit volume), as in Cosserat (polar) continua. Otherwise, the stress tensor is in general not symmetric. However, in most rock mechanics problems the stress tensor is symmetric.

Derivation with the divergence theorem

The same results may be obtained more elegantly by considering an arbitrary volume $V$ of the body, bounded by the surface $S$ with outward unit normal $n_{j}$. The stress vector $T_{i}$ acting on the surface is given by Cauchy’s formula:

$$T_{i}=\sigma_{ji}n_{j} \tag{10}$$

since, according to Fig. 1, $\sigma_{ji}$ is the stress component acting in the direction $x_{i}$ on the face whose normal is parallel to $x_{j}$. We will also need the divergence theorem (Gauss’s theorem), which transforms an integral over the surface $S$ into an integral over the volume $V$. For any continuous and differentiable field $g$:

$$\int_{S}g\,n_{j}\,\mathrm{d}S=\int_{V}g_{,j}\,\mathrm{d}V \tag{11}$$

where $g$ may also carry other indices, for example $g=\sigma_{ji}$.

Equilibrium of forces

The resultant of the surface and body forces acting on the volume must vanish:

$$\int_{S}T_{i}\,\mathrm{d}S+\int_{V}f_{i}\,\mathrm{d}V=0 \tag{12}$$

By substituting Cauchy’s formula (10) and applying the divergence theorem (11) to the surface integral we get:

This must hold for every volume $V$ inside the body, no matter how small. Since the integrand is continuous, it must vanish at every point, which gives the equilibrium equations (5):

$$\sigma_{ji,j}+f_{i}=0 \tag{14}$$

Equilibrium of moments

The moment of a force $F_{k}$ acting at the point $x_{j}$ about the origin is the cross product $\varepsilon_{ijk}x_{j}F_{k}$, where $\varepsilon_{ijk}$ is the permutation symbol. The resultant moment of the surface and body forces must vanish:

$$\int_{S}\varepsilon_{ijk}x_{j}T_{k}\,\mathrm{d}S+\int_{V}\varepsilon_{ijk}x_{j}f_{k}\,\mathrm{d}V=0 \tag{15}$$

By substituting $T_{k}=\sigma_{lk}n_{l}$ and applying the divergence theorem with $g=x_{j}\sigma_{lk}$, the surface integral becomes:

where we used $x_{j,l}=\delta_{jl}$. The moment equation then becomes:

$$\int_{V}\left[\varepsilon_{ijk}\sigma_{jk}+\varepsilon_{ijk}x_{j}\left(\sigma_{lk,l}+f_{k}\right)\right]\mathrm{d}V=0 \tag{17}$$

The term in parentheses vanishes because of the equilibrium of forces, Eq. (5). Thus:

$$\int_{V}\varepsilon_{ijk}\sigma_{jk}\,\mathrm{d}V=0 \tag{18}$$

and, since the volume $V$ is arbitrary:

$$\varepsilon_{ijk}\sigma_{jk}=0 \tag{19}$$

Writing out this equation for $i=1,2,3$ gives:

$$\begin{array}{l}\varepsilon_{123}\sigma_{23}+\varepsilon_{132}\sigma_{32}=\sigma_{23}-\sigma_{32}=0\\\varepsilon_{231}\sigma_{31}+\varepsilon_{213}\sigma_{13}=\sigma_{31}-\sigma_{13}=0\\\varepsilon_{312}\sigma_{12}+\varepsilon_{321}\sigma_{21}=\sigma_{12}-\sigma_{21}=0\end{array} \tag{20}$$

which is the symmetry of the stress tensor, Eq. (9).

Two-dimensional problems

When the stresses do not depend on $x_{3}$ and $\sigma_{13}=\sigma_{23}=0$, $f_{3}=0$, as in plane stress and plane strain, the third equilibrium equation is satisfied identically and the first two simplify as follows:

$$\begin{array}{l}\dfrac{\partial \sigma_{11}}{\partial x_{1}}+\dfrac{\partial \sigma_{21}}{\partial x_{2}}+f_{1}=0\\ \dfrac{\partial \sigma_{12}}{\partial x_{1}}+\dfrac{\partial \sigma_{22}}{\partial x_{2}}+f_{2}=0\end{array} \tag{21}$$

Example

Given:

$$\begin{array}{ll}\sigma_{11}=12x_{1}x_{2}^{2}x_{3} & \sigma_{12}=4x_{1}x_{3}-12x_{2}\\ \sigma_{22}=3x_{1}^{2}x_{3}^{2}+16x_{2} & \sigma_{13}=-6x_{2}^{2}x_{3}^{2}\\ \sigma_{33}=15x_{3}-x_{1}^{3}x_{3} & \sigma_{23}=x_{1}^{3}x_{2}-2x_{3}^{2}\end{array} \tag{22}$$

where the stresses are in kPa and the coordinates in m. The body is in static equilibrium and the only body force is its weight. Taking the acceleration of gravity as $g\approx 10\ \mathrm{m/s^{2}}$, estimate the density of the material and the direction of the acceleration of gravity.

Solution

First we calculate the body forces from the equilibrium equations. Since the stress tensor is symmetric, $\sigma_{21}=\sigma_{12}$, $\sigma_{31}=\sigma_{13}$ and $\sigma_{32}=\sigma_{23}$. By using Eqs. (4) we get:

$$\begin{array}{l}\dfrac{\partial \sigma_{11}}{\partial x_{1}}+\dfrac{\partial \sigma_{21}}{\partial x_{2}}+\dfrac{\partial \sigma_{31}}{\partial x_{3}}+f_{1}=0\Rightarrow\\12x_{2}^{2}x_{3}-12-12x_{2}^{2}x_{3}+f_{1}=0\Rightarrow\\f_{1}=12\ \mathrm{kN/m^{3}}\end{array} \tag{23}$$ $$\begin{array}{l}\dfrac{\partial \sigma_{12}}{\partial x_{1}}+\dfrac{\partial \sigma_{22}}{\partial x_{2}}+\dfrac{\partial \sigma_{32}}{\partial x_{3}}+f_{2}=0\Rightarrow\\4x_{3}+16-4x_{3}+f_{2}=0\Rightarrow\\f_{2}=-16\ \mathrm{kN/m^{3}}\end{array} \tag{24}$$ $$\begin{array}{l}\dfrac{\partial \sigma_{13}}{\partial x_{1}}+\dfrac{\partial \sigma_{23}}{\partial x_{2}}+\dfrac{\partial \sigma_{33}}{\partial x_{3}}+f_{3}=0\Rightarrow\\0+x_{1}^{3}+15-x_{1}^{3}+f_{3}=0\Rightarrow\\f_{3}=-15\ \mathrm{kN/m^{3}}\end{array} \tag{25}$$

The body force is constant, as expected for the weight of a homogeneous material. The weight per unit volume is $f_{i}=\rho g_{i}$, where $g_{i}$ is the acceleration of gravity, whose magnitude is $g$. Thus, the magnitude of the body force is the unit weight of the material:

and the density is:

$$\rho=\dfrac{25\ \mathrm{kN/m^{3}}}{10\ \mathrm{m/s^{2}}}=2.5\ \mathrm{t/m^{3}} \tag{27}$$

which is a typical value for rock. The acceleration of gravity has the direction of the body force, i.e. of the unit vector:

so that $g_{i}\approx(4.8,-6.4,-6.0)\ \mathrm{m/s^{2}}$.