Equilibrium equations

A solid body is in static equilibrium when the resultant force and moment on each axis is equal to zero. This can be expressed by the equilibrium equations. In this article we will prove the equilibrium equations by calculating the resultant force and moment on each axis. A more elegant solution may be derived by using Gauss’s theorem and Cauchy’s formula. This approach may be found in international bibliography.

Consider a solid body in static equilibrium that neither moves nor rotates. Surface and body forces act on this body. We cut an infinitesimal parallelepiped inside the body and we analyze the forces that act on it as shown in Fig. 1. We will assume that the stress field is continuous and differentiable inside the whole body.

Figure 1: Infinitesimal parallelepiped representing a point in a body under static equilibrium. The stresses acting on the opposite sides of the cube are slightly different.

The stress components on each side of the cube is a function of the position since we have a non uniform but continuous stress field. For example on side \( 4 \) the normal stress is \( \sigma_{11}(x_{1},x_{2},x_{3})=\sigma_{11} \). On the opposite side \( 2 \) the normal stress is \( \sigma_{11}(x_{1}+\mathrm{d}x_{1},x_{2},x_{3}) \). By taking under consideration Taylor’s theorem we may write:

\[ \sigma_{11}(x_{1}+dx,x_{2},x_{3})=\sigma_{11}+\dfrac{\partial\sigma_{11}}{\partial x_{1}}dx_{1} \] (1)

where the higher order terms have been neglected because they are relatively small. We follow the same procedure for all the components as shown if Fig. 1.







Equilibrium of the body demands that the resultant forces must vanish. By summing up the forces with direction parallel to axis \( x_{1} \) we get:

\[ \begin{array}{l}\left(\sigma_{11}+\dfrac{\partial\sigma_{11}}{\partial x_{1}}\mathrm{d}x_{1}\right)\mathrm{d}x_{2}\mathrm{d}x_{3}-\sigma_{11}\mathrm{d}x_{2}\mathrm{d}x_{3}+\\+\left(\sigma_{21}+\dfrac{\partial\sigma_{21}}{\partial x_{2}}\mathrm{d}x_{2}\right)\mathrm{d}x_{1}\mathrm{d}x_{3}-\sigma_{21}\mathrm{d}x_{1}\mathrm{d}x_{3}+\\+\left(\sigma_{31}+\dfrac{\partial\sigma_{31}}{\partial x_{3}}\mathrm{d}x_{3}\right)\mathrm{d}x_{1}\mathrm{d}x_{2}-\sigma_{31}\mathrm{d}x_{1}\mathrm{d}x_{2}+f_{1}\mathrm{d}x_{1}\mathrm{d}x_{2}\mathrm{d}x_{3}=0\end{array} \] (2)

where \( \mathrm{d}x_{1} \), \( \mathrm{d}x_{2} \) and \( \mathrm{d}x_{3} \) are the dimensions of the parallelepiped and \( f_{1} \) is the component of the body force parallel to \( x_{1} \). By dividing with \( \mathrm{d}x_{1}\mathrm{d}x_{2}\mathrm{d}x_{3} \) we get:

\[ \dfrac{\partial\sigma_{11}}{\partial x_{1}}+\dfrac{\partial\sigma_{21}}{\partial x_{2}}+\dfrac{\partial\sigma_{31}}{\partial x_{3}}+f_{1}=0 \] (3)

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

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

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

\[ \sigma_{ji,j}+f_{i}=0 \] (5)

The resultant moment on each axis must also vanish. By taking under consideration all the forces that contribute to moment about axis \( x_{3} \) we may write:

\[ \begin{array}{l}-\left(\sigma_{11}+\dfrac{\partial\sigma_{11}}{\partial x_{1}}\mathrm{d}x_{1}\right)\mathrm{d}x_{2}\mathrm{d}x_{3}\dfrac{\mathrm{d}x_{2}}{2}+\sigma_{11}\mathrm{d}x_{2}\mathrm{d}x_{3}\dfrac{\mathrm{d}x_{2}}{2}+\\+\left(\sigma_{22}+\dfrac{\partial\sigma_{22}}{\partial x_{2}}\mathrm{d}x_{2}\right)\mathrm{d}x_{1}\mathrm{d}x_{3}\dfrac{\mathrm{d}x_{1}}{2}-\sigma_{22}\mathrm{d}x_{1}\mathrm{d}x_{3}\dfrac{\mathrm{d}x_{1}}{2}+\\+\left(\sigma_{12}+\dfrac{\partial\sigma_{12}}{\partial x_{1}}\mathrm{d}x_{1}\right)\mathrm{d}x_{2}\mathrm{d}x_{3}\mathrm{d}x_{1}-\left(\sigma_{21}+\dfrac{\partial\sigma_{21}}{\partial x_{2}}\mathrm{d}x_{2}\right)\mathrm{d}x_{1}\mathrm{d}x_{3}\mathrm{d}x_{2}-\\-\left(\sigma_{31}+\dfrac{\partial\sigma_{31}}{\partial x_{3}}\mathrm{d}x_{3}\right)\mathrm{d}x_{1}\mathrm{d}x_{2}\dfrac{\mathrm{d}x_{2}}{2}+\sigma_{31}\mathrm{d}x_{1}\mathrm{d}x_{2}\dfrac{\mathrm{d}x_{2}}{2}+\\+\left(\sigma_{32}+\dfrac{\partial\sigma_{32}}{\partial x_{3}}\mathrm{d}x_{3}\right)\mathrm{d}x_{1}\mathrm{d}x_{2}\dfrac{\mathrm{d}x_{1}}{2}-\sigma_{32}\mathrm{d}x_{1}\mathrm{d}x_{2}\dfrac{\mathrm{d}x_{1}}{2}-\\-f_{1}\mathrm{d}x_{1}\mathrm{d}x_{2}\mathrm{d}x_{3}\dfrac{\mathrm{d}x_{2}}{2}+f_{2}\mathrm{d}x_{1}\mathrm{d}x_{2}\mathrm{d}x_{3}\dfrac{\mathrm{d}x_{1}}{2}=0\end{array} \] (6)

by dividing with \( \mathrm{d}x_{1}\mathrm{d}x_{2}\mathrm{d}x_{3}\) and taking the limit \( \mathrm{d}x_{1}\to 0\), \( \mathrm{d}x_{2}\to 0\) and \( \mathrm{d}x_{3}\to 0\) we derive:

\[ \sigma_{12}=\sigma_{21} \] (7)

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

\[ \sigma_{ij}=\sigma_{ji} \] (8)

For the case of two dimensional problems, equilibrium equations simplify as follows:

\[ \begin{array}{l}\dfrac{\partial \sigma_{11}}{\partial x_{1}}+\dfrac{\partial \sigma_{21}}{\partial x_{2}}+f_{1}=0\\ \dfrac{\partial \sigma_{12}}{\partial x_{1}}+\dfrac{\partial \sigma_{22}}{\partial x_{2}}+f_{2}=0\end{array} \] (9)







Exercise

\[ \begin{array}{ccc}\sigma_{11}=12x_{1}^{2}x_{2}^{2}x_{3} & \sigma_{22}=-9x_{1}^{3}x_{2}^{2} & \sigma_{33}=4x_{2}^{2}x_{3}^{3}\\ \sigma_{12}=4x_{2}^{3} & \sigma_{13}=-12x_{1}x_{2}^{2}x_{3}^{2} & \sigma_{23}=16x_{1}^{3}x_{2}x_{3}\end{array} \] (10)

Calculate the body forces in order to achieve static equilibrium.

Show solution…