Stress (mechanics) - Equilibrium Equations and Symmetry of The Stress Tensor

Equilibrium Equations and Symmetry of The Stress Tensor

When a body is in equilibrium the components of the stress tensor in every point of the body satisfy the equilibrium equations,


\sigma_{ji,j}+ F_i = 0
\,\!

For example, for a hydrostatic fluid in equilibrium conditions, the stress tensor takes on the form:

,

where is the hydrostatic pressure, and is the kronecker delta.

Derivation of equilibrium equations
Consider a continuum body (see Figure 4) occupying a volume, having a surface area, with defined traction or surface forces per unit area acting on every point of the body surface, and body forces per unit of volume on every point within the volume . Thus, if the body is in equilibrium the resultant force acting on the volume is zero, thus:

By definition the stress vector is, then

Using the Gauss's divergence theorem to convert a surface integral to a volume integral gives

For an arbitrary volume the integral vanishes, and we have the equilibrium equations

At the same time, equilibrium requires that the summation of moments with respect to an arbitrary point is zero, which leads to the conclusion that the stress tensor is symmetric, i.e.

Derivation of symmetry of the stress tensor
Summing moments about point O (Figure 4) the resultant moment is zero as the body is in equilibrium. Thus,
\begin{align}
M_O &=\int_S (\mathbf{r}\times\mathbf{T})dS + \int_V (\mathbf{r}\times\mathbf{F})dV=0 \\
0 &= \int_S\varepsilon_{ijk}x_jT_k^{(n)}dS + \int_V\varepsilon_{ijk}x_jF_k dV \\
\end{align}\,\!

where is the position vector and is expressed as

Knowing that and using Gauss's divergence theorem to change from a surface integral to a volume integral, we have

\begin{align}
0 &= \int_S \varepsilon_{ijk}x_j\sigma_{mk}n_m\, dS + \int_V\varepsilon_{ijk}x_jF_k\, dV \\
&= \int_V (\varepsilon_{ijk}x_j\sigma_{mk})_{,m} dV + \int_V\varepsilon_{ijk}x_jF_k\, dV \\
&= \int_V (\varepsilon_{ijk}x_{j,m}\sigma_{mk}+\varepsilon_{ijk}x_j\sigma_{mk,m}) dV + \int_V\varepsilon_{ijk}x_jF_k\, dV \\
&= \int_V (\varepsilon_{ijk}x_{j,m}\sigma_{mk}) dV+ \int_V \varepsilon_{ijk}x_j(\sigma_{mk,m}+F_k)dV \\
\end{align}
\,\!

The second integral is zero as it contains the equilibrium equations. This leaves the first integral, where, therefore

For an arbitrary volume V, we then have

which is satisfied at every point within the body. Expanding this equation we have

, and

or in general

This proves that the stress tensor is symmetric

However, in the presence of couple-stresses, i.e. moments per unit volume, the stress tensor is non-symmetric. This also is the case when the Knudsen number is close to one, or the continuum is a non-Newtonian fluid, which can lead to rotationally non-invariant fluids, such as polymers.

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