path: root/source/know/concept/cauchy-stress-tensor/index.md

---
title: "Cauchy stress tensor"
sort_title: "Cauchy stress tensor"
date: 2021-03-31
categories:
- Physics
- Continuum physics
layout: "concept"
---

Roughly speaking, **stress** is the solid equivalent of fluid pressure:
it describes the net force acting on an imaginary partition surface inside a solid.
However, unlike fluids at rest,
where the pressure is always perpendicular to such a surface,
solid stress is usually much more complicated.

Formally, the concept of stress can be applied to any continuum
(not just solids), including fluids,
but it is arguably most intuitive for solids.


## Definition

In the solid, imagine an infinitesimal cube
whose sides, $$\dd{S}_x$$, $$\dd{S}_y$$ and $$\dd{S}_z$$,
are orthogonal to the $$x$$, $$y$$ and $$z$$ axes, respectively.
There is a force $$\dd{\va{F}}_1$$ acting on $$\dd{S}_x$$,
$$\dd{\va{F}}_2$$ on $$\dd{S}_y$$, and $$\dd{\va{F}}_3$$ on $$\dd{S}_z$$.
Then we can decompose each of these forces, for example:

$$\begin{aligned}
    \dd{\va{F}}_1
    = \va{e}_x F_{x1} + \va{e}_y F_{y1} + \va{e}_z F_{z1}
\end{aligned}$$

Where $$\va{e}_x$$, $$\va{e}_y$$ and $$\va{e}_z$$ are the basis unit vectors.
If we divide each of the force components by the area $$\dd{S}_x$$
(like in a fluid, in order to get the pressure),
we find the stresses $$\sigma_{xx}$$, $$\sigma_{yx}$$ and $$\sigma_{zx}$$
that are being "felt" by the $$x$$ surface element $$\dd{S}_x$$:

$$\begin{aligned}
    \dd{\va{F}}_1
    = \big( \va{e}_x \sigma_{xx}  + \va{e}_y \sigma_{yx} + \va{e}_z \sigma_{zx} \big) \dd{S}_x
\end{aligned}$$

The perpendicular component $$\sigma_{xx}$$ is called a **tensile stress**,
and its sign is always chosen so that a positive value corresponds to a tension,
i.e. the $$x$$-side is pulled away from the rest of the cube.
The tangential components $$\sigma_{yx}$$ and $$\sigma_{zx}$$
are called **shear stresses**.

Evidently, the other two forces $$\dd{\va{F}}_2$$ and $$\dd{\va{F}}_3$$
can be decomposed in the exact same way,
yielding nine stress components in total:

$$\begin{aligned}
    \dd{\va{F}}_2
    &= \va{e}_x F_{x2} + \va{e}_y F_{y2} + \va{e}_z F_{z2}
    = \big( \va{e}_x \sigma_{xy}  + \va{e}_y \sigma_{yy} + \va{e}_z \sigma_{zy} \big) \dd{S}_y
    \\
    \dd{\va{F}}_3
    &= \va{e}_x F_{x3} + \va{e}_y F_{y3} + \va{e}_z F_{z3}
    = \big( \va{e}_x \sigma_{xz}  + \va{e}_y \sigma_{yz} + \va{e}_z \sigma_{zz} \big) \dd{S}_z
\end{aligned}$$

The total force $$\dd{\va{F}}$$ on the entire infinitesimal cube
is simply the sum of the previous three:

$$\begin{aligned}
    \dd{\va{F}}
    = \dd{\va{F}}_1 + \dd{\va{F}}_2 + \dd{\va{F}}_3
\end{aligned}$$

We can then decompose $$\dd{\va{F}}$$ into its net components
along the $$x$$, $$y$$ and $$z$$ axes:

$$\begin{aligned}
    \dd{\va{F}}
    = \va{e}_x \dd{F}_x + \va{e}_y \dd{F}_y + \va{e}_z \dd{F}_z
\end{aligned}$$

From the preceding equations, we find that these components are given by:

$$\begin{aligned}
    \dd{F}_x
    &= \sigma_{xx} \dd{S}_x + \sigma_{xy} \dd{S}_y + \sigma_{xz} \dd{S}_z
    \\
    \dd{F}_y
    &= \sigma_{yx} \dd{S}_x + \sigma_{yy} \dd{S}_y + \sigma_{yz} \dd{S}_z
    \\
    \dd{F}_z
    &= \sigma_{zx} \dd{S}_x + \sigma_{zy} \dd{S}_y + \sigma_{zz} \dd{S}_z
\end{aligned}$$

We can write this much more compactly using index notation,
where $$i, j \in \{x, y, z\}$$:

$$\begin{aligned}
    \boxed{
        \dd{F}_i
        = \sum_{j} \sigma_{ij} \dd{S}_j
    }
\end{aligned}$$

The stress components $$\sigma_{ij}$$ can be written as a second-rank tensor
(i.e. a matrix that transforms in a certain way),
called the **Cauchy stress tensor** $$\hat{\sigma}$$:

$$\begin{aligned}
    \boxed{
        \hat{\sigma} \equiv
        \{ \sigma_{ij} \} =
        \begin{pmatrix}
            \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\
            \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\
            \sigma_{zx} & \sigma_{zy} & \sigma_{zz}
        \end{pmatrix}
    }
\end{aligned}$$

Then $$\dd{\va{F}}$$ is written even more compactly
using the dot product, with $$\dd{\va{S}} = (\dd{S}_x, \dd{S}_y, \dd{S}_z)$$:

$$\begin{aligned}
    \boxed{
        \dd{\va{F}}
        = \hat{\sigma} \cdot \dd{\va{S}}
    }
\end{aligned}$$

All forces on the cube's sides can be written in this form.
**Cauchy's stress theorem** states that the force on *any*
surface element inside the solid can be written like this,
simply by projecting it onto the $$x$$, $$y$$ and $$z$$ zero-planes
to get the areas $$\dd{S}_x$$, $$\dd{S}_y$$ and $$\dd{S}_z$$.

Note that for fluids, the pressure $$p$$ was defined
such that $$\dd{\va{F}} = - p \dd{\va{S}}$$.
If we wanted to define $$p$$ for solids in the same way,
we would need $$\hat{\sigma}$$ to be diagonal *and*
all of its diagonal elements to be identical.
Since this is almost never the case,
the scalar pressure is ill-defined in solids.


## Equilibrium

The total force $$\va{F}$$ acting on a (non-infinitesimal) volume $$V$$ of the solid
is given by the sum of the total body force $$\va{F}_b$$ and total surface force $$\va{F}_s$$,
where $$\vec{f}$$ is the body force density:

$$\begin{aligned}
    \va{F}
    = \va{F}_b + \va{F}_s
    = \int_V \va{f} \dd{V} + \oint_S \hat{\sigma} \cdot \dd{\va{S}}
\end{aligned}$$

We can rewrite the surface term using the divergence theorem,
where $$\top$$ is the transpose:

$$\begin{aligned}
    \va{F}_s
    = \oint_S \hat{\sigma} \cdot \dd{\va{S}}
    = \int_V \nabla \cdot \hat{\sigma}^{\top} \dd{V}
\end{aligned}$$

For some people, this equation may be more enlightening in index notation,
where $$\nabla_j \equiv \ipdv{}{x_j}$$ is the partial derivative with respect to the $$j$$th coordinate:

$$\begin{aligned}
    F_{s, i}
    = \oint_S \sum_j \sigma_{ij} \dd{S_j}
    = \int_V \sum_{j} \nabla_{\!j} \sigma_{ij} \dd{V}
\end{aligned}$$

In any case, the total force $$\va{F}$$ can then be expressed
as a single volume integral over $$V$$:

$$\begin{aligned}
    \va{F}
    = \int_V \va{f} \dd{V} + \int_V \nabla \cdot \hat{\sigma}^{\top} \dd{V}
    = \int_V \va{f^*} \dd{V}
\end{aligned}$$

Where we have defined the **effective force density** $$\va{f^*}$$ as follows:

$$\begin{aligned}
    \boxed{
        \va{f^*}
        = \va{f} + \nabla \cdot \hat{\sigma}^{\top}
    }
\end{aligned}$$

The volume $$V$$ is in **mechanical equilibrium** if the net force acting on it amounts to zero:

$$\begin{aligned}
    \va{F}
    = 0
\end{aligned}$$

However, because $$V$$ is abritrary, the equilibrium condition for the whole solid is in fact:

$$\begin{aligned}
    \boxed{
        \va{f^*}
        = 0
    }
\end{aligned}$$

This is reminiscent of the equilibrium condition of a fluid
(see [hydrostatic pressure](/know/concept/hydrostatic-pressure/)).
Note that it is a set of coupled differential equations,
which needs boundary conditions at the object's surface.
Newton's third law states that the two sides of the boundary
exert opposite forces on each other,
so the boundary condition is continuity of the **stress vector**
$$\hat{\sigma} \cdot \va{n}$$:

$$\begin{aligned}
    \boxed{
        \hat{\sigma}_{\mathrm{outer}} \cdot \va{n}
        = - \hat{\sigma}_{\mathrm{inner}} \cdot \va{n}
    }
\end{aligned}$$

Where the normal of the outer surface is $$\va{n}$$,
and the normal of the inner surface is $$-\va{n}$$.
Note that the above equation does *not* mean
that $$-\hat{\sigma}_{\mathrm{inner}}$$ equals $$\hat{\sigma}_{\mathrm{outer}}$$:
the tensors are allowed to be very different,
as long as the stress vector's three components are equal.


## References
1.  B. Lautrup,
    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
    CRC Press.