Categories: Continuum physics, Physics.

Cauchy stress tensor

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.


In the solid, imagine an infinitesimal cube whose sides, dSx\dd{S}_x, dSy\dd{S}_y and dSz\dd{S}_z, are orthogonal to the xx, yy and zz axes, respectively. There is a force dF1\dd{\va{F}}_1 acting on dSx\dd{S}_x, dF2\dd{\va{F}}_2 on dSy\dd{S}_y, and dF3\dd{\va{F}}_3 on dSz\dd{S}_z. Then we can decompose each of these forces, for example:

dF1=exFx1+eyFy1+ezFz1\begin{aligned} \dd{\va{F}}_1 = \va{e}_x F_{x1} + \va{e}_y F_{y1} + \va{e}_z F_{z1} \end{aligned}

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

dF1=(exσxx+eyσyx+ezσzx)dSx\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 σxx\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 xx-side is pulled away from the rest of the cube. The tangential components σyx\sigma_{yx} and σzx\sigma_{zx} are called shear stresses.

Evidently, the other two forces dF2\dd{\va{F}}_2 and dF3\dd{\va{F}}_3 can be decomposed in the exact same way, yielding nine stress components in total:

dF2=exFx2+eyFy2+ezFz2=(exσxy+eyσyy+ezσzy)dSydF3=exFx3+eyFy3+ezFz3=(exσxz+eyσyz+ezσzz)dSz\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 dF\dd{\va{F}} on the entire infinitesimal cube is simply the sum of the previous three:

dF=dF1+dF2+dF3\begin{aligned} \dd{\va{F}} = \dd{\va{F}}_1 + \dd{\va{F}}_2 + \dd{\va{F}}_3 \end{aligned}

We can then decompose dF\dd{\va{F}} into its net components along the xx, yy and zz axes:

dF=exdFx+eydFy+ezdFz\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:

dFx=σxxdSx+σxydSy+σxzdSzdFy=σyxdSx+σyydSy+σyzdSzdFz=σzxdSx+σzydSy+σzzdSz\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{x,y,z}i, j \in \{x, y, z\}:

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

The stress components σij\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 σ^\hat{\sigma}:

σ^{σij}=(σxxσxyσxzσyxσyyσyzσzxσzyσzz)\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 dF\dd{\va{F}} is written even more compactly using the dot product, with dS=(dSx,dSy,dSz)\dd{\va{S}} = (\dd{S}_x, \dd{S}_y, \dd{S}_z):

dF=σ^dS\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 xx, yy and zz zero-planes to get the areas dSx\dd{S}_x, dSy\dd{S}_y and dSz\dd{S}_z.

Note that for fluids, the pressure pp was defined such that dF=pdS\dd{\va{F}} = - p \dd{\va{S}}. If we wanted to define pp for solids in the same way, we would need σ^\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.


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

F=Fb+Fs=VfdV+Sσ^dS\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:

Fs=Sσ^dS=Vσ^dV\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 j/xj\nabla_j \equiv \ipdv{}{x_j} is the partial derivative with respect to the jjth coordinate:

Fs,i=SjσijdSj=Vj ⁣jσijdV\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 F\va{F} can then be expressed as a single volume integral over VV:

F=VfdV+Vσ^dV=VfdV\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 f\va{f^*} as follows:

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

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

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

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

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

This is reminiscent of the equilibrium condition of a fluid (see 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 σ^n\hat{\sigma} \cdot \va{n}:

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


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