From 07a63237de774b3a57a0975e03cf2c6b68f165b5 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Wed, 31 Mar 2021 19:57:54 +0200
Subject: Expand knowledge base

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+---
+title: "Cauchy stress tensor"
+firstLetter: "C"
+publishDate: 2021-03-31
+categories:
+- Physics
+- Continuum physics
+
+date: 2021-03-31T09:43:33+02:00
+draft: false
+markup: pandoc
+---
+
+# 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.
+
+
+## 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-order tensor
+(i.e. a matrix that transforms in a certain way),
+called the **Cauchy stress tensor** $\hat{\sigma}$:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\sigma} =
+        \{ \sigma_{ij} \} =
+        \begin{pmatrix}
+            \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\
+            \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\
+            \sigma_{zx} & \sigma_{zy} & \sigma_{zz}
+        \end{pmatrix}
+    }
+\end{aligned}$$
+
+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 $\vec{\sigma}_i$ is the $i$th row of the tensor (implicitly transposed):
+
+$$\begin{aligned}
+    F_{s, i}
+    = \oint_S \sum_j \sigma_{ij} \dd{S_j}
+    = \int_V \sum_{j} \nabla_{\!j} \sigma_{ij} \dd{V}
+    = \int_V \nabla \cdot \vec{\sigma}_i \dd{V}
+\end{aligned}$$
+
+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.
+
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