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diff --git a/content/know/concept/cauchy-stress-tensor/index.pdc b/content/know/concept/cauchy-stress-tensor/index.pdc new file mode 100644 index 0000000..6c7c97a --- /dev/null +++ b/content/know/concept/cauchy-stress-tensor/index.pdc @@ -0,0 +1,245 @@ +--- +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. + |