From 6ce0bb9a8f9fd7d169cbb414a9537d68c5290aae Mon Sep 17 00:00:00 2001 From: Prefetch Date: Fri, 14 Oct 2022 23:25:28 +0200 Subject: Initial commit after migration from Hugo --- source/know/concept/hookes-law/index.md | 239 ++++++++++++++++++++++++++++++++ 1 file changed, 239 insertions(+) create mode 100644 source/know/concept/hookes-law/index.md (limited to 'source/know/concept/hookes-law') diff --git a/source/know/concept/hookes-law/index.md b/source/know/concept/hookes-law/index.md new file mode 100644 index 0000000..57ab27f --- /dev/null +++ b/source/know/concept/hookes-law/index.md @@ -0,0 +1,239 @@ +--- +title: "Hooke's law" +date: 2021-04-02 +categories: +- Physics +- Continuum physics +layout: "concept" +--- + +In its simplest form, **Hooke's law** dictates that +changing the length of an elastic object requires +a force that is proportional the desired length difference. +In its most general form, it gives a linear relationship +between the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$ +to the [Cauchy strain tensor](/know/concept/cauchy-strain-tensor/) $\hat{u}$. + +Importantly, all forms of Hooke's law are only valid for small deformations, +since the stress-strain relationship becomes nonlinear otherwise. + + +## Simple form + +The simple form of the law is traditionally quoted for springs, +since they have a spring constant $k$ giving the ratio +between the force $F$ and extension $x$: + +$$\begin{aligned} + \boxed{ + F + = k x + } +\end{aligned}$$ + +In general, all solids are elastic for small extensions, +and therefore also obey Hooke's law. +In light of this fact, we replace the traditional spring +with a rod of length $L$ and cross-section $A$. + +The constant $k$ depends on, among several things, +the spring's length $L$ and cross-section $A$, +so for our generalization, we want a new parameter +to describe the proportionality independently of the rod's dimensions. +To achieve this, we realize that the force $F$ is spread across $A$, +and that the extension $x$ should be take relative to $L$. + +$$\begin{aligned} + \frac{F}{A} + = \Big( k \frac{L}{A} \Big) \frac{x}{L} +\end{aligned}$$ + +The force-per-area $F/A$ on a solid is the definition of **stress**, +and the relative elongation $x/L$ is the defintion of **strain**. +If $F$ acts along the $x$-axis, we can then write: + +$$\begin{aligned} + \boxed{ + \sigma_{xx} + = E u_{xx} + } +\end{aligned}$$ + +Where the proportionality constant $E$, +known as the **elastic modulus** or **Young's modulus**, +is the general material parameter that we wanted: + +$$\begin{aligned} + E + = k \frac{L}{A} +\end{aligned}$$ + +Due to the microscopic structure of some (usually crystalline) materials, +$E$ might be dependent on the direction of the force $F$. +For simplicity, we only consider **isotropic** materials, +which have the same properties measured from any direction. + +However, we are still missing something. +When a spring is pulled, +it becomes narrower as its coils move apart, +and this effect is also seen when stretching solids in general: +if we pull our rod along the $x$-axis, we expect it to deform in $y$ and $z$ as well. +This is described by **Poisson's ratio** $\nu$: + +$$\begin{aligned} + \boxed{ + \nu + \equiv - \frac{u_{yy}}{u_{xx}} + } +\end{aligned}$$ + +Note that $u_{yy} = u_{zz}$ because the material is assumed to be isotropic. +Intuitively, you may expect that the volume of the object is conserved, +but for most materials that is not accurate. + +In summary, for our example case with a force $F = T A$ pulling at the rod +along the $x$-axis, the full stress and strain tensors are given by: + +$$\begin{aligned} + \hat{\sigma} = + \begin{bmatrix} + T & 0 & 0 \\ + 0 & 0 & 0 \\ + 0 & 0 & 0 + \end{bmatrix} + \qquad + \hat{u} = + \begin{bmatrix} + T/E & 0 & 0 \\ + 0 & -\nu T/E & 0 \\ + 0 & 0 & -\nu T/E + \end{bmatrix} +\end{aligned}$$ + + +## General isotropic form + +The general form of Hooke's law is a linear relationship +between the stress and strain tensors: + +$$\begin{aligned} + \boxed{ + \hat{\sigma} + = 2 \mu \: \hat{u} + \lambda \Tr(\hat{u}) \: \hat{1} + } +\end{aligned}$$ + +Where $\Tr{}$ is the trace. +This is often written in index notation, +with the Kronecker delta $\delta_{ij}$: + +$$\begin{aligned} + \boxed{ + \sigma_{ij} + = 2 \mu u_{ij} + \lambda \delta_{ij} \sum_{k} u_{kk} + } +\end{aligned}$$ + +The constants $\mu$ and $\lambda$ are called the **Lamé coefficients**, +and are related to $E$ and $\nu$ in a way we can derive +by returning to the example with a tension $T = F/A$ along $x$. +For $\sigma_{xx}$, we have: + +$$\begin{aligned} + T + = \sigma_{xx} + &= 2 \mu u_{xx} + \lambda (u_{xx} + u_{yy} + u_{zz}) + \\ + &= \frac{2 \mu}{E} T + \frac{\lambda}{E} T - \frac{\nu \lambda}{E} (T + T) + \\ + &= \frac{T}{E} \Big( 2 \mu + \lambda (1 - 2 \nu) \Big) +\end{aligned}$$ + +Meanwhile, the other diagonal stresses $\sigma_{yy} = \sigma_{zz}$ +are expressed in terms of the strain like so: + +$$\begin{aligned} + 0 + = \sigma_{yy} + &= 2 \mu u_{yy} + \lambda (u_{xx} + u_{yy} + u_{zz}) + \\ + &= - \frac{2 \nu \mu}{E} T + \frac{\lambda}{E} T - \frac{\nu \lambda}{E} (T + T) + \\ + &= \frac{E}{T} \Big( \!-\! 2 \nu \mu + \lambda (1 - 2 \nu) \Big) +\end{aligned}$$ + +After dividing out superfluous factors from the two preceding equations, +we arrive at: + +$$\begin{aligned} + E + = 2 \mu + \lambda (1 - 2 \nu) + \qquad \quad + 2 \nu \mu + = \lambda (1 - 2 \nu) +\end{aligned}$$ + +Solving this system of equations for the Lamé coefficients +yields the following result: + +$$\begin{aligned} + \boxed{ + \lambda + = \frac{E \nu}{(1 - 2 \nu)(1 + \nu)} + \qquad \quad + \mu + = \frac{E}{2 (1 + \nu)} + } +\end{aligned}$$ + +Which can straightforwardly be inverted +to express $E$ and $\nu$ as a function of $\mu$ and $\lambda$: + +$$\begin{aligned} + \boxed{ + E + = \mu \frac{3 \lambda + 2 \mu}{\lambda + \mu} + \qquad \quad + \nu + = \frac{\lambda}{2 (\lambda + \mu)} + } +\end{aligned}$$ + +Hooke's law itself can also be inverted, +i.e. we can express the strain as a function of stress. +First, observe that the trace of the stress tensor satisfies: + +$$\begin{aligned} + \Tr(\hat{\sigma}) + = \sum_{i} \sigma_{ii} + = 2 \mu \sum_{i} u_{ii} + \lambda \sum_{i} \sum_{k} u_{kk} + = (2 \mu + 3 \lambda) \sum_{i} u_{ii} +\end{aligned}$$ + +Inserting this into Hooke's law +yields an equation that only contains one strain component $u_{ij}$: + +$$\begin{aligned} + \sigma_{ij} + = 2 \mu u_{ij} + \frac{\lambda}{2 \mu + 3 \lambda} \delta_{ij} \sum_{k} \sigma_{kk} +\end{aligned}$$ + +Which is therefore trivial to isolate for $u_{ij}$, +leading us to Hooke's inverted law: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + u_{ij} + &= \frac{\sigma_{ij}}{2 \mu} - \frac{\lambda}{2 \mu (3 \lambda + 2 \mu)} \delta_{ij} \sum_{k} \sigma_{kk} + \\ + &= \frac{1 + \nu}{E} \sigma_{ij} - \frac{\nu}{E} \delta_{ij} \sum_{k} \sigma_{kk} + \end{aligned} + } +\end{aligned}$$ + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. -- cgit v1.2.3