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diff --git a/source/know/concept/hookes-law/index.md b/source/know/concept/hookes-law/index.md index 87e04de..292b735 100644 --- a/source/know/concept/hookes-law/index.md +++ b/source/know/concept/hookes-law/index.md @@ -12,8 +12,8 @@ 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/) $\hat{u}$. +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. @@ -22,8 +22,8 @@ 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 $x$: +since they have a spring constant $$k$$ giving the ratio +between the force $$F$$ and extension $$x$$: $$\begin{aligned} \boxed{ @@ -35,23 +35,23 @@ $$\begin{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 $A$. +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 $A$, +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 $L$. +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: +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: $$\begin{aligned} \boxed{ @@ -60,7 +60,7 @@ $$\begin{aligned} } \end{aligned}$$ -Where the proportionality constant $E$, +Where the proportionality constant $$E$$, known as the **elastic modulus** or **Young's modulus**, is the general material parameter that we wanted: @@ -70,7 +70,7 @@ $$\begin{aligned} \end{aligned}$$ Due to the microscopic structure of some (usually crystalline) materials, -$E$ might be dependent on the direction of the force $F$. +$$E$$ might be dependent on the direction of the force $$F$$. For simplicity, we only consider **isotropic** materials, which have the same properties measured from any direction. @@ -78,8 +78,8 @@ However, we are still missing something. When a spring is pulled, it becomes narrower as its coils move apart, and this effect is also seen when stretching solids in general: -if we pull our rod along the $x$-axis, we expect it to deform in $y$ and $z$ as well. -This is described by **Poisson's ratio** $\nu$: +if we pull our rod along the $$x$$-axis, we expect it to deform in $$y$$ and $$z$$ as well. +This is described by **Poisson's ratio** $$\nu$$: $$\begin{aligned} \boxed{ @@ -88,12 +88,12 @@ $$\begin{aligned} } \end{aligned}$$ -Note that $u_{yy} = u_{zz}$ because the material is assumed to be isotropic. +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: +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: $$\begin{aligned} \hat{\sigma} = @@ -124,9 +124,9 @@ $$\begin{aligned} } \end{aligned}$$ -Where $\Tr{}$ is the trace. +Where $$\Tr{}$$ is the trace. This is often written in index notation, -with the Kronecker delta $\delta_{ij}$: +with the Kronecker delta $$\delta_{ij}$$: $$\begin{aligned} \boxed{ @@ -135,10 +135,10 @@ $$\begin{aligned} } \end{aligned}$$ -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: +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: $$\begin{aligned} T @@ -150,7 +150,7 @@ $$\begin{aligned} &= \frac{T}{E} \Big( 2 \mu + \lambda (1 - 2 \nu) \Big) \end{aligned}$$ -Meanwhile, the other diagonal stresses $\sigma_{yy} = \sigma_{zz}$ +Meanwhile, the other diagonal stresses $$\sigma_{yy} = \sigma_{zz}$$ are expressed in terms of the strain like so: $$\begin{aligned} @@ -188,7 +188,7 @@ $$\begin{aligned} \end{aligned}$$ Which can straightforwardly be inverted -to express $E$ and $\nu$ as a function of $\mu$ and $\lambda$: +to express $$E$$ and $$\nu$$ as a function of $$\mu$$ and $$\lambda$$: $$\begin{aligned} \boxed{ @@ -212,14 +212,14 @@ $$\begin{aligned} \end{aligned}$$ Inserting this into Hooke's law -yields an equation that only contains one strain component $u_{ij}$: +yields an equation that only contains one strain component $$u_{ij}$$: $$\begin{aligned} \sigma_{ij} = 2 \mu u_{ij} + \frac{\lambda}{2 \mu + 3 \lambda} \delta_{ij} \sum_{k} \sigma_{kk} \end{aligned}$$ -Which is therefore trivial to isolate for $u_{ij}$, +Which is therefore trivial to isolate for $$u_{ij}$$, leading us to Hooke's inverted law: $$\begin{aligned} |