Categories: Continuum physics, Physics.

# Hooke’s law

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 $\hat{\sigma}$ to the 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}
1. B. Lautrup, Physics of continuous matter: exotic and everyday phenomena in the macroscopic world, 2nd edition, CRC Press.