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 σ^\hat{\sigma} to the Cauchy strain tensor u^\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 kk giving the ratio between the force FF and extension xx:

F=kx\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 LL and cross-section AA.

The constant kk depends on, among several things, the spring’s length LL and cross-section AA, 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 FF is spread across AA, and that the extension xx should be take relative to LL.

FA=(kLA)xL\begin{aligned} \frac{F}{A} = \Big( k \frac{L}{A} \Big) \frac{x}{L} \end{aligned}

The force-per-area F/AF/A on a solid is the definition of stress, and the relative elongation x/Lx/L is the defintion of strain. If FF acts along the xx-axis, we can then write:

σxx=Euxx\begin{aligned} \boxed{ \sigma_{xx} = E u_{xx} } \end{aligned}

Where the proportionality constant EE, known as the elastic modulus or Young’s modulus, is the general material parameter that we wanted:

E=kLA\begin{aligned} E = k \frac{L}{A} \end{aligned}

Due to the microscopic structure of some (usually crystalline) materials, EE might be dependent on the direction of the force FF. 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 xx-axis, we expect it to deform in yy and zz as well. This is described by Poisson’s ratio ν\nu:

νuyyuxx\begin{aligned} \boxed{ \nu \equiv - \frac{u_{yy}}{u_{xx}} } \end{aligned}

Note that uyy=uzzu_{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=TAF = T A pulling at the rod along the xx-axis, the full stress and strain tensors are given by:

σ^=[T00000000]u^=[T/E000νT/E000νT/E]\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:

σ^=2μu^+λTr(u^)1^\begin{aligned} \boxed{ \hat{\sigma} = 2 \mu \: \hat{u} + \lambda \Tr(\hat{u}) \: \hat{1} } \end{aligned}

Where Tr\Tr{} is the trace. This is often written in index notation, with the Kronecker delta δij\delta_{ij}:

σij=2μuij+λδijkukk\begin{aligned} \boxed{ \sigma_{ij} = 2 \mu u_{ij} + \lambda \delta_{ij} \sum_{k} u_{kk} } \end{aligned}

The constants μ\mu and λ\lambda are called the Lamé coefficients, and are related to EE and ν\nu in a way we can derive by returning to the example with a tension T=F/AT = F/A along xx. For σxx\sigma_{xx}, we have:

T=σxx=2μuxx+λ(uxx+uyy+uzz)=2μET+λETνλE(T+T)=TE(2μ+λ(12ν))\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 σyy=σzz\sigma_{yy} = \sigma_{zz} are expressed in terms of the strain like so:

0=σyy=2μuyy+λ(uxx+uyy+uzz)=2νμET+λETνλE(T+T)=ET( ⁣ ⁣2νμ+λ(12ν))\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:

E=2μ+λ(12ν)2νμ=λ(12ν)\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:

λ=Eν(12ν)(1+ν)μ=E2(1+ν)\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 EE and ν\nu as a function of μ\mu and λ\lambda:

E=μ3λ+2μλ+μν=λ2(λ+μ)\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:

Tr(σ^)=iσii=2μiuii+λikukk=(2μ+3λ)iuii\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 uiju_{ij}:

σij=2μuij+λ2μ+3λδijkσkk\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 uiju_{ij}, leading us to Hooke’s inverted law:

uij=σij2μλ2μ(3λ+2μ)δijkσkk=1+νEσijνEδijkσkk\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.