# Linear elasticity¶

For the linear elasticity model, the strain energy density is given by

$\psi = \frac{\firstlame}{2} (\trace \bm{\varepsilon})^2 + \secondlame \bm{\varepsilon} \tcolon \bm{\varepsilon} .$

The constitutive law (stress-strain relationship) is therefore given by its gradient,

(42)$\bm\sigma(\bm{u}) = \frac{\partial \psi}{\partial \bm{\varepsilon}} = \firstlame (\trace \bm\varepsilon) \bm{I} + 2 \secondlame \bm\varepsilon,$

where the colon represents a double contraction (over both indices of $$\bm{\varepsilon}$$), $$\bm{\varepsilon}$$ is (small/infintesimal) strain tensor defined by

(43)$\bm{\varepsilon} = \dfrac{1}{2}\left(\nabla \bm{u} + \nabla \bm{u}^T \right),$

and the Lamé parameters are given in terms of Young’s modulus $$E$$, and Poisson’s ratio $$\nu$$ by

\begin{aligned} \firstlame &= \frac{E \nu}{(1 + \nu)(1 - 2 \nu)}, \\ \secondlame &= \frac{E}{2(1 + \nu)}. \end{aligned}

The constitutive law (stress-strain relationship) can also be written as

(44)$\bm{\sigma} = \mathsf{C} \tcolon \bm{\varepsilon}.$

For notational convenience, we express the symmetric second order tensors $$\bm \sigma$$ and $$\bm \varepsilon$$ as vectors of length 6 using the Voigt notation. Hence, the fourth order elasticity tensor $$\mathsf C$$ (also known as elastic moduli tensor or material stiffness tensor) can be represented as

(45)$\mathsf C = \begin{pmatrix} \firstlame + 2\secondlame & \firstlame & \firstlame & & & \\ \firstlame & \firstlame + 2\secondlame & \firstlame & & & \\ \firstlame & \firstlame & \firstlame + 2\secondlame & & & \\ & & & \secondlame & & \\ & & & & \secondlame & \\ & & & & & \secondlame \end{pmatrix}.$

Note that the incompressible limit $$\nu \to \frac 1 2$$ causes $$\firstlame \to \infty$$, and thus $$\mathsf C$$ becomes singular.

## Incompressibility¶

One can see from the above equations that as $$\firstlame \to \infty$$, it is necessary that $$\nabla\cdot \bm{u} \longrightarrow 0$$ which gives an idea of alternative strategies. One approach is to define an auxiliary variable $$p$$, and rewrite constitutive equation (42) as

(46)\begin{aligned} \bm \sigma(\bm u, p) &= -p \, \bm{I} + 2 \mu \bm \varepsilon, \\ p &= - \firstlame \trace \bm \varepsilon. \end{aligned}

Alternatively, we can use the definition of hydrostatic pressure i.e., $$p_{\text{hyd}} = - \frac{\trace \bm \sigma}{3}$$ and arrive at

(47)\begin{aligned} \bm \sigma(\bm u, p_{\text{hyd}}) &= -p_{\text{hyd}} \, \bm{I} + 2 \mu \bm \varepsilon_{\text{dev}}, \\ p_{\text{hyd}} &= -\bulk \trace \bm \varepsilon. \end{aligned}

where $$\bm \varepsilon_{\text{dev}} = \bm \varepsilon - \frac{1}{3} \trace \bm \varepsilon ~ \bm{I}$$ is the deviatoric part of the linear strain tensor and $$\bulk$$ is the bulk modulus. We present a general constitutive equation as

(48)\begin{aligned} \bm \sigma(\bm u, p) &= \left(\bulk_p \trace \bm \varepsilon -p \right) \bm{I} + 2 \mu \bm \varepsilon_{\text{dev}}, \\ p &= -\left(\bulk - \bulk_p \right) \trace \bm \varepsilon. \end{aligned}

where

(49)$\bulk_p = \frac{2 \mu \left(1 + \nu_p \right)}{3 \left(1 - 2 \nu_p \right)},$

is the primal portion of the bulk modulus, defined in terms of $$\nu_p$$ with $$-1 \leq \nu_p < \nu$$, where $$\nu$$ is the physical Poisson’s ratio. The standard full-train formulation (46) is obtained using $$\nu_p = 0$$, and the deviatoric formulation (47) with $$\nu_p = -1$$.