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,

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

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

(53)\[ \bm{\varepsilon} = \dfrac{1}{2}\left(\nabla \bm{u} + \nabla \bm{u}^T \right), \]

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

(54)\[ \bm{\sigma} = \mathsf{C} \tcolon \bm{\varepsilon}. \]

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

(55)\[ \mathsf C = \begin{pmatrix} \firstlame + 2\secondlame & \firstlame & \firstlame & & & \\ \firstlame & \firstlame + 2\secondlame & \firstlame & & & \\ \firstlame & \firstlame & \firstlame + 2\secondlame & & & \\ & & & \secondlame & & \\ & & & & \secondlame & \\ & & & & & \secondlame \end{pmatrix}. \]

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 (52) as

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

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

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

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

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

where

(59)\[ \bulk_p = \frac{2 \mu \left(1 + \nu_p \right)}{3 \left(1 - 2 \nu_p \right)}, \]

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 (56) is obtained using \(\nu_p = 0\), and the deviatoric formulation (57) with \(\nu_p = -1\).