Chapter 7
Hooke’s law
7.1 General form
In order to compute the deformation of an elastic body, we need a constitutive relation (material law) to close the equations of elastostatic equilibrium, \begin {equation} \nabla \cdot \t {\sigma } = 0 \quad \text {and} \quad \t {\varepsilon }(\v {r}) = \frac {1}{2} \left (\nabla \v {u} + (\nabla \v {u})^T\right ). \end {equation} Since we will be working in linear elasticity, the constitutive equation is a linear relationship between \(\t {\sigma }\) and \(\t {\varepsilon }\). The most general form of this linear relationship is \begin {equation} \label {eqn:Hookeslaw} \t {\sigma } = \tt {C} : \t {\varepsilon }, \text {or using Einstein summation} \ \sigma _{ij} = C_{ijkl} \varepsilon _{kl}. \end {equation} It is called Hooke’s law. The quantity \(\tt {C}\) is a fourth order symmetric tensor of elastic constants that contains at most 21 independent elastic moduli. To see that there are only \(21\) independent coefficients, it is useful to remove the symmetric entries from \(\t {\sigma }\) and \(\t {\varepsilon }\) and express them as 6-vectors in what is called Voigt notation, \begin {equation} \label {eq:Voigtstress} \v {\sigma } = (\sigma _{xx},\sigma _{yy},\sigma _{zz},\sigma _{yz},\sigma _{xz},\sigma _{xy}) \end {equation} and \begin {equation} \label {eq:Voigtstrain} \v {\varepsilon } = (\varepsilon _{xx},\varepsilon _{yy},\varepsilon _{zz},2\varepsilon _{yz},2\varepsilon _{xz},2\varepsilon _{xy}). \end {equation} Then \(\sigma =\t {C} \cdot \v {\varepsilon }\) where \(\t {C}\) is a \(6\times 6\) symmetric matrix called the stiffness matrix containing the above-mentioned 21 independent elastic constants. (There are \(6\cdot 6=36\) components, but the matrix is symmetric.)
Note that the off-diagonal components of \(\t {\sigma }\) are often denoted by \(\tau _{xy} \equiv \sigma _{xy}, \tau _{xz} \equiv \sigma _{xz}\ \text {and}\ \tau _{yz} \equiv \sigma _{yz}\). The off-diagonal components of \(\t {\varepsilon }\) are often denoted by \(\gamma _{xy} \equiv 2\varepsilon _{xy}, \gamma _{xz} \equiv 2\varepsilon _{xz}\ \text {and}\ \gamma _{yz} \equiv 2\varepsilon _{yz}\) and absorb the factor of 2 that occurs in Eq. \eqref{eq:Voigtstrain}. Voigt notation then becomes \begin {equation} \label {eq:Voigtstressgamma} \v {\sigma } = (\sigma _{xx},\sigma _{yy},\sigma _{zz},\tau _{yz},\tau _{xz},\tau _{xy}) \end {equation} and \begin {equation} \label {eq:Voigtstraingamma} \v {\varepsilon } = (\varepsilon _{xx},\varepsilon _{yy},\varepsilon _{zz},\gamma _{yz},\gamma _{xz},\gamma _{xy}). \end {equation}
Note: It is important to keep in mind that the \(\gamma \)’s contain a factor 2 but the \(\tau \)’s do not. The factor of 2 ensures that \(\v {\sigma }=\t {C} \cdot \v {\varepsilon }\) and \(\t {\sigma } = \tt {C} : \t {\varepsilon }\) are the same constitutive law.
7.2 Isotropic solids
For isotropic elasticity, the total 21 independent elastic constants reduce to two. The constitutive equation for isotropic elasticity is \begin {equation} \label {eq:isotropicelasticity} \sigma _{ij} = \lambda \delta _{ij} \varepsilon _{kk} + 2\mu \varepsilon _ {ij} \end {equation} or its inverse \begin {equation} \label {eq:isotropicelasticityinverse} \varepsilon _{ij} = \frac {1}{2G} \sigma _{ij} - \frac {\nu }{E} \delta _{ij} \delta _{kk} = \frac {1}{E} [(1+\nu )\sigma _{ij} - \nu \delta _{ij} \sigma _{kk}], \end {equation} where \(\delta _{ij}\) is the Kronecker-Delta. These expressions have been conveniently written in their most simple form. The constants that show up in Eqs. \eqref{eq:isotropicelasticity} and \eqref{eq:isotropicelasticityinverse} are the shear modulus \(\mu \), Lamé’s first constant \(\lambda \), Young’s modulus \(E\) and Poisson number \(\nu \). Both \(\lambda \) and \(\nu \) are often called Lamé’s constants. Note that \(\sigma _{kk} = 3\sigma _h\) (Einstein summation!) where \(\sigma _h\) is the hydrostatic stress.
The four moduli are not independent (only two are), and the following expressions relate the pairs \(\lambda \), \(\mu \) and \(E\), \(\nu \) to each other: \begin {align} \label {eqn:Relation_Lame} \lambda &= \frac {E\nu }{(1+\nu )(1-2\nu )} \\ \mu &= \frac {E}{2(1+\nu )} \\ \lambda + \mu &= \frac {E}{2(1+\nu )(1-2\nu )} \\ E &= \frac {\mu (3\lambda + 2\mu )}{\lambda + \mu } \\ \nu &= \frac {\lambda }{2(\lambda + \mu )} \end {align}
Note that another common symbol for the shear modulus \(\mu \) is the Latin letter \(G\).
The volumetric strain \(\varepsilon _h = \frac {1}{3}\ {\rm tr} \ \t {\varepsilon } = \frac {1}{E} [(1 + \nu )\sigma _h - 3\, \nu \sigma _h] = \frac {1}{E} (1-2\nu )\sigma _h\) vanishes at \(\nu = 1/2\). In this case, \(\sigma _{ij} = 2\sigma _h \varepsilon _{ij}\) because the \(\varepsilon _h = \varepsilon _{kk}\) must vanish.
We can also write down a free energy functional (often also called a hyperelastic energy density), which is quadratic in the strain \(\t {\varepsilon }\), \begin {equation} \label {eqn:Free_energy} W = \frac {1}{2} \lambda \varepsilon _{ii}^2 + \mu \varepsilon _{ij}^2 \end {equation} Using \(\sigma _{ij} = \partial W / \partial \varepsilon _{ij}\) recovers the above constitutive expression Eq. \eqref{eq:isotropicelasticity}. From the free energy functional we see that any isotropic material must have \(\lambda > 0\) and \(\mu > 0\), otherwise the energy could be made arbitrarily small by increasing the deformation of the solid. This limits the Poisson number to the range \( -1 < \nu < 1/2 \). Note that \(\nu <0\) is typically only achieved for architectured materials such as foams or metamaterials.