Weilin Deng | Miscellaneous notes on linear elasticity

Tensor

Some good introductions to tensors can be found in

Javier Bonet and Richard D. Wood, Nonlinear Continuum Mechanics for Finite Element Analysis, Chapter 2
Piaras Kelly, Mechanics Lecture Notes Part III: Foundations of Continuum Mechanics

Principle of virtual work

For a static, linear elastic boundary value problem, the governing equation reads

$\begin{align} \nabla \cdot \boldsymbol{\sigma} + \boldsymbol{b} = \boldsymbol{o}, \quad \text{in} \quad \Omega \label{eq:equilbrium} \end{align}$

with the boundary conditions

$\begin{equation} \begin{aligned} \boldsymbol{u} &= \boldsymbol{u}^*, \quad \text{on} \quad \Gamma_u, \\ \boldsymbol{\sigma} \cdot \boldsymbol{n} &= \boldsymbol{t}^*, \quad \text{on} \quad \Gamma_s. \end{aligned} \label{eq:BC} \end{equation}$

Consider a virtual displacement filed $\delta\boldsymbol{u}$ which satisfies $\delta\boldsymbol{u} = \boldsymbol{o}$ on $\Gamma_u$. Multiply $\delta\boldsymbol{u}$ on both side of \eqref{eq:equilbrium} and integrate over $\Omega$, we obtain

$\begin{equation} \begin{aligned} & \int_{\Omega} (\nabla \cdot \boldsymbol{\sigma}) \cdot \delta\boldsymbol{u} \, dV + \int_{\Omega} \boldsymbol{b} \cdot \delta\boldsymbol{u} \, dV \\ &= \int_{\Omega} \nabla \cdot (\boldsymbol{\sigma} \cdot \delta\boldsymbol{u}) \, dV - \int_{\Omega} \boldsymbol{\sigma}:\nabla \delta \boldsymbol{u} \, dV + \int_{\Omega} \boldsymbol{b} \cdot \delta\boldsymbol{u} \, dV \\ &= \int_{\Gamma} \boldsymbol{n} \cdot (\boldsymbol{\sigma} \cdot \delta\boldsymbol{u}) \, dS - \int_{\Omega} \boldsymbol{\sigma}:\nabla\delta\boldsymbol{u} \, dV + \int_{\Omega} \boldsymbol{b} \cdot \delta\boldsymbol{u} \, dV \\ &= \int_{\Gamma_s} \boldsymbol{t}^* \cdot \delta\boldsymbol{u} \, dS - \int_{\Omega} \boldsymbol{\sigma}:\nabla\delta\boldsymbol{u} \, dV + \int_{\Omega} \boldsymbol{b} \cdot \delta\boldsymbol{u} \, dV \\ &= 0, \end{aligned} \label{eq:virtualwork} \end{equation}$

which can be rewrite as

$\begin{align} \int_{\Omega} \boldsymbol{\sigma}:\nabla\delta\boldsymbol{u} \, dV &= \int_{\Gamma_s} \boldsymbol{t}^* \cdot \delta\boldsymbol{u} \, dS + \int_{\Omega} \boldsymbol{b} \cdot \delta\boldsymbol{u}\, dV, \label{eq:virtualwork1} \end{align}$

$\begin{align} \int_{\Omega} \boldsymbol{\sigma}:\delta\boldsymbol{\epsilon} \, dV &= \int_{\Gamma_s} \boldsymbol{t}^* \cdot \delta\boldsymbol{u} \, dS + \int_{\Omega} \boldsymbol{b} \cdot \delta\boldsymbol{u}\, dV. \label{eq:virtualwork2} \end{align}$

In arriving \eqref{eq:virtualwork} we invoked the Green’s theorem, and in arriving \eqref{eq:virtualwork2} from \eqref{eq:virtualwork1}, we used the property that $\boldsymbol{\sigma}$ is symmetric and $\boldsymbol{\sigma}:\nabla\delta\boldsymbol{u}$ = $\boldsymbol{\sigma}:(\text{sym}(\nabla\delta\boldsymbol{u}) + \text{skewsym}(\nabla\delta\boldsymbol{u}))$ = $\boldsymbol{\sigma}:\text{sym}(\nabla\delta\boldsymbol{u})$ = $\boldsymbol{\sigma}:\delta\boldsymbol{\epsilon}$.

The \eqref{eq:virtualwork2} has clear physical meaning, that is, the internal work stored in the solid (LHS) is equal to the external work done by the surface traction and body force (RHS).

Constitutive relation for isotropic solid

The constitutive relation, or the stress–strain ($\boldsymbol{\sigma}–\boldsymbol{\epsilon}$) relation, for the isotropic solid is

$\begin{align} \boldsymbol{\sigma} = \boldsymbol{C} : \boldsymbol{\epsilon}, \end{align}$

where $C_{ijkl} = \lambda \delta_{ij} \delta_{kl} + \mu (\delta_{ik} \delta_{jl} + \delta_{il} \delta_{jk})$. In index from, $\begin{align} \sigma_{ij} = 2\mu\epsilon_{ij} + \lambda \epsilon_{kk} \delta_{ij}. \end{align}$

Betti reciprocal theorem

Consider two different equilibrium states of a solid $\boldsymbol{u}^{(1)}$, $\boldsymbol{\epsilon}^{(1)}$, $\boldsymbol{\sigma}^{(1)}$ and $\boldsymbol{u}^{(2)}$, $\boldsymbol{\epsilon}^{(2)}$, $\boldsymbol{\sigma}^{(2)}$. The two states could be under different boundary conditions. We can show that

$\begin{equation} \begin{aligned} & \int_{\Gamma} (\boldsymbol{n} \cdot \boldsymbol{\sigma}^{(1)}) \cdot \boldsymbol{u}^{(2)} \, dS + \int_{\Omega} \boldsymbol{b}^{(1)} \cdot \boldsymbol{u}^{(2)} \, dV \\ &= \int_{\Gamma} (\boldsymbol{n} \cdot \boldsymbol{\sigma}^{(2)}) \cdot \boldsymbol{u}^{(1)} \, dS + \int_{\Omega} \boldsymbol{b}^{(2)} \cdot \boldsymbol{u}^{(1)} \, dV \\ &= \int_{\Omega} \boldsymbol{\sigma}^{(1)}:\boldsymbol{\epsilon}^{(2)} \, dV \\ &= \int_{\Omega} \boldsymbol{\sigma}^{(2)}:\boldsymbol{\epsilon}^{(1)} \, dV. \end{aligned} \end{equation}$

To prove it, first observe that $\boldsymbol{\sigma}^{(1)}:\boldsymbol{\epsilon}^{(2)}$ = $\boldsymbol{\epsilon}^{(2)}:\boldsymbol{C}:\boldsymbol{\epsilon}^{(1)}$ = $\boldsymbol{\epsilon}^{(1)}:\boldsymbol{C}:\boldsymbol{\epsilon}^{(2)}$ = $\boldsymbol{\sigma}^{(2)}:\boldsymbol{\epsilon}^{(1)}$ . Then note that

$\begin{equation} \begin{aligned} & \int_{\Gamma} (\boldsymbol{n} \cdot \boldsymbol{\sigma}^{(1)}) \cdot \boldsymbol{u}^{(2)} \, dS + \int_{\Omega} \boldsymbol{b}^{(1)} \cdot \boldsymbol{u}^{(2)} \, dV \\ &= \int_{\Omega} \nabla \cdot (\boldsymbol{\sigma}^{(1)} \cdot \boldsymbol{u}^{(2)}) \, dV + \int_{\Omega} \boldsymbol{b}^{(1)} \cdot \boldsymbol{u}^{(2)} \, dV \\ &= \int_{\Omega} (\nabla \cdot \boldsymbol{\sigma}^{(1)} + \boldsymbol{b}^{(1)}) \, dV + \int_{\Omega} \boldsymbol{\sigma}^{(1)} : \nabla\boldsymbol{u}^{(2)} \, dV \\ &= \int_{\Omega} \boldsymbol{\sigma}^{(1)} : \boldsymbol{\epsilon}^{(2)} \, dV. \end{aligned} \end{equation}$

Similarly, we have

$\begin{equation} \begin{aligned} \int_{\Gamma} (\boldsymbol{n} \cdot \boldsymbol{\sigma}^{(2)}) \cdot \boldsymbol{u}^{(1)} \, dS + \int_{\Omega} \boldsymbol{b}^{(2)} \cdot \boldsymbol{u}^{(1)} \, dV = \int_{\Omega} \boldsymbol{\sigma}^{(2)} : \boldsymbol{\epsilon}^{(1)} \, dV. \end{aligned} \end{equation}$

This completes the proof of the Betti reciprocal theorem.

Linear elasticity

Notes: Relations between elastic constants

Curvilinear coordinates

Notes: Linear elastic equations in curvilinear coordinates