I am trying to identify the weak form of PDE in structural mechanics. I read a lot of papers where they are using the elliptic boundary value problem

\begin{equation} \int\limits_{\Omega} \delta \epsilon : \sigma \,d\Omega - \Big[\int\limits_{\Omega} \delta u \cdot b \, d\Omega + \int\limits_{\Gamma_N} \delta u \cdot t \, d\Gamma\Big] = 0 \end{equation} where $b$ represent the body forces, $t$ is the traction vector acting on the Neumann boundary, stress $\sigma$ and $u$ the displacement. With $\delta \epsilon$ and $\delta u$ they define virtual strain/displacement.

To be honest, I am not used to mechanics. For example I am using the general elliptic PDE:

\begin{align}\left\{\begin{aligned} div(A\nabla u) = & \,f && \text{ in } \Omega \\ u = & \,g_D && \text{ on }\Gamma_D \\ A \nabla_n u = &\, g_N && \text{ on }\Gamma_N \\ \end{aligned}\right. \end{align} with $A$ a bounded self-adjoint and elliptic linear operator. The weak form this PDE reads: find $u\in H^1_{\Gamma_D}(\Omega)$ for given functions $f\in L^2(\Omega)$ and $g_N \in L^2(\Gamma_N)$ such that

\begin{equation} \int\limits_{\Omega} \nabla u A \nabla v \, d\Omega = \int\limits_{\Omega} v^T f \, d\Omega + \int\limits_{\Gamma_N} v^T g_N \,d\Gamma \qquad \forall v \in H^1_{\Gamma_D}(\Omega). \end{equation}

My goal now is to identify the weak form in structural mechanics such that my general weak form will be the same. As I can see, first I need to define $v^T = \delta u$, $t = g_N$ and $b = f$. But I am not sure about the other terms. Is it correct that $\delta \epsilon = \nabla v = \nabla \delta u$ and $\nabla u = \sigma$? I cannot see the connection between these definitions. And how do I use the elliptic operator $A$ in this context? Can someone bring light into the darkness?

My new Idea is: Use that strain is equal to the derivatives of the displacement such that $\epsilon = \nabla u$. Next is to define the virtual strain as the derivative of the virtual displacement such that $\nabla v = \nabla \delta u = \delta \epsilon$. By Hookes law it is now:

\begin{equation} \nabla v : A \nabla u = \nabla \delta u : A \nabla u = \delta \epsilon : A \epsilon = \delta \epsilon : \sigma \end{equation}

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    $\begingroup$ Are you familiar with variational calculus or virtual work / virtual energy principles? Strain energy? $\endgroup$ – Paul Dec 9 '18 at 1:27
  • $\begingroup$ Just a little bit. I edited my answer. $\endgroup$ – Kerem Dec 9 '18 at 19:27
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    $\begingroup$ The stress $\sigma$ is actually $C\epsilon(u)$ where $\epsilon(u)=\frac 12 (\nabla u + \nabla u^T)$ is the symmetric gradient. $C$ is the stress strain relationship and corresponds to the coefficient $A$ in the Laplace problem you show. $\endgroup$ – Wolfgang Bangerth Dec 10 '18 at 3:05
  • $\begingroup$ $\delta\epsilon$ in the formulation is then $\varepsilon(\delta u)$. $\endgroup$ – Wolfgang Bangerth Dec 10 '18 at 3:06
  • $\begingroup$ @WolfgangBangerth thank you very much. But how do i work with the symmetric gradient? $\endgroup$ – Kerem Dec 10 '18 at 14:26

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