Identify the components of the (weak form) PDE in structural mechanics

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:

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

• Are you familiar with variational calculus or virtual work / virtual energy principles? Strain energy? – Paul Dec 9 '18 at 1:27
• Just a little bit. I edited my answer. – Kerem Dec 9 '18 at 19:27
• 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. – Wolfgang Bangerth Dec 10 '18 at 3:05
• $\delta\epsilon$ in the formulation is then $\varepsilon(\delta u)$. – Wolfgang Bangerth Dec 10 '18 at 3:06
• @WolfgangBangerth thank you very much. But how do i work with the symmetric gradient? – Kerem Dec 10 '18 at 14:26