Elasticity free boundary terms

Suppose I have a 2D elastic body, and $$\mathbf{u}(x,y)$$ is a displacement field of the body. I am trying to derive the equilibrium equations for linear elasticity; I define an elastic energy

$$E[\mathbf{u}] = \frac{\alpha}{2} (I : \epsilon)^2 + \beta \epsilon : \epsilon$$ for Lame parameters $$\alpha, \beta$$ and strain $$\epsilon = \nabla \mathbf{u} + \nabla\mathbf{u}^T.$$ If I apply the calculus of variations to $$E$$, I get the expected equilibrium equations $$(\alpha+\beta)\nabla\left(\nabla\cdot\mathbf{u}\right) + \beta \Delta\mathbf{u} = 0$$ on the interior of the body; however I am getting strange boundary conditions: on the free boundary, my calculations are that $$\alpha (\nabla \cdot \mathbf{u}) \mathbf{n} + \beta\left[\nabla \mathbf{u} + \nabla \mathbf{u}^T\right]\mathbf{n}=0$$ where $$\mathbf{n}$$ is the body's boundary normal.

Are my calculations correct? I was expecting some kind of simple natural/Neumann boundary conditions on the free boundary---are there simplifications I can apply here?

• Does the 2nd term imply the dot product between the bracketed term and n? – Chet Miller May 19 at 23:12
• @ChetMiller I'm treating the bracketed term as a matrix that's multiplying the normal vector, but since the term is symmetric, indeed it's also equal to $$\left[\nabla \mathbf{u} + \nabla \mathbf{u}^T\right]\cdot \mathbf{n}.$$ – user2617 May 20 at 4:10
• In my judgment, your relationship is correct. The equation would simplify in a specific problems like tension on a rod. If you dot it with n again, you will get the normal component of the traction, which is also zero, and will be simpler in many cases. – Chet Miller May 20 at 11:55