Traction -> stress; stress->displacement gradient

Question

If given a displacement gradient tensor, we can easily obtain the stress tensor (using Hooke's law and the strain-displacement relationship), as well as the traction vector.

If given a traction, and knowing that traction $\boldsymbol{T} = \boldsymbol{\sigma n}$ ,where $\boldsymbol{n}$ is the normal vector, we can not uniquely determine the 6 unique components of the stress tensor without some additional condition(s). The same scenario exists for going from stress to displacement gradients, where we know 6 unique stress components, but cannot uniquely determine the 9 displacement gradients without some other condition.

In terms of solving the linear elasticity equations, Neumann boundary conditions are typically imposed in terms of tractions. Would it make the problem more well-posed if Neumann boundary conditions were imposed in terms of stress, or even better, in terms of a displacement gradient?

Answer 1

In short, no. Neumann boundary conditions should be specified in terms of traction. This is clear when you move from a strong-form statement of the boundary value problem to a weak-form.

Strong form: \begin{align} div(\sigma(\mathbf{u})) + \mathbf{b} &= \mathbf{0} &&\forall \mathbf{x} \in \Omega \\ \mathbf{u} &= \hat{\mathbf{u}}(\mathbf{x}) &&\forall \mathbf{x} \in \partial\Omega_D \\ \sigma \mathbf{n} &= \mathbf{g}(\mathbf{x}) &&\forall \mathbf{x} \in \partial\Omega_N \\ \partial\Omega_N \cup\partial\Omega_D &= \partial\Omega&& \\ \partial\Omega_N \cap\partial\Omega_D &= \emptyset&& \end{align}

Taking $\mathbf{w}$ to be an appropriate test function that vanishes on $\partial\Omega_D$, the weak form is written as

\begin{align} \int_\Omega \left(\mathbf{w}\cdot\mathbf{b} - \nabla\mathbf{w}:\sigma \right)\,\mathrm{d}V + \int_{\partial\Omega_N}\mathbf{w}\cdot(\sigma\mathbf{n}) \,\mathrm{d}A = \mathbf{0} \end{align}

As you can see, the Neumann boundary condition appears directly as a traction in the weak form, so it is quite natural to specify it in terms of traction rather than a displacement gradient. In fact, it is not possible to impose the entire displacement gradient or stress tensor at a boundary, only the traction.