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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?

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    $\begingroup$ Fear -> hate; hate -> anger; ... $\endgroup$ – kηives Apr 6 '18 at 15:20
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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.

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  • $\begingroup$ I see. So in terms of a "free surface" boundary condition, this implies zero traction, but you could get a non-zero stress and still satisfy the zero traction BC, meaning that "free surface" doesn't imply zero stress? $\endgroup$ – David Apr 6 '18 at 21:24
  • $\begingroup$ That's exactly right. Take the simple example of a bar in uniaxial tension, with one end held fixed and the other end with an applied traction. The lateral surfaces are free surfaces (zero traction applied), but the material at those surfaces will still have non-zero stress ($\sigma \neq \mathbf{0}$, but $\sigma\mathbf{n} = \mathbf{0}$). $\endgroup$ – Tyler Olsen Apr 6 '18 at 21:29
  • $\begingroup$ Yeah. Is it the same with the classical cantilever problem, where the lateral surfaces are "free," but may exhibit a non-zero stress? I think the cantilever and uniaxial tension problem are the exact same except that one is compressed and the other is sheared. $\endgroup$ – David Apr 6 '18 at 21:42
  • $\begingroup$ Yes, it's the same in the cantilever problem. In general, any problem where you have a non-zero stress state at a free surface is an example of what we're talking about. This occurs in most problems, and is a consequence of Neumann boundary conditions specifying only traction, not the full stress state. $\endgroup$ – Tyler Olsen Apr 6 '18 at 21:46
  • $\begingroup$ Ah got it. Is it even possible to enforce a "stress" boundary condition, without turning it into traction? It seems the problem would be ill-posed. $\endgroup$ – David Apr 6 '18 at 21:50

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