This is a follow up question to my question yesterday Structural mechanics traction boundary condition question

Does a traction free boundary condition also mean that the displacement gradient is 0 where the boundary condition is applied? My thinking is that it does not because it seems you can have traction free (zero stress) surface with a non-zero displacement gradient.


The traction is defined as

$$ \mathbf{t} = \mathbf{n} \cdot \boldsymbol{\sigma} $$

In terms of components, the zero-traction condition is

$$ t_j = \sum_i n_i \sigma_{ij} = 0 $$

From the above you can see that the stress components don't necessarily have to be zero for the traction to be zero.

For a linear elastic material,

$$ \boldsymbol{\sigma} = \mathsf{C}:\nabla{\mathbf{u}} $$

In component form,

$$ \sigma_{ij} = \sum_k \sum_l C_{ijkl} \frac{\partial u_k}{\partial x_l} $$

Once again, since there is no requirement that all the $C_{ijkl}$ values have to be positive, you don't need all the displacement gradients to be zero for the stresses to be zero (which is not strictly necessary for zero-tractions anyway).

However, zero displacement gradients will lead to zero tractions.

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  • $\begingroup$ Hmm. In my previous question, I concluded that traction free=stress free=free surface. Does traction free mean zero traction? Does stress free mean zero stress? If so, then doesn't this mean that traction free=zero stress? $\endgroup$ – David Jan 2 '18 at 22:22
  • $\begingroup$ Stress free = traction free but not vice versa. It's more intuitive if you think in terms of balance of forces on a small element. Note that the momentum PDE is typically defined on an open set; so there are some subtle issues here, e.g., St. Venant principle. $\endgroup$ – Biswajit Banerjee Jan 3 '18 at 0:38
  • $\begingroup$ Ahh got it. So then what is free surface? Is it zero stress? When I think of free surface, I think of the classical cantilever beam problem, where one end is fixed and the other surfaces are "free" $\endgroup$ – David Jan 3 '18 at 5:29
  • $\begingroup$ I believe these "free surfaces" exhibit zero stress (tangential and normal). $\endgroup$ – David Jan 3 '18 at 5:47
  • 1
    $\begingroup$ @David Yes, but just on that surface. $\endgroup$ – nicoguaro Jan 3 '18 at 18:22

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