1
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.