# Structural mechanics - traction free = Zero displacement gradient?

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.

## 1 Answer

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.

• 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? – David Jan 2 '18 at 22:22
• 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. – Biswajit Banerjee Jan 3 '18 at 0:38
• 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" – David Jan 3 '18 at 5:29
• I believe these "free surfaces" exhibit zero stress (tangential and normal). – David Jan 3 '18 at 5:47
• @David Yes, but just on that surface. – nicoguaro Jan 3 '18 at 18:22