I saw that,

$$dF=\sigma \cdot dS$$

Where $dF$ is the differential force, $\sigma$ is the stress tensor, and $dS$ is the differential surface. This equation confuses me a bit. I'm under the impression that stress acts over a volume rather than a surface, so rather than $dS$ we should have.

$$dF=\sigma \cdot dV$$

Does stress act over a surface or a volume element?


1 Answer 1



You can have the standard

$$\sigma_{ij} = C_{ijkl} \epsilon_{kl}$$

in the bulk, but you can also have a reduced tensor, a surface stress. This surface stress is the projection of the regular stress

$$\sigma_{\alpha \beta}^s = {\boldsymbol P} \sigma {\boldsymbol P}$$

where ${\boldsymbol P} = 1 - {\boldsymbol n}({\boldsymbol r}) \otimes {\boldsymbol n}({\boldsymbol r})$ is the projection operator constructed from the orthonormal basis using the tangent vectors from the surface normal with $\alpha, \beta = 1,2$.

Typically in large systems, the surface stress contribution to the total energy of the system is negligible. However, as the size of the system is brought smaller, the surface plays more of a role in determining the overall elastic configuration of the system and can modulate some of the elastic dependent properties.


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.