Using Lame's constants, the Hooke's law of isotropic materials in 3-dimensions can be written as: $$ \begin{aligned} \sigma_{ij}&=\lambda \varepsilon_{kk}\delta_{ij}+2\mu\varepsilon_{ij}=c_{ijk\ell}\varepsilon_{k\ell}\\ c_{ijk\ell}&=\lambda\delta_{ij}\delta_{k\ell}+\mu(\delta_{ik}\delta_{j\ell}+\delta_{i\ell}\delta_{jk}) \end{aligned}\tag{1} $$

While in plane stress problems, Hooke's law seems to be different from the 3D case.

So my question is: Does the relation $(1)$ still hold in plane stress problem? Any help will be greatly appriciated!

  • $\begingroup$ How is it different? Can you provide more details. $\endgroup$ May 12 '16 at 16:09
  • $\begingroup$ It's in the name. Plane stress means all stresses have 0 in the z direction, but strains don't (due to Poisson's ratio). Plane strain is the opposite. They have equivalents in 3D where either parts are very thin or very thick in the z direction. $\endgroup$ Aug 25 '19 at 1:07

It still holds in plane stress situations. It is just that, in such situations, one of the principal stresses (the one in the thickness direction) is equal to zero. An, of course, the shear stresses involving that direction are also equal to zero. So the 6 stress equations reduce to 3. And the zero stress condition in the thickness direction results in a kinematic constraint on the strains.


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