# Plane stress and incompressibility

The relation between bulk modulus ($$K$$) and Young's modulus ($$E_Y$$) and Poisson's ratio ($$\nu$$) is given by:

$$\begin{equation} K = \left\{ \begin{array}{ll} \frac{E_Y}{3(1-2\nu)} & \mbox{for 3D} \\[10pt] \frac{E_Y}{2(1+\nu)(1-2\nu)} & \mbox{for 2D plane strain} \\[10pt] \frac{E_Y}{2(1-\nu)} & \mbox{for 2D plane stress} \end{array} \right. \end{equation}$$

Under incompressibility, the Poisson's ratio is $$\nu$$ = 1/2, so the bulk modulus would be:

$$\begin{equation} K = \left\{ \begin{array}{ll} \infty & \mbox{for 3D} \\[10pt] \infty & \mbox{for 2D plane strain} \\[10pt] E_Y & \mbox{for 2D plane stress} \end{array} \right. \end{equation}$$

The 2D plane strain assumption recovers the value of 3D, but the 2D plane stress does not. So, my question is: is it correct to say that plane stress assumption does not model incompressibility?

PS: my original question has an error in equations that led to an incorrect interpretation of physical behavior. The correct equation for 2D plane stress is:

$$\begin{equation} K = \begin{array}{ll} \frac{E_Y}{2(1-2\nu)} & \mbox{for 2D plane stress} \end{array} \end{equation}$$

With this correction $$K$$ = $$\infty$$ for 2D plane stress too.

If a compressive plane stress is applied, the material gets thinner in the $$z$$ direction but expands in $$x$$ and $$y$$ directions (as given by Poisson's ratio) so that its volume remains constant.
Under plane strain, the expansion in $$x$$ and $$y$$ is prevented by definition of what plane strain means. Therefore its thickness can't change, and the material is not only incompressible but also perfectly rigid.