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.


This is not correct. An incompressible material can not change its total volume, but it can change its shape.

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.

  • $\begingroup$ Thank you for helping me with this doubt. After thinking about your answer I went back to the equations and I was able to find the error. $\endgroup$
    – Eduardo
    Sep 3 '19 at 1:42

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