What is the relationship between the strain in different directions?

Question

A cube made of an isotropic material is subjected to a tensile stress as shown in Figure 2. What is the relationship between the strain in different directions ?

There is an elongation in $z$ direction and contraction in $x, y$ directions. So $\epsilon_x, \epsilon_y < \epsilon_z$. Is my reasoning correct ? Since they say material is isotropic, can I say anything else about $\epsilon_x, \epsilon_y$ ? (Like $\epsilon_x = \epsilon_y$). Please explain how someone can arrive at a conclusion about $\epsilon_x, \epsilon_y, \epsilon_z$.

Answer 1

In this case, we say that $\epsilon_{axial}$ ($z$-direction in your diagram) is positive by convention. Then, for a normal material with a Poisson's ratio $\nu \ge 0$, $\epsilon_{transverse}$ ($x$- and $y$-directions in your diagram) will be negative. The fact that the material is isotropic means we have no way of distinctly labeling one direction $x$ and one direction $y$, so it must be the case that $\epsilon_x = \epsilon_y$.

So in summary, yes, assuming $\nu \ge 0$ you are justified in saying $\epsilon_x = \epsilon_y < \epsilon_z$.

Note, however, that even though almost all materials will have $\nu \ge 0$, there does exist a class of materials known as auxetics which have $\nu \lt 0$. A common naturally occurring example is $\alpha$-cristobalite silica. In this case, the relationship between the strains is not so immediately obvious, and would depend on the exact value of $\nu$ (since the strain would be positive in all directions). These materials are not usually isotropic though, so $\nu \ge 0$ is probably a safe assumption.