# What is the relationship between the strain in different directions?

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 $$\epsilon_x, \epsilon_y, \epsilon_z$$.

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.