Why are five independent slip systems required for plasticity to occur at the macroscopic level? According to the von Mises criteria, we often encounter that five independent slip system is required for plasticity to occur at macroscopic level. Can somebody give an explain briefly about this proof or argument?
 A: Since arbitrary deformation involves 6 strains, and if you have a constant volume which gives you 1 constraint, you end up with 6 - 1 = 5 independent strains. Now, analogous to needing 5 pieces of data to solve a simple system of equations with 5 unknown variables, you need exactly 5 slip systems to determined the 5 independent strains.
More mathematically, we have 6 different strains labeled as $\varepsilon_{ij}$, since $\varepsilon$ is a symmetric tensor that can be described by only:
$$
\tag{1}
\varepsilon_{11},~~\varepsilon_{12},~~\varepsilon_{13},~~\varepsilon_{22},~~\varepsilon_{23},~~\varepsilon_{33},
$$
and then we have the one constraint due to enforcing the volume to be constant:
\begin{eqnarray}
\tag{2}
\textrm{Tr}\left( \varepsilon \right) &=0, \\ \varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33}&= 0.\tag{3}\\
\end{eqnarray}
A: But what happens if we are considering the deformation of a single crystal? And there are a lot of papers showing that only one or two slip systems are activated in a grain.
