Aluminum has preferred direction of and closed packed plane (111), therefore, it has 12 active slip systems. Is it possible that aluminum has more than 12 slip systems, and how?
There are four sets of planes of the type {111} and each one contains three vectors of the type , so there are 12 physically distinct slip systems of the type {111}.
Note, however, that these 12 systems are NOT all independent. Some of them can be produced by linear combinations of slip on other systems of the same type. In fact, it's possible to show that there can never be more than 5 independent systems. For full details, I recommend the book by Kelly and Knowles, but in brief: the action of a slip system induces a strain in the material, which can be represented by a strain tensor. In general, this tensor has 6 independent components: 3 normal and 3 shear. However, plastic shear induces no volume change, and so the 3 normal components must sum to zero. The result is 5 independent components, so that any arbitrary change of shape can be accommodated by 5 independent slip systems.
There are, of course, other types of slip system available, which would increase the total number of systems available. There's no reason these couldn't be activated, except that the {111} systems are much softer. In fact, it can be shown that {111} systems provide the full 5 independent systems, so that any arbitrary change of shape can be accommodated by slip on this type of system. It would therefore be extremely unlikely to find slip on any other type of system.
There are four sets of planes of the type {111} and each one contains three vectors of the type , so there are 12 physically distinct slip systems of the type {111}.
Note, however, that these 12 systems are NOT all independent. Some of them can be produced by linear combinations of slip on other systems of the same type. In fact, it's possible to show that there can never be more than 5 independent systems. For full details, I recommend the book by Kelly and Knowles, but in brief: the action of a slip system induces a strain in the material, which can be represented by a strain tensor. In general, this tensor has 6 independent components: 3 normal and 3 shear. However, plastic shear induces no volume change, and so the 3 normal components must sum to zero. The result is 5 independent components, so that any arbitrary change of shape can be accommodated by 5 independent slip systems.
There are, of course, other types of slip system available, which would increase the total number of systems available. There's no reason these couldn't be activated, except that the {111} systems are much softer. In fact, it can be shown that {111} systems provide the full 5 independent systems, so that any arbitrary change of shape can be accommodated by slip on this type of system. It would therefore be extremely unlikely to find slip on any other type of system.
For aluminum Seeger in 1958 reported that {1 0 0} slip systems can also be activated above 450 C.
Slip in face centered cubic (fcc) crystals occurs along the close packed plane. Specifically, the slip plane is of type {111}, and the direction is of type .
Hii, the close packed planes and atomically dense directions make the active systems which means that the atomic motion needs less energy on these slip systems.
As Philip explained, these systems are not really independent, the strain in one system generates a strain into another.
Other systems who need higher activation energy could exist also but these systems can not be activated before activation of other slip systems who need less activation energy. It is also possible that different mechanisms exist in different conditions. The case that Kamyar cited for aluminum is a high temperature (T>= 2/3 Tmelt) case where different mechanism can play a role (creep!!)...
PS. Volker,I am really interested in the references that you cited. Could you send me a copy? Thank you.
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