|
|
||||
| DoITPoMS > TLP Library > Slip in Single Crystals > Slip in HCP metals 3 | ||
|
|
|
Having determined which slip system will operate first, it should now be possible to calculate the minimum force needed to cause plastic flow during the application of a stress to the crystal. The calculation proceeds as follows. The following diagram shows the orientation of the [021] tensile axis with respect to the unit cell vectors b and c (parallel to [010] and [001] respectively). These three vectors all lie in the (100) plane.
Identification of the angles l and f. The diagram shows coplanar vectors on the (100) plane. The (001) slip plane lies horizontal and extends perpendicular to the screen. The initial angle between the tensile axis and the slip plane normal, f0, is
and the angle between the tensile axis and the slip plane normal, l0, is l0 = 90° - f0 = 43.3° The Schmid factor for the [010](001) slip system is therefore cos f0 cos l0 = cos 46.7° x cos 43.3° = 0.499 If the critical resolved shear stress for cadmium is 0.15 MPa, and the initial crystal diameter is 3 mm, then the force required to cause slip can be calculated:
Now consider what happens when the crystal is plastically extended. If the original length of crystal was l0 = 50 mm, and the crystal is extended by 50 % to l1 = 75 mm, then:
The force required to cause further deformation of the crystal in this condition can be calculated as before:
The force required to cause slip is lower after the crystal has been deformed. This phenomenon is known as geometric softening - once deformation has started, less load is required to further deform the crystal. Geometric softening depends heavily on the orientation of the tensile axis within the crystal. For some orientations, no geometric softening is observed. There are other factors that control slip, which will not be discussed here, and these can dominate over the geometric factor. |
|
Except where otherwise noted, content on this page is licensed under a |
|