Slip in HCP metals 3: calculation of forces
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.
The initial angle between the tensile axis and the slip plane normal, φ_{0}, is
\[{\varphi _0} = {\tan ^{  1}}\left( {\frac{{2a}}{c}} \right) = {\tan ^{  1}}\left( {\frac{{2 \times 2.98}}{{5.62}}} \right) = 46.7^\circ \]
and the angle between the tensile axis and the slip direction, λ_{0}, is
λ_{0} = 90°  φ_{0} = 43.3°
The Schmid factor for the [010](001) slip system is therefore
cos φ_{0} cos λ_{0} = cos 46.7° × 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:
\[{\tau _c} = {\sigma _y}\cos {\varphi _0}\cos {\lambda _0} = \frac{F}{A}\cos {\varphi _0}\cos {\lambda _0}\]
\[F = \frac{{{\tau _c}A}}{{\cos {\varphi _0}\cos {\lambda _0}}} = \frac{{0.15 \times {{10}^6} \times \pi \times {{(1.5 \times {{10}^{  3}})}^2}}}{{0.499}} = \underline{\underline {2.12{\rm{ N}}}} \]
Now consider what happens when the crystal is plastically extended. If the original length of crystal was l_{0} = 50 mm, and the crystal is extended by 50 % to l_{1} = 75 mm, then:

The diameter of the crystal will decrease as volume is conserved:
l_{0}d_{0}^{2} = l_{1}d_{1}^{2}
\[{\rm{thus}}\;{d_1} = \sqrt {\frac{{{l_0}{d_0}^2}}{{{l_1}}}} = \sqrt {\frac{{50 \times {3^2}}}{{75}}} = \underline{\underline {2.45{\rm{ mm}}}} \]

The geometry of the slipped crystal changes according to:
l_{0} cos φ_{0} = l_{1} cos φ_{1} and l_{0} sin λ_{0} = l_{1} sin λ_{1}
hence:
φ_{1} = 62.8° and λ_{1} = 27.2°
and so the new Schmid factor for the extended crystal is:
cos φ_{1} cos λ_{1} = 0.407
The force required to cause further deformation of the crystal in this condition can be calculated as before:
\[F = \frac{{{\tau _c}A}}{{\cos {\varphi _1}\cos {\lambda _1}}} = \frac{{0.15 \times {{10}^6} \times \pi \times {{(1.225 \times {{10}^{  3}})}^2}}}{{0.407}} = \underline{\underline {1.{\rm{74 N}}}} \]
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.