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DoITPoMS Teaching & Learning Packages Slip in Single Crystals Slip in HCP metals 3: calculation of forces
Slip in Single Crystals AimsBefore you startIntroductionSlip geometry: the critical resolved shear stressGeometry during slipSlip in HCP metals 1: slip systemsSlip in HCP metals 2: application of Schmid's LawSlip in HCP metals 3: calculation of forcesSlip in HCP metals 4: observing slip in cadmiumVideo clips of slip in a single cadmium crystalExercise: Determination of the critical resolved shear stress for slip in cadmiumSlip in CCP metalsSummaryQuestionsGoing further
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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.

Diagram showing orientation of the [021] tensile axis with respect to the unit cell vectors b and c

Identification of the angles λ and φ. 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, φ0, 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, λ0, is

λ0 = 90° - φ0 = 43.3°

The Schmid factor for the [010](001) slip system is therefore

cos φ0 cos λ0 = 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 l0 = 50 mm, and the crystal is extended by 50 % to l1 = 75 mm, then:

  • The diameter of the crystal will decrease as volume is conserved:

    l0d02 = l1d12

    \[{\rm{thus}}\;{d_1} = \sqrt {\frac{{{l_0}{d_0}^2}}{{{l_1}}}} = \sqrt {\frac{{50 \times {3^2}}}{{75}}} = \underline{\underline {2.45{\rm{ mm}}}} \]

  • The geometry of the slipped crystal changes according to:

    l0 cos φ0 = l1 cos φ1 and l0 sin λ0 = l1 sin λ1

    hence:

    φ1 = 62.8° and λ1 = 27.2°

    and so the new Schmid factor for the extended crystal is:

    cos φ1 cos λ1 = 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.

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