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 furtherTLP creditsTLP contentsShow all contentViewing and downloading resourcesAbout the TLPsTerms of useFeedbackCredits Print this page

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# Geometry as slip proceeds

Let N = number of slip planes in the sample, d = perpendicular spacing of slip planes. During slip, both N and d remain constant.

d = (l/N) cos φ \ N d = l cos φ

N and d are constant, so lcos φ must also be constant throughout slip. Hence

l_{0} cos φ_{0} = l_{1} cos φ_{1}

The same can be done for λ:

d= (l/N) cos (90º - λ) \ N d = l sin λ

N and d are constant, so l sin λ must also be constant throughout slip. Hence

l_{0} sin λ_{0} = l_{1} sin λ_{1}