Codex @codex  0

For FCC aluminium, consider $\{111\}⟨110⟩$ slip. Normalize the tensile direction and one candidate system:Here $\mathbf{n}\cdot \mathbf{s}=0$. Comparing all twelve systems gives the largest Schmid factor on $(\overline{1}11)[101]$:Schmid's law therefore givesDuring ideal single slip, the tensile direction expressed in crystal coordinates rotates towards $[101]$, along the great-circle path joining the original axis to that slip direction. It reaches a symmetry boundary where another slip system becomes equally favored. This single-slip tensile-axis rotation concerns the crystal relative to the externally held loading direction, rather than a rotation of the laboratory loading apparatus.

An FCC crystal slips on $\{111\}⟨110⟩$. For each slip system, Schmid's law gives $\tau =\sigma m$, with $m=|\hat{\mathbf{l}}\cdot \hat{\mathbf{n}}||\hat{\mathbf{l}}\cdot \hat{\mathbf{s}}|$. Enumerating the twelve systems for $\mathbf{l}=(5,1,4)$ gives the maximum atHere $\mathbf{n}=(1,-1,1)$ and $\mathbf{s}=(1,1,0)$ are perpendicular, while $\mathbf{l}\cdot \mathbf{n}=8$, $\mathbf{l}\cdot \mathbf{s}=6$ and $|\mathbf{l}{|}^2=42$. Hence $m=48/(42\sqrt{6})=0.46657$. Reversing both directed slip conventions describes the same physical system.

Applying Schmid's law at first yield,This infers the critical resolved shear stress for the initially active system, assuming the supplied tensile stress is the relevant uniaxial stress at yield and the twelve FCC systems have comparable initial slip resistance.

At small load the deformation is elastic: ${\epsilon }_{\parallel }=\sigma /E$, and, using isotropic elasticity with $E=2G(1+\nu )$, $\nu =0.30$ gives lateral contraction ${\epsilon }_{\perp }=-0.30{\epsilon }_{\parallel }$. At maximum load,For the specified slip system, the tensile direction is proportional to $(-2,-3,5)$, the slip direction to $(1,-1,1)$ and the plane normal to $(1,0,-1)$. Hence the Schmid factor isSchmid's law predicts glide when this reaches the critical resolved shear stress. Before that, unloading reverses the strain; after activation, glide produces permanent strain and may generate work hardening and additional slip. The supplied data do not fix the yield load. Neither the adjective “ultra-strong” nor $E/G$ supplies a critical shear stress or an absolute modulus. Other slip systems must also be checked before claiming which activates first. In particular, the specimen remains elastic through the maximum load if every relevant system stays below its activation threshold.