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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.

A second slip system begins to operate as the tensile-axis rotation makes its resolved shear stress comparable to that on the primary system. Dislocations on intersecting planes interact much more strongly, giving the transition from easy single slip to rapid work hardening. The standard geometric estimate neglects differences between the evolving critical stresses of the two systems.

The primary direction is $\mathbf{s}\parallel [101]$. Single-slip shear carries the line joining the specimen ends, expressed in crystal coordinates, alongThe initial fundamental sector has $x<y<z$. Its first reached boundary is $x=y$, at $t=1$, giving $\mathbf{r}=[225]$. At that orientation $(\overline{1}11)[101]$ and $(1\overline{1}1)[011]$ have equal Schmid factors. Since the component perpendicular to the primary slip direction is unchanged by that shear, the length ratio follows directly:ThereforeThe corresponding logarithmic strain is $\ln (L/L_0)=0.2260$. This is an ideal equal-critical-stress geometric onset estimate. Real secondary slip can start earlier or later because of latent hardening, nonuniform deformation and grip constraints; the problem supplies no hardening law that would determine that shift.^[ref]

Initially, one slip system carries most of the deformation. Easy dislocation glide gives the nearly flat stage-I response, although lattice rotation and gradual dislocation interactions can already change the required tensile stress. In tension the loading direction expressed in the rotating crystal approaches the primary slip direction. Its path has the form $(5+t,1+t,4)$ until it reaches the symmetry boundary $y=z$, at $[844]=[211]$.

At that orientation, $(1\bar{1}1)[110]$ and $(11\bar{1})[101]$ have equal Schmid factors, $1/\sqrt{6}$. Conjugate slip therefore becomes competitive. Intersections between the two dislocation families generate obstacles, junctions and forest hardening, giving the sharp stage-II rise in flow stress. This is the ideal single-slip-to-conjugate-slip construction; real activation stresses can be shifted by latent hardening and imperfections.

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.

In an ideal single-slip tensile test neglecting elastic stretch, the specimen end-to-end vector expressed in crystal coordinates changes as $\mathbf{r}={\mathbf{r}}_0+\gamma (\mathbf{n}\cdot {\mathbf{r}}_0)\mathbf{s}$, with unit slip direction $\mathbf{s}$ and slip-plane normal $\mathbf{n}$. It moves towards the slip direction along a great circle. The perpendicular component is unchanged, so $L\sin \lambda =L_0\sin {\lambda }_0$. Secondary slip changes this kinematics; equal Schmid factors determine onset only if the systems have equal current critical stresses.

An obstacle calculation produces a resolved shear-stress increment. A single crystal needs its Schmid factor, while a polycrystal needs a deformation-geometry or Taylor factor. The relation $\Delta \sigma =M\Delta \tau$ therefore requires an explicit orientation assumption rather than a universal tensile prefactor.