= Solution
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, along
$$
\mathbf r(t)=(1,2,4)+t(1,0,1)=(1+t,2,4+t).
$$
The initial fundamental sector has $x<y<z$. Its first reached boundary is $x=y$, at $t=1$, giving $\mathbf r=[225]$. At that orientation $(\overline111)[101]$ and $(1\overline11)[011]$ have equal <Schmid factors>. Since the component perpendicular to the primary slip direction is unchanged by that shear, the length ratio follows directly:
$$
\frac{L}{L_0}=\frac{|(2,2,5)|}{|(1,2,4)|}
=\sqrt{\frac{33}{21}}=\sqrt{\frac{11}7}.
$$
Therefore
$$
\boxed{\varepsilon_{\rm eng}=\sqrt{11/7}-1=0.2536\simeq25.4\%.}
$$
The corresponding logarithmic <strain> is $\ln(L/L_0)=0.2260$. This is an \b[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.<Applied Mechanics of Solids by A. F. Bower>[§3.12.2, points10–12: tensile-axis rotation, secondary slip and stages of hardening]{ref}
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