Past exam of the materials science course of the University of Cambridge / 2016 / Paper 1 / 5

The alloy shear yield stress can be roughly $70/0.78\simeq 90$ times the stated pure-aluminium critical resolved shear stress. Strengthening raises the stress required for dislocation glide rather than requiring homogeneous shearing of a perfect crystal.

These mechanisms can coexist. The numerical comparison is an order-of-magnitude comparison of shear strengths; it is not a conversion of the alloy's shear yield stress directly into a tensile yield stress.

For a coarse-grained, approximately untextured polycrystal undergoing general plastic flow, use the conventional orientation-and-compatibility estimateWith the supplied ${\tau }_c=27.5\mathrm{MPa}$,The round $M=3$ result is a conventional estimate for an unspecified coarse polycrystal, not a material-independent exact conversion. The most favorably oriented individual grains can first slip at $2{\tau }_c=55\mathrm{MPa}$; that is not the same event as compatible macroscopic yielding throughout the aggregate.

There is no single crystallographic tensile axis for all grains. Individual grains rotate and can develop crystallographic texture, while the macroscopic bar remains loaded along its tensile axis. A specified initial texture or yield criterion would be needed for a more exact numerical prediction.

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.

Take ordinary basal slip in magnesium, $(0001)⟨11\overline{2}0⟩$. In three-index hexagonal notation, $[101]$ denotes a vector proportional to ${\mathbf{a}}_1+\mathbf{c}$, not a cubic Cartesian $(1,0,1)$ with equal axis lengths. The basal slip direction closest to the tensile axis is along ${\mathbf{a}}_1$; its four-index representative is $[2\overline{1}\overline{1}0]$.

For $r=c/a$,Thus the algebraic answer isAn external crystallographic reference gives $a=0.320927\mathrm{nm}$ and $c=0.521033\mathrm{nm}$, hence $r\simeq 1.624$ and ${\sigma }_y\simeq 1.79\mathrm{MPa}$. This is a conditional external-data estimate, not a fabricated value attributed to the supplied Data Book.^[ref]

The tensile axis rotates towards the selected basal slip direction during ideal single slip. As the axis approaches the basal plane, the resolved basal shear eventually becomes small; other modes, hardening and instability can interrupt the single-slip description. TODO: verify the exact magnesium lattice ratio supplied in the historical examination Data Book. Taking a $45^{\circ }$ angle solely from the indices would incorrectly use a cubic metric.

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]

The printed numerical ratios are about $35$, not about $10$:The qualitative contrast is nevertheless meaningful. FCC aluminium has close-packed glide planes and relatively mobile dislocations. Magnesium also has easy close-packed basal glide, although it lacks enough equivalently easy slip systems for arbitrary room-temperature polycrystal deformation.

BCC iron has no close-packed plane. Its screw-dislocation cores and lattice-friction barrier give much stronger resistance to glide; thermally activated kink motion makes the critical stress particularly temperature- and rate-sensitive. This is the relevant Peierls-stress explanation. A high polycrystal constraint factor cannot explain the supplied contrast, because the data are already critical resolved shear stresses. Neither the precise factor35 nor a universal BCC/FCC ratio follows from crystal packing alone: purity, dislocation structure, temperature and testing conditions also matter.