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A. F. Bower is the author of Applied Mechanics of Solids by A. F. Bower, an online text whose title page gives his name as Allan F. Bower. The text connects elasticity, plasticity in continuum mechanics and fracture mechanics to explicit models of how solids deform. Its treatment of single-crystal plastic deformation follows slip systems and the rotation of the tensile axis, showing why a single crystal's response depends on its orientation.^[ref]

Erich Schmid studied the plastic deformation of single crystals, connecting applied loading to the resolved shear stress acting on a slip system. Schmid's law expresses the geometric projection that underlies this connection. The University of Vienna's account of his scientific work describes his research on crystal deformation and his jointly authored treatment of crystal plasticity.^[ref]

Forest hardening is resistance to dislocation glide caused by dislocations that cut across a glide plane. An advancing dislocation must negotiate their intersections, form or break junctions, or bow between obstacles. The resulting obstacle network is called a forest because it obstructs motion across the plane, even though it is made of line defects rather than solid inclusions.

The mechanism connects dislocation density with work hardening: a shorter spacing between effective obstacles generally demands greater applied shear stress. It also explains why different slip systems need not evolve independently; slip on one system changes the obstacles encountered by another, contributing to latent hardening. Comparing a specimen before and after controlled plastic deformation tests the increase in flow resistance, while microstructural observation tests whether an assumed obstacle mechanism is plausible. Not every change of yield stress is forest hardening; other microstructural changes can also strengthen a material.

Latent hardening is increased resistance on one crystal slip system caused by plastic slip on another slip system. The word latent emphasizes that the affected system need not have been active during the earlier loading. It is therefore a statement about interactions between deformation mechanisms, rather than simply about the total amount of deformation.

A simple crystal-plasticity model writes ${\dot{g}}_{\alpha }={\sum }_{\beta }{h}_{\alpha \beta }|{\dot{\gamma }}_{\beta }|$, where $g_{\alpha }$ is the critical resolved shear stress for system $\alpha$ and ${\gamma }_{\beta }$ is slip on system $\beta$. Off-diagonal coefficients represent cross-system hardening, while diagonal coefficients describe self-hardening.^[ref] Changing loading direction after predeformation can activate previously less-favoured slip systems and reveal this history dependence. Predicting the next system to move therefore requires both resolved stress geometry and the evolving resistances.

Plastic flow involving more than one active slip system. Interactions between dislocations on intersecting systems can give substantially faster work hardening than easy single slip.

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]

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.

The nickel-rich $\gamma$ matrix is a disordered substitutional solid solution on a face-centered cubic lattice. The ordered ${\gamma }^{\prime }$ phase has the L12 structure, approximately $\mathrm{N}{\mathrm{i}}_3\mathrm{Al}$: aluminium occupies cube corners and nickel occupies face centres. Its occupied-site geometry is often called ordered FCC, but its chemically labelled Bravais lattice is primitive cubic, with a four-atom basis. A face-centre translation exchanges chemical sublattices and is not a symmetry of the ordered crystal.

Each Al atom has twelve nearest-neighbour Ni atoms at Euclidean distance $a/\sqrt{2}$. Each Ni atom has twelve nearest neighbours too, but these are four Al and eight Ni. The chemical distinction matters for slip. In disordered $\gamma$, the shortest perfect Burgers vector isFor the ordered primitive cubic translation lattice, the shortest order-preserving perfect vector isThis does not imply that $⟨100⟩$ slip must dominate. A vector must lie in a favourable glide plane and have an accessible core structure and activation stress. Matrix dislocations commonly glide on $\{111\}⟨110⟩$ systems. A single $a⟨110⟩/2$ passage through ${\gamma }^{\prime }$ leaves an antiphase boundary; a second matching passage restores order, giving total vector $a⟨110⟩$. Such a superdislocation has a longer total vector than $a⟨100⟩$, yet its constituent dislocations can move on the favourable close-packed planes and interact with the matrix. Paired motion balances line and antiphase-boundary energies, so the shortest abstract lattice translation alone does not select the active slip system.^[ref]

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.

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.

The low-temperature zirconium matrix has a hexagonal close packing structure. Its crystallographic $a$ and $c$ directions therefore have different elasticity, thermal expansion and available slip systems. Producing tubing also develops a crystallographic texture, so averaging over grains does not necessarily remove these directional effects.

Different expansion along the crystal axes produces intergranular thermal stresses during temperature changes. Under load, easy dislocation glide modes may accommodate some strains much more readily than others; additional slip or crystal twinning is needed for general compatibility. Consequently strength and ductility can depend on the axial, circumferential and radial orientation of the tube.

Irradiation creates vacancies and self-interstitials that cluster into dislocation loops with different crystallographic orientations. Their anisotropic accumulation can produce irradiation growth even without external stress. Different directions can elongate or contract while the net volume change remains small. Texture turns these grain-scale strains into macroscopic length and diameter changes, bowing and stresses between constrained components. This mechanism is distinct from the added cavity volume of void swelling.

The crystallographic character of irradiation loops matters: atomistic calculations by Rory Hulse and Christopher P. Race examine several possible loop types and show that loop energy and shape depend on orientation and size. A single generic spherical-defect description therefore misses important aspects of zirconium's dimensional response.^[ref]

Hydrogen also produces direction-sensitive damage. Zirconium hydrides are brittle precipitates whose preferred orientation depends on crystal texture and stress history. Radially oriented hydrides can offer crack paths across a thin tube wall under circumferential tension. Controlling texture, hydrogen pickup and hydride orientation helps manage that risk; changing the hexagonal structure's orientation cannot by itself eliminate corrosion or irradiation damage.

Single-crystal plastic deformation is irreversible shape change of a single crystal, often through dislocation motion on particular slip systems. A slip system specifies a crystallographic plane and a direction within it. Applied loading resolves into a shear stress that drives slip, making the response depend on crystal orientation.^[ref]

Under uniaxial stress $\sigma$, the resolved stress on a slip system is $\tau =\sigma \cos \varphi \cos \lambda$, where $\varphi$ is the angle between the loading axis and plane normal and $\lambda$ the angle to the slip direction. For compatible $45$-degree angles, $\tau =\sigma /2$. Slip begins when the relevant resistance is reached; subsequent deformation can rotate the lattice and change which systems carry further strain.^[ref]

Slip-line field theory solves certain two-dimensional problems of plastic deformation by constructing two orthogonal families of characteristic curves, called slip lines. Its classical assumptions include plane strain, negligible inertia and body force, and a rigid, perfectly plastic material. The curves are aligned with the directions of maximum in-plane shear stress, rather than the principal stress directions.^[ref]

The curve network encodes both the stress field and constraints on admissible velocity. Indentation of a thick workpiece by a rigid flat punch provides a motivating application: the construction predicts a plastic flow region and the load needed to sustain that flow. It is useful because geometry can turn coupled field equations into relations integrated along curves. A slip line is a continuum characteristic, however, not necessarily a microscopic crystal slip system or the actual path of a moving material particle.

The ratio $M=\sigma /{\tau }_c$ connecting a macroscopic tensile flow stress to a reference critical resolved shear stress in a specified polycrystal model. Texture, slip families and compatibility assumptions matter. An order-three factor is a conventional estimate for randomly oriented cubic grains, distinct from the factor two for a perfectly oriented single slip system.