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An interface between ordered domains related by a displacement that is a translation of the underlying disordered lattice but not of the ordered chemical structure. Slip by that displacement can leave an antiphase boundary behind a dislocation; a second appropriate displacement can restore the order.

The lattice translation represented by closure failure of a circuit around a dislocation, with a consistent circuit convention. It is a vector, so dislocation reactions conserve its direction as well as its magnitude.

A feature that absorbs mobile point defects, such as a free surface, grain boundary, dislocation or cavity. Capture efficiency and relative preference for defect types are as important as visible size or number.

A curved dislocation can be approximated as a line with tension $T_{\ell }$ of order $G{b}^2$. The coefficient depends on elastic constants, character and logarithmic length scales. The simple estimate $T_{\ell }\simeq G{b}^2/2$ gives a pinning bypass stress $Gb/\ell$ for a semicircular segment of free span $\ell$.

A closed dislocation line. Vacancy or interstitial clustering can form a platelet bounded by a loop; its Burgers vector, habit plane and possible stacking fault specify its crystallographic character.

Transactions of the Japan Institute of Metals 29, 970–976. The model and comparison with experimental specimens distinguish grain size, aspect ratio and dislocation density.

Resistance to plastic flow associated with grain boundaries obstructing or transmitting dislocations. In the ordinary Hall-Petch regime, smaller grains raise the yield stress, but at high temperature the same boundaries can accelerate diffusion creep.

A dislocation that accommodates some lattice mismatch at an interface. An initially coherent epitaxial layer or island can instead store elastic deformation; introducing a misfit dislocation trades some strain energy for the defect's formation energy. A projected image must be indexed before a particular spacing discrepancy is assigned to misfit strain.

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.

Cambridge institutional account with a transmission-electron micrograph supplied by Yanhui Zhang, showing paired dislocation loops after creep testing. It explains order restoration by two matrix dislocations. brearley.msm.cam.ac.uk/2002/Zhang.html

Resistance to dislocation motion caused by the energetic cost of disrupting chemical order, for example creating an antiphase boundary in a gamma prime phase precipitate.

A dislocation unable to shear an obstacle can bow around it and leave a loop. Balancing the Peach-Koehler force against the tension of a curved dislocation gives the critical scale $\Delta \tau \simeq 2T_{\ell }/(b\ell )$, where $\ell$ is the free obstacle span on the slip plane. Centre pitch and free span are not interchangeable when particle size is appreciable.

A dislocation whose Burgers vector is not a full translation of the perfect crystal, so its glide can leave a planar stacking fault. Vector conservation and the actual slip plane constrain reactions and dissociation.

A Shockley partial dislocation in an FCC slip plane has Burgers vector of the family $a⟨112⟩/6$. Its vector must lie in the specified plane, and the two vectors must sum to the perfect dislocation vector.

For the first plane normal ${\mathbf{n}}_1=[\bar{1}11]$, one allowed dissociation isThe vectors add to $a[303]/6$, and ${\mathbf{n}}_1\cdot [211]=-2+1+1=0$, ${\mathbf{n}}_1\cdot [1\bar{1}2]=-1-1+2=0$.

For ${\mathbf{n}}_2=[111]$,Their sum is $a[\bar{3}30]/6$, and both have zero dot product with $[111]$. Interchanging leading and trailing labels does not change these vector identities.

Each perfect vector has squared magnitude $a^2/2$, whereas the sum of the two partial squared magnitudes is $a^2/3$. The line-energy estimate proportional to $G{b}^2$ therefore favors separation, opposed by the stacking-fault energy of the ribbon between the partials.

Use the two partials on each plane as obtained above. The four pairwise sums, in units of $a/6$, areThe shortest resultant gives the largest line-energy reduction in the $G{b}^2$ approximation:The initial summed squared magnitude is $a^2/3$ and the final value is $a^2/18$, a sixfold reduction in that part of the energy estimate.

The line of intersection of the two original planes is parallel toIts dot product with the resultant Burgers vector is zero, so the junction is edge in character. A plane containing both line and Burgers vector has normal $[100]$, rather than an FCC close-packed $\{111\}$ glide normal. The resulting stair-rod dislocation is therefore sessile for ordinary glide and can form a Lomer-Cottrell lock. It obstructs further dislocation motion and promotes work hardening.

This energetic comparison assumes comparable line lengths and elastic prefactors. Geometry, line character and stacking-fault geometry affect a complete junction-energy calculation, but the small resultant identifies the favorable candidate requested by the vector model.

Orowan bypass bows a dislocation around an unshearable obstacle and leaves a loop. A dislocation line-tension estimate for a segment with tension $T_{\ell }$ gives the limiting shear-stress scalewhere $\ell$ is a free obstacle gap on the slip plane. For an FCC perfect dislocation,Using $T_{\ell }\simeq G{b}^2/2$ and the simple gap estimate $\ell \simeq L-d\simeq 0.34\mu \mathrm{m}$ givesConverting to tensile stress requires an orientation factor. With an explicitly assumed polycrystalline FCC Taylor factor $M\simeq 3$,The graph and spacing approximations do not justify a precise last digit. A favorably oriented single crystal with Schmid factor1/2 would instead give $\Delta {\sigma }_y\simeq 2\Delta \tau$, around110MPa.

This is a line-tension estimate. A true slip-plane obstacle distribution, line character, logarithmic elastic corrections and precipitate size distribution are not specified. If the quoted pitch is used directly as an effective free spacing, neglecting diameter, the same simplified calculation gives $\Delta \tau \simeq 19\mathrm{MPa}$ and $M\Delta \tau \simeq 57\mathrm{MPa}$. The larger gap-corrected estimate explicitly uses the size inferred above; the tensile factor and spacing convention must not be silently treated as provided measurements.

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.

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.

Material creep is time-dependent permanent deformation under sustained stress. Its rate is often appreciable at a large homologous temperature $T/T_m$, because diffusion and obstacle-escape processes become rapid. In an activated regime a useful form iswith additional elastic, geometric and temperature prefactors absorbed into $A$ for the chosen approximation. The activation energy belongs to the rate-controlling process: bulk diffusion, grain-boundary diffusion and dislocation motion need not have the same value.

Boundary sliding can accommodate grain-shape changes during diffusional material creep; it is not a reason to identify either diffusion field as dislocation glide. The supplied map's A/B/C placement, including the boundary between B and C, determines these assignments. Its numerical contours apply to the material state used to construct that map, not to arbitrary grain sizes.