Past exam of the materials science course of the University of Cambridge / 2022 / Paper 1 / 5

The exponential in the Peierls stress makes a large core-width-to-Burgers vector ratio $w/b$ favourable for easy dislocation glide. Densely packed slip planes often have wider cores because the shear distortion can spread over more atomic rows, reducing the lattice-periodic barrier. Close-packed directions provide short lattice translations and hence small allowed Burgers vectors; their line energies, approximately proportional to $G{b}^2$, are also relatively low. These effects favour dense planes and directions, but the relation between packing and core width is a physical argument rather than a consequence of substituting arbitrary independent $w,b$ into the formula. Body-centred cubic metals have densely packed directions without a truly close-packed plane, and their screw-core behaviour can require additional qualifications.

The cosine of the angle between Burgers vector and line direction is $1/3$, so the angle is $70.53^{\circ }$: the dislocation is mixed. Both vectors must lie in its dislocation glide plane. Their cross product is proportional toso the plane is $(10\bar{1})$, a member of $\{110\}$. The screw component magnitude is $a/(2\sqrt{3})$ and the edge component magnitude is $a\sqrt{2/3}$.

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.

Burgers vector conservation gives two candidate reactions, depending on the relative signs of the approaching segments:Initially $b_1^2+b_2^2=3a^2/2$. The sum product has $b^2=a^2$, while the difference product has $b^2=2a^2$. Thus the $a[001]$ product is favoured by Frank's rule; the $a[1\bar{1}0]$ product is disfavoured. The result is not annihilation, because the two supplied vectors are not opposite. The same ordering survives a simple character correction: with $\nu =0.30$, line energy proportional to $G{b}^2[{\cos }^2\chi +{\sin }^2\chi /(1-\nu )]$ gives the two parent segments a combined coefficient $1.786a^2$, whereas any character of the sum product has coefficient between $a^2$ and $1.429a^2$ and any character of the difference product at least $2a^2$, comparing equal lengths and common logarithmic factors. A favourable dislocation junction can obstruct further glide and harden the crystal. Its actual mobility requires the junction line and allowed glide plane: the second screw can occupy several $\{110\}$ planes containing $[\bar{1}11]$, so geometry is not unique. For instance, its $(101)$ plane intersects $(10\bar{1})$ along $[010]$; an $a[001]$ junction on that line would glide geometrically on $(100)$, not either original plane. Dislocation cross-slip, junction unzipping or bypass can compete under load. An energy inequality alone is not a universal proof of a permanently sessile product.