Question 15.7: Slip in MgO and Mg Based on the principles of the bonding of......
Slip in MgO and Mg
Based on the principles of the bonding of MgO and Mg and the Peierls-Nabarro stress, explain why slip in MgO is much more difficult than in Mg. For the purpose of this discussion, consider the {0001}<100> slip system in Mg.
As determined in Example 15-6, both the spacing between slip planes and b are 0.280 nm for MgO. For the {0001}<100> slip system in Mg, the interplanar spacing is half of the height of the HCP unit cell or 0.5209 nm/2 = 0.260 nm, and the Burgers vector is the distance between atoms in the close-packed direction or a/2 = 0.32087 nm/2 5 0.160 nm. The Peierls-Nabarro stress is given by Equation 4-2,
τ = c \exp ( – \frac{kd}{ b} )where τ is the shear stress required to move a dislocation, d is the interplanar spacing between slip planes, b is the magnitude of the Burgers vectors, and c and k are material specific constants. Ignoring differences in c and k, we see that the stress required to propagate a dislocation increases as the ratio of d/b decreases. Therefore, we can reasonably expect that MgO with a d/b ratio of 1.0 will require a larger stress to cause dislocations to propagate than Mg (with a d/b ratio of 1.62), thereby making slip more difficult. Additionally, as discussed in Section 4-3, dislocation propagation in ionically bonded materials requires that cations move past other cations and anions move past other anions, creating electronic repulsion that resists dislocation motion.