Mechanical Properties and Deformation

FCC slip systems

In face-centred cubic metals, slip occurs on the four close-packed \{111\} planes along the three close-packed \langle 110\rangle directions, giving 4 \times 3 = 12 slip systems. A slip system is a combination of a slip plane and a slip direction, and the most readily activated system is one whose direction lies inside a close-packed plane. A direction [uvw] lies in a plane (hkl) when hu + kv + lw = 0, so the test is a simple dot product between the plane normal and the direction. For example, (111) with [1\bar{1}0] gives 1 - 1 + 0 = 0 and is a valid close-packed system, whereas [110] on (111) gives 2 \neq 0 and is not. Plane indices such as (110) are not close-packed in FCC and do not act as slip planes, and directions such as \langle 112\rangle are not close-packed slip directions.

A screw dislocation has its line parallel to its Burgers vector, so it is not confined to a single plane and can glide on any plane that contains the Burgers vector direction. For a Burgers vector along [110], the \{111\} planes that satisfy hu + kv + lw = 0 are (\bar{1}11) and (11\bar{1}), while (111) gives a sum of 2 and is excluded.

A mixed dislocation has both edge and screw character, so its slip plane must contain both the Burgers vector \mathbf{b} and the line direction \mathbf{u}. The plane normal is therefore the cross product \mathbf{n} = \mathbf{b} \times \mathbf{u}. For \mathbf{b} \propto [110] and \mathbf{u} = [011], the cross product gives \mathbf{n} \propto (1, -1, 1), so the slip plane is (1\bar{1}1).

Vickers hardness testing

Vickers hardness uses a square-base diamond pyramid whose opposite faces meet at an included angle of 136^\circ, so the indentation is geometrically similar at all loads. The hardness number is HV = \dfrac{2F\sin(\theta/2)}{d^{2}}, where F is the load, \theta is the included angle and d is the mean diagonal length. For \theta = 136^\circ this reduces to HV \approx 1.8544\,F/d^{2}, which gives kg mm^{-2} when F is in kgf and d is in mm. As a short example, a load of 5 kgf producing a mean diagonal of 0.4 mm gives HV \approx 1.8544 \times 5 / 0.16 \approx 58 kg mm^{-2}. Students often forget that d is the average of the two diagonals and that the load must be in kgf.

For a ductile metal, Vickers hardness is roughly proportional to yield strength, H_V \approx 3\sigma_y, so a harder ductile metal has the higher yield strength. A harder surface also resists scratching and plastic penetration more effectively, so it has the greater scratch resistance. Hardness is not a measure of stiffness, because Young's modulus depends on interatomic bonding and is not changed much by the hardness of the material, so a harder metal does not necessarily have a higher modulus. Higher hardness generally goes with reduced ductility, so a harder metal should not be assumed to be more ductile.

Work hardening and hardness

Work hardening, or strain hardening, raises the flow stress and the hardness of a metal because plastic deformation multiplies dislocations, which then impede one another. A specimen that has been plastically strained beyond the offset yield point therefore has a higher hardness than an annealed specimen, while a specimen unloaded below yield keeps its original microstructure and hardness. When comparing two specimens unloaded from different points on the stress-strain curve, the one with the larger plastic prestrain is harder.

For a material obeying the Hollomon relation \sigma = K\epsilon^{n}, the work-hardening rate is d\sigma/d\epsilon = nK\epsilon^{n-1}. The maximum load is reached when the Considère criterion d\sigma/d\epsilon = \sigma is satisfied, and this happens at a true strain equal to n. At that point the work-hardening rate equals the flow stress, \sigma = Kn^{n}, so the rate can be found directly from K and n. As an example, with K = 400 MPa and n = 0.2, the stress at maximum load is 400 \times 0.2^{0.2} \approx 290 MPa, which is also the work-hardening rate at that point.