Slip: Why Metals Are Soft and Shapeable

The paradox: metals should be far stronger than they are

The last two guides built the character. Guide 1 introduced the dislocation — an edge is an extra half-plane of atoms wedged into the crystal, a screw is a spiral ramp — and guide 2 gave you the Burgers vector b that measures exactly how much slip a dislocation carries, read off from a Burgers circuit drawn around its line. Now we cash all of that in for the single most important consequence in the mechanics of metals: why you can bend a paperclip, hammer gold into leaf, or roll a steel ingot into foil, when the bonds holding those atoms together are individually stiff and strong.

Start by asking how strong a metal ought to be. To shear a perfect crystal you would have to slide one whole plane of atoms bodily over the plane beneath it, snapping every bond across that plane at the same instant and reforming it one atom-spacing along. Model the resistance as a gentle sine wave that rises and falls as the top plane rides over the bumps of the bottom one, match its initial slope to the ordinary elastic shear law, and out drops a clean estimate for this theoretical shear strength: about tau_max = G / (2 x pi), where G is the shear modulus. Refined models soften it to roughly G/30 to G/10, but the headline stays: a perfect crystal should yield at something like a tenth of its shear modulus.

Put copper's numbers in. Its shear modulus is about G = 48 GPa, so tau_max is roughly 48 / (2 x pi) = 7.6 GPa — call it a few billion pascals. Now measure a real, well-annealed copper single crystal: it begins to slip at about 1 MPa. That is a factor of several thousand too weak. Even a tough engineering alloy, work-hardened and full of obstacles, yields at a few hundred MPa — still 10 to 100 times below the theoretical figure. Something is letting these crystals shear at a tiny fraction of the force a rigid plane-over-plane slide would demand. That something is the dislocation.

Slip one row at a time: walking a ruck across a rug

Here is the easy way, and it is exactly the rug trick from guide 1. Suppose you want to shift a heavy rug a few centimetres across the floor. Dragging the whole thing at once means overcoming friction under every square inch simultaneously — the perfect-crystal move, and it is brutally hard. Instead you kick a small ruck, a wrinkle, into one end and walk that ruck across to the far side. At any moment only the little strip of rug under the wrinkle is off the floor, so the force is tiny — yet when the ruck exits the far edge, the entire rug has advanced by one wrinkle's width. Dislocation glide is that ruck. The dislocation is the wrinkle; as it sweeps across a plane, only the handful of bonds right at its core break and reform, one row of atoms at a time, so the whole upper half of the crystal slides over the lower half at a cost thousands of times below the rigid slide.

WHY SLIP IS CHEAP: move the rug by walking a ruck across it

  DRAG THE WHOLE RUG                WALK A SMALL RUCK ACROSS
  (perfect-crystal picture)         (real crystal: dislocation glide)
  every bond on the plane           only the few bonds at the core
  breaks at ONCE                    break & remake, ONE row at a time
  --> needs ~G/2pi  (~GPa)          --> needs ~G/1000 (~MPa)

  an edge dislocation gliding left-to-right on its slip plane:

    step 1         step 2         step 3         exited
    | | |_|        | |_| |        |_| | |        | | | |
    | | |T|   -->  | |T| |   -->  |T| | |   -->  | | | |__ slip = b
    =========      =========      =========      =========
    | | | |        | | | |        | | | |        | | | |

  T (the extra half-plane) is the edge dislocation; each hop shifts
  the top half over the bottom by exactly ONE Burgers vector b.

Two things are worth pinning down. First, one dislocation sweeping all the way across leaves a permanent step of exactly one Burgers vector — an atomic staircase, sub-nanometre high. Visible, macroscopic deformation is the sum of enormous numbers of dislocations gliding, one atomic step each. Second, this glide is not free to happen anywhere: a dislocation moves within the slip plane that contains both its line and its Burgers vector, and only shear resolved onto that plane drives it. An edge dislocation glides in the direction of b; a screw dislocation has b parallel to its line, so it can glide on any plane containing that line and even cross-slip from one to another — but the result is the same, the crystal shears by b.

Where slip happens: close-packed planes and slip systems

Not all planes are equally easy to glide on, and the winners are the ones you met when we stacked oranges. Dislocations move most readily on the crystal's close-packed planes and along its close-packed directions. The reason is geometric: close-packed planes are the most widely spaced planes in the crystal (largest interplanar spacing d), so they present the smoothest, gentlest corrugation to a passing dislocation — the lattice friction it must overcome, which the next guide names the Peierls-Nabarro stress, is lowest there. A specific slip plane paired with a specific slip direction lying in it is a slip system, and a crystal deforms by activating its slip systems.

Count them and you can predict ductility. In face-centred cubic metals (copper, aluminium, gold, nickel) the close-packed planes are the four {111} planes, each carrying three <110> close-packed directions: 4 x 3 = 12 slip systems, all with low friction. That abundance is why FCC metals are so gloriously ductile — you can draw copper into wire and beat gold into leaf a few atoms thick. Hexagonal close-packed metals (magnesium, zinc, titanium) have only the single basal (0001) plane close-packed, giving just three easy systems; that is too few to accommodate arbitrary shape change, so HCP metals are more anisotropic, deform partly by twinning, and are harder to work. Body-centred cubic metals (iron, tungsten) have no truly close-packed plane at all; they slip on {110}, {112} and {123} in <111> — many systems, but with higher lattice friction, which is why steel can turn brittle in the cold.

How hard you must pull: resolved shear stress and Schmid's law

A dislocation feels shear, not the full pull you apply. When you stretch a single crystal along some axis, only the fraction of that stress that resolves as shear ONTO a given slip system actually pushes its dislocations. That fraction is the resolved shear stress, and geometry hands you the conversion cleanly. Slip begins on a system the moment its resolved shear stress reaches a fixed threshold, the critical resolved shear stress (CRSS) — a near-constant property of the material and its slip system, which is exactly the content of Schmid's law.

1. Take the applied stress along the load axis, sigma = F / A (force over cross-sectional area).
2. Measure phi, the angle between the load axis and the NORMAL to the slip plane.
3. Measure lambda, the angle between the load axis and the slip DIRECTION.
4. The resolved shear stress is tau = sigma x cos(phi) x cos(lambda). The product cos(phi) x cos(lambda) is the Schmid factor m, and it never exceeds 0.5 (its maximum, when phi = lambda = 45 degrees).
5. The crystal yields on that system when tau reaches its CRSS. Because CRSS is fixed, the orientation that gives the largest m yields at the smallest applied sigma.

Try the arithmetic. Orient a copper single crystal so its best slip system sits at phi = lambda = 45 degrees, the ideal angle, giving the maximum Schmid factor m = cos(45) x cos(45) = 0.5. With CRSS about 1 MPa, slip starts when sigma x 0.5 = 1 MPa, that is at sigma = 2 MPa — a feather-light pull. Now rotate the same crystal so the slip plane is nearly parallel or nearly perpendicular to the load; m plummets toward zero, and you must haul far harder to reach the same 1 MPa on the system. Identical crystal, identical CRSS, wildly different yield stress — all bundled into the orientation factor m. (One honest caveat: Schmid's law is a fine approximation for FCC but only rough for BCC, whose screw-dislocation cores give extra, non-Schmid orientation effects.)

The honest headline — and the irony of making metals strong

So here is the honest headline of this whole rung, stated plainly: gliding dislocations let a crystal shear one row of atoms at a time, so real metals are typically 10 to 100 times weaker than a flawless lattice — and a soft, pure single crystal can be nearly a thousand times weaker. Softness and shapeability, the properties that make metals the workhorse of engineering, are a direct gift of the line defect. Without dislocations you would have hard, glass-brittle metals you could never forge, roll, draw, or stamp.

Which sets up the beautiful irony of metallurgy. If dislocations make metals weak, then to make a metal STRONG you do not try to remove them — that is nearly impossible in bulk — you make them harder to MOVE. Every strengthening trick is an obstacle course for dislocations: shrink the grains so grain boundaries block their glide (the Hall-Petch effect), dissolve solute atoms whose strain fields snag them (the solid-solution strengthening from the last guide), sprinkle in hard precipitates they must bow around, or, cleverest of all, let the dislocations get in each other's way. Cold-work a metal and its dislocation density — the total length of dislocation line packed into a cubic metre — climbs from about 10^10 per m^2 in the annealed state to 10^15 per m^2 or more, a dense thicket in which each dislocation trips over the next. That is work hardening, and it is why bending a paperclip back and forth makes it stiffer and finally snaps it.

That raises two loose threads the rest of this rung ties off. If dislocations obstruct each other and get consumed at surfaces, why does bending a metal not simply run out of them? Because they MULTIPLY — a single pinned segment can spin off loop after loop from a Frank-Read source, the multiplication engine of guide 5, which also opens up the world of partial dislocations, the stacking-fault ribbons they trail, and why some metals harden far faster than others. And before that, guide 4 asks what a dislocation costs to carry at all: its long-range elastic stress field, a line energy that scales as the square of b (which is why nature keeps b as short as possible), and the Peierls stress — the very lattice friction we have been leaning on this whole guide.