Archard Wear Depth

The volume form of Archard's equation answers a question nobody asks. Nobody sets a maintenance limit in cubic millimetres. What a drawing says is that the bore may open up by so many microns, or that the wear strip may lose so many thousandths before it is replaced — so the useful form divides the volume by the apparent contact area and gives a depth: $h = K W s / (H A)$.

Written that way the equation reorganises itself usefully, because $W/A$ is the nominal contact pressure $p$, and the relation becomes $h = K p s / H$. Wear depth per unit sliding distance is the wear coefficient times the ratio of contact pressure to hardness. That single ratio, $p/H$, is the most informative number on the page — it is roughly the fraction of the apparent area that is in real contact. At $p/H$ of 0.001, a thousandth of the footprint is actually carrying load. As it climbs past a few per cent, the asperities are running out of room to spread, the contact is approaching fully plastic, and this is the neighbourhood where mild wear commonly gives way to severe. Watch that ratio more closely than you watch the depth.

The area in the denominator is the APPARENT area — the projected footprint you could measure with a rule — and not the real contact area, which is smaller by two or three orders of magnitude. That is worth being clear about, because it looks like an inconsistency and is not. The real area is what governs the physics, and it cancelled out of the derivation before we got here. The apparent area reappears at this stage purely as bookkeeping: a way of spreading a volume over a footprint to express it as a depth.

Which is also the form's weakness. This depth is an average. It assumes the load is spread evenly over the whole footprint and that the surface recedes uniformly, and real worn parts almost never look like that. A bushing on a slightly misaligned shaft carries its load on one edge and wears there at several times this rate; a wear strip loaded through a cocked slider does the same; a pin in an oscillating joint wears an oval, not a circle. When a part fails on clearance, the measurement that condemns it is almost always the deepest point rather than the average, and the average is what this equation gives. Design with that gap in mind, and if the geometry is at all uncertain, put the margin in rather than the precision.

For a plain bearing there is one more convention to get right. The apparent area of a journal bearing is the PROJECTED area, $d \times L$ — diameter times length, the shadow the shaft casts — and not the wrapped inner surface of the bore, which is $\pi$ times larger. Using the wrapped area understates the pressure by a factor of about three and the wear depth with it. The projected convention is not arbitrary: the pressure distribution around a loaded bore is far from uniform, and its resultant is what carries the load, so the projected area is the one that makes different bearings comparable.

The design lever the depth form makes obvious is area. Spread the load over more of it and the depth falls in proportion, while the lower nominal pressure also moves the contact further from its mild-to-severe transition — two benefits from one change. It is not free. A longer bushing is harder to align, and a misaligned long bushing loads one edge, which puts you back where you started with a more expensive part.