Surface treatment and tribology
In a pin-on-disk test, a pin of hardness H = 2.0 GPa is loaded with a normal force of 50 N and slid over a distance of 1.0 km. Using the Archard wear equation V = KWL/H with a dimensionless wear coefficient K = 1.0×10⁻⁴, which correctly pairs the wear volume of the pin with the change in wear volume when the load is doubled?
- AAbout 2.5×10⁻³ mm³, and doubling the load roughly doubles it
- BAbout 2.5 mm³, and doubling the load roughly doubles it
- CAbout 2.5 mm³, and doubling the load leaves it unchanged as long as the apparent contact area is the same
Show answer and explanation
Answer: B. About 2.5 mm³, and doubling the load roughly doubles it
V = KWL/H = 1.0×10⁻⁴×50×1000/(2.0×10⁹) = 2.5×10⁻⁹ m³ = 2.5 mm³. The Archard equation assumes that a number of contact spots proportional to the real area of contact W/H produce wear debris a number of times proportional to the sliding distance, so the wear volume is proportional to the load and the sliding distance and inversely proportional to the hardness. Doubling the load doubles the real area of contact even if the apparent contact area stays the same, so the wear volume also roughly doubles. About 2.5×10⁻³ mm³ is the value calculated with the sliding distance taken as 1 instead of 1000 m. K varies widely, from about 10⁻⁸ to 10⁻², with the wear mechanism and conditions, so it is determined by experiment.