In a rolling bearing, the minimum oil film thickness is h = 0.40 µm and the root-mean-square roughnesses of the two surfaces are Rq1 = 0.15 µm and Rq2 = 0.20 µm. Find the film thickness ratio λ = h/√(Rq1² + Rq2²) and, taking λ < 1 as boundary lubrication, 1–3 as mixed lubrication and above 3 as hydrodynamic (full-film) lubrication, choose the correct pairing.

Answer: C. λ ≈ 1.6, mixed lubrication

The composite roughness is √(0.15² + 0.20²) = √0.0625 = 0.25 µm, so λ = 0.40/0.25 = 1.6, which by the guideline is mixed lubrication. When the film thickness is about 1–3 times the composite roughness, asperities touch here and there, and the load is shared between film pressure and asperity contact. λ ≈ 6.4 is the value obtained by dividing by 0.0625 without taking the square root, and λ ≈ 0.63 is the value with numerator and denominator swapped, dividing the composite roughness by the film thickness. λ can be raised by using a more viscous oil or by finishing the surfaces more smoothly to reduce roughness, which is effective in extending rolling bearing life. The boundary values are empirical guidelines and differ somewhat between sources.