Service life, nominal

The nominal service life of a rolling bearing is determined solely by its load rating, actual load, rotational speed, and the type of rolling elements.

Here is a practical example of calculating the basic rating life of a rolling bearing according to ISO 281, based on a typical application (e.g., a deep groove ball bearing in an electric motor).

Given Data

• Bearing: Deep groove ball bearing (e.g., SKF 6208)
  • Dynamic load rating $C = 32 \, \text{kN}$C=32kN
  • Static load rating $C_0 = 18.6 \, \text{kN}$C0​=18.6kN
• Load:
  • Radial force $F_r = 3 \, \text{kN}$Fr​=3kN
  • Axial force $F_a = 1 \, \text{kN}$Fa​=1kN
• Rotational speed: $n = 1,500 \, \text{min}^{-1}$n=1,500min−1
• Operating conditions:
  • Room temperature, clean lubricating grease ($a_{ISO} = 1$aISO​=1)
  • No extreme environmental conditions ($a_2 = 1$a2​=1, $a_3 = 1$a3​=1)

Equivalent Dynamic Bearing Load $P$P

For deep groove ball bearings with $F_a / F_r \leq 0.5$Fa​/Fr​≤0.5:

$$P = X \cdot F_r + Y \cdot F_a$$

P=X⋅Fr​+Y⋅Fa​

• Factors $X$X and $Y$Y: From the bearing catalog (for $F_a / C_0 = 1/18.6 \approx 0.054$Fa​/C0​=1/18.6≈0.054): $X = 1$X=1, $Y = 0$Y=0 (since $F_a / F_r = 1/3 \approx 0.33 < 0.5$Fa​/Fr​=1/3≈0.33<0.5) → Simplification: $P = F_r = 3 \, \text{kN}$P=Fr​=3kN

Basic Rating Life $L_{10}$L10​ (in millions of revolutions)

$$L_{10} = \left( \frac{C}{P} \right)^p$$

L10​=(PC​)p

• Exponent $p$p: For ball bearings: $p = 3$p=3
• Substituting the values:

$$L_{10} = \left( \frac{32}{3} \right)^3 = (10.\overline{6})^3 \approx 1,200 \, \text{million revolutions}$$

L10​=(332​)3=(10.6)3≈1,200 million revolutions

Basic Rating Life $L_{10h}$L10h​ (in operating hours)

$$L_{10h} = \frac{1,000,000}{60 \cdot n} \cdot L_{10}$$

L10h​=60⋅n1,000,000​⋅L10​

• Substituting the values:

$$L_{10h} = \frac{1,000,000}{60 \cdot 1,500} \cdot 1,200 = \frac{1,000,000}{90,000} \cdot 1,200 \approx 13,333 \, \text{hours}$$

L10h​=60⋅1,5001,000,000​⋅1,200=90,0001,000,000​⋅1,200≈13,333 hours