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Life of a deep groove ball bearing

Short answer

Fa/(V Fr)=0.344 > e, use X=0.56, Y=1.64. P=0.56*3.2+1.64*1.1=3.596 kN; with service factor 1.3, Pe=4.675 kN. L10=(29.5/4.675)^3=251 million rev; L10h=251e6/(60*1450)=2888 h.

A deep groove ball bearing has basic dynamic load rating C = 29.5 kN and static rating C0 = 15 kN. It carries a radial load of 3.2 kN and an axial load of 1.1 kN at 1450 rpm, with the inner ring rotating (V = 1). Take e = 0.26; for Fa/(V Fr) > e use X = 0.56 and Y = 1.64. The service factor for moderate shock is 1.3. Find the equivalent load, the L10 life in million revolutions and in hours.

Check the load ratio

FaVFr=1.11×3.2=0.344>e=0.26\frac{F_a}{V F_r} = \frac{1.1}{1 \times 3.2} = 0.344 > e = 0.26

So X=0.56X = 0.56 and Y=1.64Y = 1.64.

Equivalent load

P=XVFr+YFa=0.56×1×3.2+1.64×1.1=1.792+1.804=3.596 kN\begin{aligned} P &= X V F_r + Y F_a = 0.56 \times 1 \times 3.2 + 1.64 \times 1.1 \\ &= 1.792 + 1.804 = 3.596\ \text{kN} \end{aligned}

With the service factor 1.3:

Pe=1.3×3.596=4.675 kNP_e = 1.3 \times 3.596 = 4.675\ \text{kN}

Life

For a ball bearing k=3k = 3:

L10=(CPe)3=(29.54.675)3=(6.310)3=251.3 million revolutionsL_{10} = \left(\frac{C}{P_e}\right)^3 = \left(\frac{29.5}{4.675}\right)^3 = (6.310)^3 = 251.3\ \text{million revolutions} L10h=251.3×10660×1450=2888 hoursL_{10h} = \frac{251.3 \times 10^6}{60 \times 1450} = 2888\ \text{hours}

The life is about 2900 hours (about 120 days of continuous running), which is short for a continuous-duty machine. A bearing of higher rating would be selected for 10000 hours or more.

Answer: P = 3.6 kN (4.68 kN with service factor); L10 = 251 million revolutions = about 2890 hours.

✓ Answer checked by Claude Opus 5.5 (high effort)

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