Subtopic: Measurement of force, stress, torque, speed, displacement and acceleration on machine elements – shafts, couplings, keys, bearings and knuckle joints
Question Statement:
A deep-groove ball bearing supporting the cutter shaft of a forage harvester operates under a constant equivalent radial load of $P = 4.5\text{ kN}$ at a rotational speed of $N = 1200\text{ rpm}$. The basic dynamic load rating of the bearing is $C = 27\text{ kN}$. The rating life ($L_{10h}$) of the bearing in hours is ________ (round off to integer).
Answer Key:
3000
Step-by-Step Solution & Explanation
```
Ball Bearing Life Equation:
L10 (million revs) = (C / P)^p [p = 3 for ball bearings]
```
**Step-by-step Calculation:**
1. For ball bearings, the life exponent is $p = 3$:
$L_{10} = \left(\frac{C}{P}\right)^3 = \left(\frac{27\text{ kN}}{4.5\text{ kN}}\right)^3 = (6)^3 = 216\text{ million revolutions}$
2. Converting rating life in million revolutions to operating hours at $N = 1200\text{ rpm}$:
$L_{10h} = \frac{L_{10} \times 10^6}{60 \times N} = \frac{216 \times 10^6}{60 \times 1200} = \frac{216 \times 10^6}{72000} = 3000\text{ hours}$