Question

If a radial ball bearing has a basic load rating of \(50 \text{ kN}\) and desired rating life is \(6000 \text{ hours}\), then the equivalent radial load that the bearing can carry at \(500 \text{ rpm}\) is

• A. \(18.85 \text{ kN}\)
• B. \(8.85 \text{ kN}\)
• C. \(12 \text{ kN}\)
• D. \(10 \text{ kN}\)

Hint:

Always double check the exponent value matching your bearing profile type: - Ball Bearing: \(L_{10} = (C/P)^3 \implies P = C / \sqrt[3]{L_{10}}\) - Roller Bearing: \(L_{10} = (C/P)^{10/3} \implies P = C / (L_{10})^{0.3}\) Converting hours to millions of revs uses the factor \(\frac{60 \cdot N \cdot L_H}{10^6}\).

Answer 1

The Correct Option is B

Concept: The operational engineering life of a rolling element bearing is modeled by the standard load-life empirical equation defined by the catalog standards of the anti-friction bearing manufacturers (SKF/ISO): \[ L_{10} = \left(\frac{C}{P}\right)^k \] Where:
• \(L_{10}\) = Nominal rating life expressed in millions of revolutions (\(10^6\) revs).
• \(C\) = Basic dynamic load rating of the bearing (given here as \(50 \text{ kN}\)).
• \(P\) = Equivalent dynamic radial load acting on the bearing (the unknown variable to solve for).
• \(k\) = Load-life exponent, which depends on the geometric shape of the rolling elements:
• \(k = 3\) for ball bearings (point contact configurations).
• \(k = \frac{10}{3}\) for roller bearings (line contact configurations). Since the problem specifies a radial ball bearing, we use \(k = 3\).

Step 1: Convert the specified lifetime from hours into millions of revolutions.
The relation linking life in hours (\(L_H\)) and rotational velocity in revolutions per minute (\(N\)) to life in millions of revolutions (\(L_{10}\)) is: \[ L_{10} = \frac{60 \times N \times L_H}{10^6} \] Given information:
• Desired rating life, \(L_H = 6000 \text{ hours}\)
• Rotational speed, \(N = 500 \text{ rpm}\) Substituting these values: \[ L_{10} = \frac{60 \times 500 \times 6000}{10^6} \] Multiplying the numerator constants out step-by-step: \[ 60 \times 500 = 30,000 \] \[ 30,000 \times 6000 = 180,000,000 = 180 \times 10^6 \] Dividing by the denominator: \[ L_{10} = \frac{180 \times 10^6}{10^6} = 180 \text{ million revolutions} \]

Step 2: Isolate the equivalent dynamic load parameter \(P\).
Using the load-life formula with \(k = 3\): \[ 180 = \left(\frac{C}{P}\right)^3 \] Taking the cube root on both sides: \[ (180)^{1/3} = \frac{C}{P} \quad \Rightarrow \quad P = \frac{C}{(180)^{1/3}} \]

Step 3: Perform numerical calculation.
First, compute the cube root of 180: \[ 5^3 = 125, \quad 6^3 = 216 \] Approximating further: \[ (5.646)^3 \approx 180 \quad \Rightarrow \quad (180)^{1/3} \approx 5.646 \] Substitute \(C = 50 \text{ kN}\) and this root into the isolated equation: \[ P = \frac{50}{5.646} \approx 8.8556 \text{ kN} \] Rounding to two decimal places yields \(8.85 \text{ kN}\), which matches Option (2).