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

Step 1: Write the rating life directly in millions of revolutions.
The bearing runs at \(N = 500\) rpm for a design life of \(L_H = 6000\) hours, so the number of revolutions it must survive is \(60 \times N \times L_H\) converted to millions: \[ L_{10} = \frac{60 \times 500 \times 6000}{10^6} = 180 \text{ million revolutions} \]

Step 2: Apply the load life relation for ball bearings.
For ball bearings the load life index is \(k=3\), so \(L_{10} = (C/P)^3\), which gives \(C/P = (L_{10})^{1/3}\). Taking logarithms is quicker than guessing cube roots: \(\log_{10}(180) = 2.2553\), so \(\dfrac{1}{3}\log_{10}(180) = 0.7518\), and \(10^{0.7518} \approx 5.65\). So \(C/P \approx 5.65\).

Step 3: Solve for the equivalent load.
With \(C = 50\) kN, \[ P = \frac{C}{5.65} = \frac{50}{5.65} \approx 8.85 \text{ kN} \]
This matches option (2).