A self-aligning ball bearing has a basic dynamic load rating (C10, for 106 revolutions) of 35 kN. If the equivalent radial load on the bearing is 45 kN, the expected life (in 10⁶ revolutions) is

The correct option is below 0.5.

To determine the expected life of the self-aligning ball bearing, we use the load-life relationship equation for rolling contact bearings:

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

Where:
• $L_{10}$ is the rated bearing life (in million revolutions or ${10}^6$ revolutions).
• $C$ is the basic dynamic load rating = 35 kN.
• $P$ is the equivalent radial load = 45 kN.
• $p$ is the life formula exponent, which is equal to 3 for ball bearings (and 10/3 for roller bearings).

Since the question specifies a self-aligning ball bearing, we substitute $p=3$ into the equation:

$L_{10}={\left(\frac{35}{45}\right)}^3$

Simplify the fraction inside the parentheses:

$\frac{35}{45}\approx 0.7778$

Now, calculate the cube of this value to find the life:

$L_{10}\approx {\left(0.7778\right)}^3\approx 0.47$

The expected life of the bearing is approximately 0.47 million revolutions. This value is below 0.5, corresponding to the correct option.