In certain applications, ball bearings are subjected to cyclic loads and speeds. As an example, consider a ball bearing operating under the following conditions:

(i) radial load 2500 N at 700 rpm for 25% of the time

(ii) radial load 5000 N at 900 rpm for 50% of the time

(iii) radial load 1000 N at 750 rpm for the remaining 25% of the time. Under these circumstances, it is necessary to consider the complete work cycle while finding out the dynamic load capacity of the bearing.

The procedure consists of dividing the work cycle into a number of elements, during which the operating conditions of load and speed are constant.

Suppose that the work cycle is divided into x elements. Let $\mathrm{P}_1, \mathrm{P}_2, \ldots \mathrm{P_x}$ be the loads and $n_1, n_2, \ldots, n x$ be the speeds during these elements. During the first element, the life $\mathrm{L}_{1}$ corresponding to $\operatorname{load} \mathrm{P}_{1},$ is given by $\mathrm{L}_{1}=\left(\mathrm{C} / \mathrm{P}_{1}\right)^{3} \times 10^{6}$ rev.

$\therefore$ Fraction of life consumed by the first element with load $\mathrm{P}_{1}$ acting for $\mathrm{n}_{1}$ number of revolutions is, $\mathrm{n}_{1} / \mathrm{L}_{1}=\mathrm{n}_{1} \cdot\left(\mathrm{P}_{1} / \mathrm{C}\right)^{3} / 10^{6}$ and second element is $\mathrm{n}_{2} \cdot \mathrm{P}_{2} / \mathrm{C} )^{3} / 10^{6}$

Also, $\mathrm{n}_{1} / \mathrm{L}_{1}+\mathrm{n}_{2} / \mathrm{L}_{2}+\ldots = 1$ -------- Equation 1

If $P_e$ is the equivalent load for the complete work cycle, the life consumed by the work cycle is given by,

$\mathrm{n} \cdot\left(\mathrm{P}_{\mathrm{e}} / \mathrm{C}\right)^{3} / 10^{6}$ -------- Equation 2

where $n=n_{1}+n_{2}+\ldots$

equating equations 1 and 2

$\mathrm{n} \cdot\left(\mathrm{P}_{e} / \mathrm{C}\right)^{3} / 106=\mathrm{n} 1 .\left(\mathrm{P}_{1} / \mathrm{C}\right)^{3} / 10^{6}+\mathrm{n} 1 . \mathrm{P}_{1} / \mathrm{C} )^{3} / 10^{6}$

$\mathrm{n} \mathrm{P}_{\mathrm{e}}^{3}=\mathrm{n}_{1} \mathrm{P}_{1}^{3}+\mathrm{n}_{2} \mathrm{P}_{2}^{3}+\mathrm{n}_{3} \mathrm{P}_{3}^{3}$

$\therefore P_{e}=\sqrt[3]{\frac{n_{1} P_{1}^{3}+n_{2} P_{2}^{3}+n_{3} P_{3}^{3}----}{n}}$

Also, the equivalent load is called a cubic mean load ($Fm$).

$\therefore F_{m}=\sqrt[3]{\frac{n_{1} F_{1}^{3}+n_{2} F_{2}^{3}+n_{3} F_{3}^{3}--}{\Sigma n}}$ OR $\therefore F_{m}=\sqrt[3]{\frac{t_{1} P_{1}^{3}+t_{2} P_{2}^{3}+t_{3} P_{3}^{3}----}{\Sigma t}}$ --- PSG4.2

While dealing with varying load and speed condition prepare the table as below and determine the cubic mean load.

Then the most general form to determine the cubic load is given below

$P_{e}=\sqrt[K]{\frac{\sum t i N i P e i^{k}}{\sum t i N i}}$