Required rating of a tapered roller bearing

Equivalent load

$$\frac{F_a}{V F_r} = \frac{2}{1 \times 5} = 0.40 > e = 0.35$$

So $X = 0.4$ and $Y = 1.7$.

$$\begin{aligned} P &= X V F_r + Y F_a = 0.4 \times 5 + 1.7 \times 2 = 5.4\ \text{kN} \\ P_e &= 1.2 \times 5.4 = 6.48\ \text{kN} \end{aligned}$$

Required life in million revolutions

$$L_{10} = \frac{60\, n\, L_{10h}}{10^6} = \frac{60 \times 600 \times 10\,000}{10^6} = 360\ \text{million revolutions}$$

Dynamic rating

For a roller bearing the exponent is $k = 10/3$:

$$C = P_e\, L_{10}^{1/k} = 6.48 \times 360^{0.3} = 6.48 \times 5.846 = 37.9\ \text{kN}$$

Choosing the bearing

From the manufacturer's catalogue, select a tapered roller bearing whose bore equals the shaft diameter and whose basic dynamic rating is not less than 37.9 kN. Then check that:

• the static rating $C_0$ exceeds the maximum static load;
• the limiting speed is above 600 rpm;
• the $X$ and $Y$ values and $e$ of the selected bearing match those assumed (if they differ, recompute $P$ and $C$).

Tapered bearings are normally mounted in opposed pairs, and each radial load induces an axial thrust that must be added to the external thrust when the loads are split between the two bearings.

Answer: Required C = 37.9 kN (P = 5.4 kN, 6.48 kN with service factor).