Rolling contact bearing

A rolling contact (anti-friction) bearing supports a shaft through balls or rollers placed between an inner and an outer ring, so that sliding friction is replaced by rolling friction.

Main types

Roller bearings have line contact and carry higher loads than ball bearings, which have point contact and run faster with less friction.

Rolling and sliding bearings

Definitions

• Bearing life: the number of revolutions (or hours at constant speed) that a bearing completes before the first sign of fatigue (spalling) of the rolling surfaces.
• Rated life $L_{10}$: the life, in million revolutions, that 90 percent of a large group of identical bearings will reach or exceed under the same conditions. The median life is about five times $L_{10}$.
• Basic dynamic load rating $C$: the constant radial load that a group of bearings can carry for a rated life of one million revolutions with 90 percent reliability.
• Equivalent dynamic load $P$: the single radial load that would give the same life as the actual combination of radial and axial loads.

Relations

$$L_{10} = \left(\frac{C}{P}\right)^k \quad \text{million revolutions}$$

with $k = 3$ for ball bearings and $k = 10/3$ for roller bearings. Life in hours:

$$L_{10h} = \frac{10^6\, L_{10}}{60\, n}$$

Equivalent load:

$$P = X V F_r + Y F_a$$

where $V = 1$ when the inner ring rotates (1.2 when the outer ring rotates), and $X$ and $Y$ are radial and axial factors from the catalogue, depending on $F_a/F_r$ compared with $e$. If $F_a/(VF_r) \le e$, then $X = 1$ and $Y = 0$. A service (shock) factor multiplies $P$ for rough loads.

Selection steps

1. Find the radial and axial loads, speed and required life in hours.
2. Choose the type of bearing from the load direction, speed and space.
3. Assume $X$, $Y$ and calculate the equivalent load $P$ including the service factor.
4. Calculate $L_{10}$ in million revolutions and the required $C = P\,L_{10}^{1/k}$.
5. Choose from the catalogue a bearing of suitable bore with $C$ not less than the required value; check the static rating $C_0$ and the limiting speed.
6. Recheck $X$, $Y$ with the actual $C_0$ and repeat if necessary.

Check the load ratio

$$\frac{F_a}{V F_r} = \frac{1.1}{1 \times 3.2} = 0.344 > e = 0.26$$

So $X = 0.56$ and $Y = 1.64$.

Equivalent load

$$\begin{aligned} P &= X V F_r + Y F_a = 0.56 \times 1 \times 3.2 + 1.64 \times 1.1 \\ &= 1.792 + 1.804 = 3.596\ \text{kN} \end{aligned}$$

With the service factor 1.3:

$$P_e = 1.3 \times 3.596 = 4.675\ \text{kN}$$

Life

For a ball bearing $k = 3$:

$$L_{10} = \left(\frac{C}{P_e}\right)^3 = \left(\frac{29.5}{4.675}\right)^3 = (6.310)^3 = 251.3\ \text{million revolutions}$$ $$L_{10h} = \frac{251.3 \times 10^6}{60 \times 1450} = 2888\ \text{hours}$$

The life is about 2900 hours (about 120 days of continuous running), which is short for a continuous-duty machine. A bearing of higher rating would be selected for 10000 hours or more.

Answer: P = 3.6 kN (4.68 kN with service factor); L10 = 251 million revolutions = about 2890 hours.

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).