Rolling contact bearings: life and selection

Rolling bearing types, rating life and dynamic load rating, equivalent load with X and Y factors, variable loads and reliability, with a catalogue-selection and a variable-load life example.

Drafted with Aria, reviewed by the AiCanCode.org team. Spotted an error? Use Give Feedback at the bottom of the page.

Why it matters

Wheel hubs, gearboxes, differentials, alternators, water pumps and clutch release mechanisms all run on rolling bearings. Bearings are bought, not designed, so the engineer's job is to calculate the equivalent load and required life, then pick the smallest catalogue bearing whose rating is enough. A wrong exponent or a units slip in the life equation can change the answer by a factor of 60.

Key ideas

Types

  • Deep-groove ball bearing: radial load plus moderate axial load in either direction; high speed; the default choice.
  • Angular-contact ball bearing: larger axial load in one direction; used in pairs.
  • Cylindrical roller bearing: high radial load, almost no axial load (line contact).
  • Tapered roller bearing: large combined radial and axial load in one direction; used in pairs in wheel hubs and differential pinions.
  • Needle roller bearing: high radial capacity in a small radial space (gearbox idlers, universal joint crosses).
  • Thrust bearings: pure axial load.

Failure and life. A properly lubricated bearing eventually fails by surface fatigue (spalling) of the races or rolling elements. Life scatters widely, so it is stated statistically. The rating life L₁₀ is the life in revolutions that 90 % of a large group of identical bearings reach or exceed.

Dynamic load rating C is the constant load under which the bearing has L₁₀ = 1 million revolutions. It is listed in the catalogue for each bearing. The static load rating C₀ limits the load on a stationary or very slowly rotating bearing (permanent indentation, brinelling).

Load-life relation: L₁₀ = (C/P)^k million revolutions, with k = 3 for ball bearings and k = 10/3 for roller bearings. Doubling the load cuts ball-bearing life to one-eighth.

Equivalent dynamic load combines radial and axial loads: P = X·V·F_r + Y·F_a. The radial factor X, thrust factor Y and the limit e (below which F_a is ignored, so X = 1, Y = 0) come from the catalogue and depend on F_a/C₀. The rotation factor V = 1 when the inner ring rotates (in many modern catalogues V is dropped). A load or service factor (about 1.0–1.5 for smooth to moderate shock; take it from the data book) multiplies P.

Variable loads: if the load P_i acts for a fraction α_i of the revolutions, use the cubic mean P_e = (Σ α_i·P_i³)^(1/3) for ball bearings.

Reliability other than 90 %: from the Weibull distribution, L/L₁₀ = [ln(1/R)/ln(1/0.9)]^(1/b), with b ≈ 1.17 in many Indian textbooks (b ≈ 1.5 in some manufacturers' data). Higher reliability means a shorter usable life for the same bearing.

Selection procedure: find F_r and F_a at each bearing; choose a type; estimate P with X, Y and the load factor; convert the required life in hours to million revolutions; compute the required C = P·L₁₀^(1/k); pick the smallest bearing from the catalogue with C at least this value and a bore that fits the shaft; recheck X and Y with the chosen bearing's C₀.

Formulas

L₁₀ = (C / P)^k (million revolutions; k = 3 ball, 10/3 roller) L₁₀h = L₁₀ × 10⁶ / (60·N) Required C = P · L₁₀^(1/k) P = (X·V·F_r + Y·F_a)·K_s P_e = (Σ α_i·P_i³)^(1/3) (ball bearings, variable load) L / L₁₀ = [ln(1/R) / ln(1/0.9)]^(1/b)

Symbols: L₁₀ = rating life (million revolutions); L₁₀h = life (hours); N = speed (rpm); C = dynamic load rating (N); P = equivalent dynamic load (N); F_r, F_a = radial and axial loads (N); X, Y = radial and thrust factors (–); V = rotation factor (–); K_s = load or service factor (–); α_i = fraction of revolutions at load P_i; R = required reliability (fraction); b = Weibull slope (–).

Worked examples

Example 1 (standard). A deep-groove ball bearing on a gearbox shaft carries F_r = 4 kN and F_a = 1.5 kN at 1000 rpm. From the catalogue, X = 0.56 and Y = 1.6 for this load ratio; load factor 1.2. Required life 10 000 h. Find the required dynamic capacity.

  1. P = (0.56 × 4000 + 1.6 × 1500) × 1.2 = (2240 + 2400) × 1.2 = 5568 N.
  2. L₁₀ = 60 × 1000 × 10 000 / 10⁶ = 600 million revolutions.
  3. C = P·L₁₀^(1/3) = 5568 × 600^(1/3) = 5568 × 8.434 = 46 960 N ≈ 47 kN.
  4. Choose the smallest bearing in the catalogue with C ≥ 47 kN and a suitable bore, then recheck X and Y with its C₀.

Example 2 (GATE level). A ball bearing with C = 40 kN runs at 750 rpm. It carries 3 kN for 50 % of the time, 5 kN for 30 % and 8 kN for 20 %. Find its life in hours.

  1. P_e = (0.5 × 3³ + 0.3 × 5³ + 0.2 × 8³)^(1/3) kN = (13.5 + 37.5 + 102.4)^(1/3) = 153.4^(1/3) = 5.353 kN.
  2. L₁₀ = (40/5.353)³ = 417.2 million revolutions.
  3. L₁₀h = 417.2 × 10⁶ / (60 × 750) = 9271 h.
  4. Note how the 8 kN load, present only 20 % of the time, contributes two-thirds of the sum: high loads dominate bearing life.

Common mistakes

  • Using k = 3 for roller bearings (it is 10/3).
  • Forgetting the 10⁶ when converting L₁₀ to hours, or dividing by N instead of 60·N.
  • Using the arithmetic mean of variable loads instead of the cubic mean.
  • Adding F_r and F_a directly instead of using X and Y.
  • Applying the load factor twice or not at all.
  • Assuming a higher reliability allows a longer life; it is the opposite.

For GATE ME

Very frequent: L₁₀ in revolutions or hours from C and P; required C for a given life; how life changes when the load or speed changes; equivalent load for variable loading; and reliability conversions. Practise ratio questions such as "load increased by 25 %, new life?" (life × 1/1.25³ = 0.512 for ball bearings).

Quick check

  1. A ball bearing has C = 20 kN and P = 4 kN. Find L₁₀ in million revolutions.
  2. The same bearing runs at 500 rpm. Life in hours?
  3. If the load on a ball bearing doubles, what happens to its life?
  4. Which bearing type suits a wheel hub with combined loads?
  5. What life exponent is used for roller bearings?

Answers: 1. 125. 2. 125 × 10⁶/(60 × 500) = 4167 h. 3. It falls to one-eighth. 4. A pair of tapered roller bearings. 5. 10/3.

Interview questions

All Design of Machine and Automotive Elements interview questions

Try answering each one aloud before you open it.

  1. 1.What is a rolling contact bearing and how does it differ from a sliding contact bearing?Concept

    A rolling contact bearing is a type of bearing that uses rolling elements, such as balls or rollers, to maintain the separation between moving parts. It reduces friction and supports radial and axial loads. In contrast, a sliding contact bearing, also known as a plain bearing, relies on sliding motion between surfaces, which typically results in higher friction compared to rolling contact bearings.

  2. 2.Explain the concept of bearing life in the context of rolling contact bearings.Concept

    Bearing life refers to the duration or number of revolutions a bearing can endure before showing signs of fatigue or failure. It is often expressed in terms of L10 life, which is the number of revolutions at which 90% of a group of identical bearings will still be operational under specified conditions. This concept helps in predicting the reliability and performance of bearings in various applications.

  3. 3.What factors influence the selection of rolling contact bearings for a specific application?Concept

    The selection of rolling contact bearings depends on several factors, including load capacity (both radial and axial), speed of operation, environmental conditions (such as temperature and contamination), space constraints, and the desired lifespan of the bearing. Additionally, cost and availability of the bearing type may also influence the selection process.

  4. 4.Why are ball bearings commonly used in automotive applications?Application

    Ball bearings are commonly used in automotive applications because they offer low friction, high-speed capability, and can support both radial and axial loads. Their design allows for smooth rotation and efficient power transmission, which is essential in automotive components like wheels, transmissions, and engines. Additionally, ball bearings are relatively easy to maintain and replace.

  5. 5.What happens if a rolling contact bearing is subjected to loads beyond its rated capacity?Application

    If a rolling contact bearing is subjected to loads beyond its rated capacity, it can lead to premature failure due to excessive stress and deformation of the rolling elements and raceways. This can cause increased friction, overheating, and ultimately, bearing seizure. It may also result in damage to the associated machinery or equipment.

  6. 6.How does lubrication affect the life of a rolling contact bearing?Application

    Lubrication plays a crucial role in extending the life of a rolling contact bearing by reducing friction and wear between the rolling elements and raceways. It also helps in dissipating heat and protecting against corrosion. Inadequate or improper lubrication can lead to increased friction, overheating, and accelerated wear, significantly reducing the bearing's lifespan.

  7. 7.What is the significance of the dynamic load rating in rolling contact bearings?Concept

    The basic dynamic load rating C is the constant radial load (axial for thrust bearings) under which a group of identical bearings reaches a rating life L₁₀ of one million revolutions. It is a catalogue value, not a load the bearing should actually carry. Life at the real equivalent load P follows L₁₀ = (C/P)^k million revolutions, with k = 3 for ball and 10/3 for roller bearings, so the required C for a given life is P·L₁₀^(1/k). It is different from the static rating C₀, which limits permanent indentation when the bearing is stationary.

  8. 8.Calculate the L10 life of a ball bearing with a dynamic load rating of 5000 N, subjected to a constant radial load of 2000 N.Numerical

    For a ball bearing the life exponent is 3, so L₁₀ = (C/P)³ = (5000/2000)³ = 2.5³ = 15.625 million revolutions. For a roller bearing the exponent would be 10/3, giving about 21.2 million revolutions. To get hours, divide the revolutions by 60 times the speed in rpm.

  9. 9.A bearing operates at 1500 RPM and has an L10 life of 20 million revolutions. Calculate its expected life in hours.Numerical

    The expected life in hours can be calculated using the formula: Life (hours) = L10 / (RPM * 60). Substituting the given values: Life (hours) = 20,000,000 / (1500 * 60) = 222.22 hours.

  10. 10.Explain why tapered roller bearings are preferred in applications with combined radial and axial loads.Application

    Tapered roller bearings are preferred in applications with combined radial and axial loads because their design allows them to handle both types of loads efficiently. The tapered shape of the rollers and raceways enables them to support axial loads in one direction while also accommodating radial loads. This makes them ideal for applications like vehicle wheel hubs, where both load types are present.

Finished this topic? Mark it so your progress, study plan and readiness keep up.

Stuck on something here?

PreviousHelical and leaf springs for vehicle suspensionNext Sliding contact bearings and hydrodynamic lubrication