Rolling and sliding contact bearings

Types and selection of rolling bearings by L10 life and equivalent load, and hydrodynamic journal bearings with the Sommerfeld number and Petroff's friction equation.

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

Why it matters

Every rotating shaft runs in bearings, and bearing failure is one of the most common reasons machines stop. A designer must choose between rolling-contact bearings (selected from a catalogue for a required life) and sliding (journal) bearings (designed for a lubricant film), and must calculate life, friction loss and heat for either.

Key ideas

Rolling-contact (anti-friction) bearings. Balls or rollers roll between inner and outer races, held apart by a cage.

  • Deep-groove ball: radial plus moderate axial load, high speed, the most common type.
  • Angular-contact ball: larger axial load in one direction; used in pairs on spindles.
  • Self-aligning ball or spherical roller: tolerate shaft misalignment.
  • Cylindrical roller: high radial load, little or no axial load.
  • Tapered roller: combined heavy radial and axial load (wheel hubs, gearboxes), mounted in pairs.
  • Needle roller: high radial capacity in a small radial space.
  • Thrust ball/roller: axial load only. They have low starting friction, need little lubrication and are standardised, but are noisier, have finite fatigue life, and are sensitive to shock and contamination.

Rating life. Rolling bearings fail by surface fatigue (spalling). Lives of identical bearings scatter widely, so the rating life L₁₀ is the life that 90 % of a large group reaches or exceeds. The basic dynamic load rating C (from the catalogue) is the load giving L₁₀ = 10⁶ revolutions. The static load rating C₀ limits permanent indentation when stationary or slowly turning.

Equivalent dynamic load. A combined radial load F_r and axial load F_a is replaced by P = X·F_r + Y·F_a (X, Y from the catalogue, depending on F_a/F_r and F_a/C₀), multiplied by a service or load factor for shock. Some catalogues also include a rotation factor V for outer-race rotation.

Sliding-contact (journal) bearings. A shaft (journal) turns in a bush. Lubrication regimes:

  • Hydrodynamic (thick film): the journal drags oil into a converging wedge, building pressure that separates the surfaces. Needs adequate speed and viscosity.
  • Mixed and boundary: partial or full metal contact at low speed or high load (high friction, wear).
  • Hydrostatic: pressure supplied by an external pump, giving a film even at zero speed. The Stribeck curve shows friction coefficient against the bearing characteristic μN/p: friction falls through boundary and mixed regimes to a minimum, then rises slowly in the hydrodynamic regime. Designers run well to the right of the minimum for stability.

Journal bearings are quiet, take shock well, have practically unlimited life with a full film, and are compact radially; they need a lubricant supply and have higher starting friction. They are preferred for very high speeds with heavy loads (turbines, large crankshafts).

Design numbers for journal bearings. Unit load p = W/(L·d); length-to-diameter ratio L/d (often 0.5 to 1.5); radial clearance c; Sommerfeld number S, a dimensionless group that fixes the operating eccentricity and friction (from Raimondi–Boyd charts). Petroff's equation gives the friction of a lightly loaded, concentric journal and is the usual GATE formula.

Formulas

Rolling bearing life: L₁₀ = (C/P)^k million revolutions; k = 3 (ball), 10/3 (roller).

Life in hours: L_h = L₁₀ × 10⁶/(60·N) (N in rev/min).

Equivalent load: P = (X·F_r + Y·F_a)·K_s (N); K_s service factor.

Required rating: C = P·L₁₀^(1/k) (L₁₀ in million revolutions).

Reliability (Weibull, typical): L/L₁₀ = [ln(1/R)/ln(1/0.9)]^(1/b) with b ≈ 1.5 for ball bearings (from your catalogue).

Journal bearing: p = W/(L·d) (Pa); S = (r/c)²·μ·n_s/p; μ absolute viscosity (Pa·s), n_s speed (rev/s), r journal radius (m), c radial clearance (m).

Petroff: f = 2π²·(μ·n_s/p)·(r/c); friction torque T_f = f·W·r (N·m); power loss P_f = 2π·n_s·T_f (W).

Worked examples

Example 1 (standard). A deep-groove ball bearing carries F_r = 4 kN and F_a = 1.5 kN at 720 rev/min, with X = 0.56 and Y = 1.6 (from the catalogue) and a service factor of 1.2. A life of 20 000 h is required. Find the dynamic load rating needed.

  1. P = (0.56 × 4000 + 1.6 × 1500) × 1.2 = (2240 + 2400) × 1.2 = 5568 N.
  2. Life in revolutions: L₁₀ = 60 × 720 × 20 000/10⁶ = 864 million rev.
  3. C = P·L₁₀^(1/3) = 5568 × 864^(1/3) = 5568 × 9.524 = 53 030 N.
  4. Select a bearing with C ≥ 53.0 kN from the catalogue (and check its X, Y values for the chosen size).

Example 2 (GATE level). A full journal bearing has d = 60 mm, L = 60 mm, radial clearance 0.03 mm and carries 6 kN at 1200 rev/min. Oil viscosity is 20 mPa·s. Using Petroff's equation, find the Sommerfeld number, coefficient of friction, friction torque and power lost.

  1. p = W/(L·d) = 6000/(0.06 × 0.06) = 1.667 × 10⁶ Pa; n_s = 1200/60 = 20 rev/s; r/c = 30/0.03 = 1000.
  2. S = (r/c)²·μ·n_s/p = 10⁶ × 0.02 × 20/1.667 × 10⁶ = 0.24.
  3. f = 2π² × (0.02 × 20/1.667 × 10⁶) × 1000 = 19.74 × 2.4 × 10⁻⁷ × 1000 = 0.00474.
  4. T_f = f·W·r = 0.00474 × 6000 × 0.03 = 0.853 N·m.
  5. P_f = 2π × 20 × 0.853 = 107 W.
  6. S = 0.24, f ≈ 0.0047, T_f ≈ 0.85 N·m, power loss ≈ 107 W (Petroff ignores eccentricity, so it slightly underestimates friction at this load).

Common mistakes

  • Treating L₁₀ = (C/P)³ as revolutions instead of millions of revolutions.
  • Using exponent 3 for roller bearings (it is 10/3).
  • Forgetting the axial load or the service factor when finding P.
  • Mixing diameter and radius, or diametral and radial clearance, in r/c.
  • Using rev/min instead of rev/s in S and Petroff's equation.
  • Assuming lower viscosity always reduces loss: too thin an oil drops the bearing into mixed lubrication and wear.

For GATE PI

Expect L₁₀ life or required C for a ball or roller bearing, life in hours, effect of doubling the load on life (it falls by 2³ = 8 for ball bearings), equivalent load with given X and Y, and Petroff friction, Sommerfeld number and power loss for journal bearings. Practise unit handling in the journal-bearing formulas.

Quick check

  1. A ball bearing has C = 5 kN and P = 1 kN. What is L₁₀?
  2. If the load on a ball bearing doubles, by what factor does its life change?
  3. Name a rolling bearing suited to heavy combined radial and axial load.
  4. Why is a hydrodynamic bearing unsuitable for frequent starts under load?

Answers: 1. 125 million revolutions; 2. it falls to 1/8; 3. tapered roller bearing; 4. at start-up there is no film, so it runs in boundary lubrication with high friction and wear.

Interview questions

All Theory of Machines and Machine Design 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, also known as an anti-friction bearing, uses balls or rollers to maintain the separation between the bearing races, reducing friction and wear. In contrast, a sliding contact bearing, or plain bearing, relies on a sliding motion between surfaces, which typically requires lubrication to minimize friction. Rolling contact bearings generally have lower friction and can handle higher speeds compared to sliding contact bearings.

  2. 2.Explain the working principle of a ball bearing.Concept

    A ball bearing consists of a series of balls placed between two races, an inner and an outer race. As the shaft rotates, the balls roll between the races, reducing friction and allowing smooth rotation. The balls carry the load and distribute it evenly across the races, which minimizes wear and extends the bearing's lifespan.

  3. 3.Why are tapered roller bearings used in vehicle wheel hubs?Application

    Tapered roller bearings are used in vehicle wheel hubs because they can handle both radial and axial loads. The tapered shape allows them to support large axial forces, which is essential for vehicle wheels that experience forces from different directions during operation. This makes them ideal for applications where stability and load-bearing capacity are critical.

  4. 4.What happens if a bearing is not properly lubricated?Application

    If a bearing is not properly lubricated, it can lead to increased friction and wear, resulting in overheating and potential failure. Lack of lubrication can cause metal-to-metal contact, leading to surface damage and reduced bearing life. Proper lubrication is essential to reduce friction, dissipate heat, and protect against corrosion.

  5. 5.Explain the concept of bearing life and the factors that affect it.Concept

    A rolling bearing eventually fails by surface fatigue of the races or rolling elements, and lives of identical bearings scatter widely. The rating life L10 is the life, usually in millions of revolutions, that 90 % of a large group reaches or exceeds; it varies as (C/P)³ for ball and (C/P)^(10/3) for roller bearings, so load is the dominant factor. Lubrication, contamination, temperature, misalignment and the required reliability also modify the life actually achieved.

  6. 6.How does misalignment affect bearing performance?Application

    Misalignment in bearings can lead to uneven load distribution, increased friction, and premature wear. It can cause excessive vibration and noise, reducing the bearing's efficiency and lifespan. Proper alignment is crucial to ensure even load distribution and optimal performance.

  7. 7.What is the purpose of a bearing cage?Concept

    A bearing cage, also known as a retainer, is used to separate and guide the rolling elements within a bearing. It helps maintain the correct spacing between the rolling elements, preventing them from coming into contact with each other. This reduces friction and wear, ensuring smooth operation and extending the bearing's life.

  8. 8.A ball bearing carries an equivalent radial load of 500 N and must have an L10 life of 100 million revolutions. What dynamic load rating is required?Numerical

    For ball bearings L10 = (C/P)³ with L10 in millions of revolutions, so C = P × L10^(1/3) = 500 × 100^(1/3) = 500 × 4.64 ≈ 2321 N. You would select from the catalogue a bearing with C of at least about 2.32 kN. For a roller bearing the exponent would be 10/3 instead of 3.

  9. 9.What are the advantages of using ceramic bearings over steel bearings?Application

    Ceramic bearings offer several advantages over steel bearings, including higher speed capabilities, lower friction, and reduced weight. They are also more resistant to corrosion and can operate at higher temperatures. However, they are generally more expensive and can be more brittle than steel bearings.

  10. 10.A cylindrical roller bearing has a dynamic load rating of 5000 N and carries an equivalent radial load of 2000 N. What is its L10 life?Numerical

    For roller bearings L10 = (C/P)^(10/3) million revolutions. Here (5000/2000)^(10/3) = 2.5^(10/3) ≈ 21.2 million revolutions. This is the life that 90 % of a large group of such bearings would reach or exceed; dividing by 60 × speed in rev/min converts it to hours.

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

Stuck on something here?

PreviousCouplings, clutches and brakes