Mechanical Engineering Design · Part 3 — Design of Mechanical Elements

Rolling-Contact Bearings

Selecting catalog bearings from load, life, and reliability requirements: the load-life power law, the Weibull reliability model, the master catalog-rating equation, equivalent loading under combined radial and thrust forces, variable loads, and the selection of ball, cylindrical, and tapered roller bearings.

11–1Bearing Types

Learning objectives

Recognize bearing families from radial load, axial load, speed, alignment, space, and service constraints.
Convert a shaft free-body diagram and duty cycle into bearing radial and thrust reactions.
Calculate equivalent load, rating life, and required dynamic capacity without confusing load and capacity.
Use reliability and application factors with their stated statistical and data-source limitations.
Iterate a catalog selection when static rating, combined-load factors, or mounting details depend on the candidate.
Verify fatigue life, static safety, speed, fits, lubrication, sealing, stiffness, and manufacturability.

A Catalog-Selection Discipline

Rolling-contact bearings are precision, mass-produced components selected from manufacturers' catalogs rather than designed from scratch. The engineer's task is to match the application's load, speed, life, and reliability to a published rating. The two broad families are ball bearings and roller bearings.

Common Configurations

Deep-groove ball bearings take radial and moderate thrust loads; angular-contact ball bearings take heavier thrust in one direction. Cylindrical roller bearings carry high radial loads but little thrust; tapered roller bearings carry heavy combined radial and thrust loads; needle bearings suit confined radial spaces; spherical bearings accommodate misalignment. Each type trades load capacity, thrust capability, speed, and misalignment tolerance.

Physical meaning

A rolling bearing redirects distributed Hertzian contacts into housing reactions. The external shaft load is not the catalog rating: equilibrium determines the applied reaction, whereas \(C_{10}\) is a manufacturer-tested fatigue-capacity statistic. Bearing stiffness, clearance, and fits also control shaft position even when fatigue strength is ample.

Problem-solving strategy

Draw the shaft free-body diagram, set positive radial and axial directions, and solve reactions at each bearing.
Classify steady, combined, or variable loading and select the matching equivalent-load model.
Convert service hours and speed to revolutions, then apply the correct ball or roller exponent.
Choose a preliminary type and catalog candidate; use only factors published for that bearing series.
Iterate combined-load factors and audit life, reliability, static rating, limiting speed, fits, and lubrication.

11–2Bearing Life

Life Is a Statistical Quantity

Bearing failure is surface fatigue (pitting/spalling), so life — the number of inner-ring revolutions (or hours at constant speed) until the first evidence of fatigue — is inherently stochastic and quoted statistically. Identical bearings under identical loads fail at widely scattered lives.

Rating Life and Median Life

The rating life \(L_{10}\) (also "minimum," "B10," or "L10" life) is the life 90% of a group will reach or exceed before failure — the 10th percentile of the revolutions-to-failure distribution. Median life (50th percentile) runs 4–5 times the \(L_{10}\) life. Manufacturers pick a rating life for their catalog ratings; the most common is \(10^6\) revolutions (Timken uses 90×10⁶ rev, i.e. 3000 h at 500 rev/min).

11–3Bearing Load Life at Rated Reliability

The Load-Life Power Law

Testing identical bearing groups at different loads and plotting load vs. \(L_{10}\) life on log-log axes gives a straight line: \(F L^{1/a} = \) constant, all at 0.90 reliability. The exponent is \(a = 3\) for ball bearings and \(a = 10/3\) for roller bearings. The catalog load rating \(C_{10}\) (Basic Dynamic Load Rating) is the radial load causing 10% failure at the rating life.

Load-Life Relation

\[ F_1 L_1^{1/a} = F_2 L_2^{1/a} \qquad (a = 3\ \text{ball},\ a = 10/3\ \text{roller}) \]Eqs. (11–1, 11–2)

Catalog Rating from Desired Load & Life

\[ C_{10} = F_D\left(\frac{L_D}{L_R}\right)^{1/a} \]Eq. (11–3)

Life in Revolutions from Hours

\[ L = 60\,\ell\,n \quad (\ell\ \text{hours},\ n\ \text{rev/min}) \]Eq. (11–4)

Assumptions and applicability

The basic rating relation models rolling-contact fatigue under properly mounted, clean, adequately lubricated service. It does not predict seizure, brinelling, cage damage, wear, corrosion, false brinelling, or failure from contamination. \(F\), \(C_{10}\), and \(L_R\) must use one manufacturer's rating basis and consistent units.

11–4Reliability versus Life — The Weibull Distribution

Why Weibull

The three-parameter Weibull distribution models bearing reliability because it flexibly fits the skewed life data. Reliability at a dimensionless life \(x = L/L_{10}\) is given by the relation below, with three parameters: \(x_0\) (the guaranteed minimum life), \(\theta\) (a characteristic value), and \(b\) (the shape parameter controlling skewness).

Weibull Reliability

\[ R = \exp\!\left[-\left(\frac{x - x_0}{\theta - x_0}\right)^{b}\right], \qquad x = \frac{L}{L_{10}} \]Eq. (11–5)

Dimensionless Life at Reliability R

\[ x = x_0 + (\theta - x_0)\left[\ln\frac{1}{R}\right]^{1/b} \]Eq. (11–8)

11–5Relating Load, Life, and Reliability

The Master Catalog-Rating Equation

Combining the load-life power law with the Weibull reliability model gives the central design equation: the required catalog rating \(C_{10}\) for a desired load \(F_D\), desired life \(x_D = L_D/L_{10}\), and reliability goal \(R_D\). An application factor \(a_f\) inflates the design load to cover shock and overload (a factor of safety on load). For \(R_D > 0.90\), the denominator drops below 1, raising the required rating.

Master Design Equation (catalog rating)

\[ C_{10} = a_f\,F_D\left[\frac{x_D}{x_0 + (\theta - x_0)\,(\ln(1/R_D))^{1/b}}\right]^{1/a} \]Eq. (11–9)

Simplified Form (R ≥ 0.90, p_f = 1 − R_D)

\[ C_{10} = a_f\,F_D\left[\frac{x_D}{x_0 + (\theta - x_0)(1 - R_D)^{1/b}}\right]^{1/a} \]Eq. (11–10)

11–6Combined Radial and Thrust Loading

The Equivalent Radial Load

A ball bearing resists radial and thrust loads simultaneously. The equivalent radial load \(F_e\) does the same fatigue damage as the combined radial \(F_r\) and axial \(F_a\) loads acting together. A rotation factor \(V\) corrects for which ring rotates: \(V = 1\) (inner ring) or \(V = 1.2\) (outer ring). The relationship between the dimensionless groups \(F_e/(VF_r)\) and \(F_a/(VF_r)\) is two straight-line segments meeting at a transition ratio \(e\).

Equivalent Radial Load

\[ F_e = X_i\,V F_r + Y_i\,F_a, \qquad i = \begin{cases} 1 & F_a/(VF_r) \le e \\ 2 & F_a/(VF_r) > e \end{cases} \]Eq. (11–12)

Damage weighting

The exponent makes brief high loads disproportionately important: doubling a ball-bearing load multiplies its fatigue contribution by eight. Fractions \(f_i\) must represent fractions of total revolutions, not merely clock time when speed changes. This is a fatigue model, not a static-impact check.

Finding X and Y

The factors \(X_1, Y_1, X_2, Y_2\) and the transition value \(e\) depend on the ratio \(F_a/C_0\), where \(C_0\) is the basic static load rating (tabulated alongside \(C_{10}\)). Below the transition (\(F_a/(VF_r) \le e\)) thrust scarcely matters (\(X_1 = 1, Y_1 = 0\)); above it, thrust dominates the equivalent load. Self-aligning bearings use \(V = 1\) for either ring.

11–7Variable Loading

Equivalent Steady Load for a Duty Cycle

When load or speed varies through a duty cycle, the fatigue damage is integrated to find a single equivalent constant load \(F_{eq}\) that produces the same damage. For a piecewise cycle in which load \(F_i\) acts for a fraction of revolutions, the cube-weighted (or \(a\)-power-weighted) average applies, mirroring Miner's cumulative-damage logic.

Equivalent Periodic Load

\[ F_{eq} = \left[\sum f_i\,(a_{f,i}\,F_i)^{a}\right]^{1/a} \]Eq. (11–17)

Integrated Form

For a continuously varying load \(F(\theta)\) over one period, the sum becomes an integral \(F_{eq} = \left[\frac{1}{\Theta}\int_0^{\Theta} F^a\,d\theta\right]^{1/a}\). The result is then used as \(F_D\) in the master rating equation, so variable-load selection reduces to the steady-load procedure once \(F_{eq}\) is known.

11–8Selection of Ball and Cylindrical Roller Bearings

The Selection Procedure

The workflow: (1) determine \(F_r\), \(F_a\), speed, desired life, and reliability; (2) compute \(F_{eq}\) for combined or variable loads; (3) evaluate the required \(C_{10}\) from the master equation; (4) choose a candidate from the catalog with \(C_{10}\) at least equal to the required value; (5) since \(F_e\) depends on \(F_a/C_0\) (and thus on the chosen bearing), iterate once or twice until the assumed and tabulated \(X, Y, e\) agree. Cylindrical roller bearings (no thrust capacity, \(a = 10/3\)) skip the equivalent-load step.

11–9Selection of Tapered Roller Bearings

Induced Thrust in Mounted Pairs

Tapered roller bearings carry combined radial and thrust loads but, because of their geometry, a radial load induces an axial reaction. They are mounted in opposed pairs (direct or indirect). The induced thrust from a radial load is \(F_i = 0.47 F_r/K\), where \(K\) is the bearing's ratio of radial to thrust rating (≈1.5 default until a specific bearing is chosen). The net thrust on each cup depends on the external thrust and the difference of the two induced thrusts, requiring a small bookkeeping procedure for which bearing carries the thrust.

Induced Thrust & Dynamic Equivalent Load

\[ F_i = \frac{0.47\,F_r}{K} \qquad\qquad F_e = 0.4\,F_r + K\,F_a \]Eqs. (11–18, 11–19)

Manufacturer-Specific Ratings

Tapered roller catalogs (e.g. Timken) rate at 90×10⁶ revolutions rather than 10⁶, so the rating life \(L_R\) in the master equation must match the manufacturer's basis. The selection otherwise follows the same load-life-reliability logic, iterating on \(K\) once a specific bearing is chosen.

11–10Design Assessment for Selected Rolling-Contact Bearings

Verifying the Chosen Bearing

After selection, confirm the bearing meets the goal: back-calculate the achieved reliability at the desired life from the chosen \(C_{10}\), or the achieved life at the desired reliability. Rearranging the Weibull relation gives the realized reliability \(R\) for the actual load and life. A design is acceptable when the achieved reliability meets or exceeds the goal with the chosen catalog bearing — accounting for the fact that catalog \(C_{10}\) values come in discrete steps, so the selected bearing usually exceeds the requirement.

Achieved Reliability (design assessment)

\[ R = \exp\!\left[-\left(\frac{x_D\,(a_f F_D/C_{10})^{a} - x_0}{\theta - x_0}\right)^{b}\right] \]Eq. (11–21)

11–11Lubrication

Purpose and Choice

Lubricant in a rolling-contact bearing separates surfaces with an elastohydrodynamic film, dissipates heat, protects against corrosion, and flushes debris. Grease suits sealed, low-to-moderate speed, low-maintenance service; oil suits high speeds and high temperatures where heat removal matters. Manufacturers specify a limiting speed (often as a \(dn\) value, bore in mm × rev/min) beyond which special lubrication or cooling is required.

11–12Mounting and Enclosure

Locating, Preload, and Sealing

A typical two-bearing shaft uses one bearing to locate the shaft axially (its rings clamped both ways) and lets the other float to accommodate thermal expansion — preventing axial binding. Press/shoulder fits, retaining rings, and locknuts hold the rings; angular-contact and tapered pairs are often preloaded to remove internal clearance and stiffen the mounting. Seals and shields (contact lip seals, non-contact labyrinths) retain lubricant and exclude contaminants — essential to reaching the rated fatigue life.

Fits are functional design variables: the ring under a rotating load normally needs interference to prevent creep, while the floating position needs a controlled sliding interface. Final shoulders, fillets, tolerances, internal clearance, preload, and installation forces must follow current manufacturer drawings and applicable ISO/ABMA practice; thermal growth and contamination testing remain system responsibilities.

EXAMPLE 1 Audit the service life of a conveyor bearing

A clean, properly lubricated ball bearing has catalog dynamic rating \(C_{10}=28.0\ \text{kN}\), carries a constant radial reaction of \(6.00\ \text{kN}\), and turns at \(1200\ \text{rev/min}\). Estimate its \(L_{10}\) life in revolutions and operating hours.

Recognition strategy

The load is steady and radial, reliability is the catalog 90-percent basis, and the target is life rather than capacity; therefore invert the ball-bearing load-life power law.

Given information

\(C_{10}=28.0\ \text{kN}\), \(F=6.00\ \text{kN}\), \(n=1200\ \text{rev/min}\), exponent \(a=3\), and catalog basis \(L_R=10^6\) revolutions.

Model and interpretation

The bearing is the system. Its solved shaft reaction is treated as a constant equivalent radial load; positive load magnitude is used, and fatigue life excludes wear or lubrication failure.

Step-by-step solution

1. Select the relation. Rearranging \(C_{10}=F(L_{10}/L_R)^{1/3}\) gives \(L_{10}=L_R(C_{10}/F)^3\).

2. Form the dimensionless ratio. The capacity-to-load ratio is \(C_{10}/F=28.0/6.00=4.6667\), with kilonewtons canceling.

3. Compute revolutions. Thus \(L_{10}=10^6(4.6667)^3=1.0163\times10^8\ \text{rev}\).

4. Convert to service time. Using \(\ell=L_{10}/(60n)\) gives \(\ell=1.0163\times10^8/(60\times1200)=1412\ \text{h}\).

Verification

Units: The load ratio is dimensionless, so multiplying by the catalog basis leaves revolutions; division by revolutions per hour correctly leaves hours.

Signs and direction: Bearing fatigue uses the positive magnitude of the radial reaction; reversing its coordinate sign would not create negative life.

Plausibility: A rating-to-load ratio greater than one must produce life above one million revolutions, and the cubic value near 102 times agrees.

Final result: \(\boxed{L_{10}=1.02\times10^8\ \text{rev}\approx1.41\times10^3\ \text{h}}\).

Engineering interpretation

This is a 90-percent survival fatigue estimate, not a guaranteed replacement interval. Contamination, fits, speed limit, lubrication condition, and the catalog's current life-adjustment guidance still require verification.

EXAMPLE 2 Reduce a two-level mixer duty cycle

A mixer ball bearing runs at \(900\ \text{rev/min}\) under \(3.00\ \text{kN}\) for 70 percent of its revolutions and \(7.00\ \text{kN}\) for the remaining 30 percent. Find the equivalent load and \(C_{10}\) required for \(18000\) hours at \(L_{10}\).

Recognition strategy

Changing load with constant speed is a variable-amplitude fatigue problem. Revolution fractions and the ball-bearing cubic exponent reduce the cycle before sizing capacity.

Given information

\(f_1=0.70\), \(F_1=3.00\ \text{kN}\), \(f_2=0.30\), \(F_2=7.00\ \text{kN}\), \(n=900\ \text{rev/min}\), \(\ell=18000\ \text{h}\), \(a_f=1.00\).

Model and interpretation

Each level is a positive equivalent radial load. The repeated cycle is stationary, speed is constant, and the basic catalog relation is used at 90-percent reliability with a \(10^6\)-rev basis.

Step-by-step solution

1. Weight fatigue damage. For a ball bearing, \(F_{eq}=[0.70(3.00)^3+0.30(7.00)^3]^{1/3}\ \text{kN}\).

2. Evaluate equivalent load. The cubic sum is \(18.9+102.9=121.8\ \text{kN}^3\), giving \(F_{eq}=4.958\ \text{kN}\).

3. Convert desired life. The service target is \(L_D=60(18000)(900)=9.72\times10^8\ \text{rev}=972L_R\).

4. Size dynamic capacity. Therefore \(C_{10}=4.958(972)^{1/3}=49.1\ \text{kN}\); selection rounds upward, never downward.

Verification

Units: Cubing loads, summing like units, and taking the cube root returns kilonewtons; the life ratio is dimensionless, so capacity is also kilonewtons.

Signs and direction: Both duty levels are radial-load magnitudes. Reaction directions belong in the shaft equilibrium model, not as negative fatigue damage.

Plausibility: The equivalent \(4.958\ \text{kN}\) lies between 3 and 7 kN but above the arithmetic average because the high-load interval dominates damage.

Final result: \(\boxed{F_{eq}=4.96\ \text{kN},\quad C_{10,\mathrm{req}}=49.1\ \text{kN}}\).

Engineering interpretation

The 30-percent overload contributes most of the damage. A catalog choice must exceed \(49.1\ \text{kN}\), while shock, nonconstant speed, static indentation, and measured duty-cycle uncertainty may justify a documented application factor.

EXAMPLE 3 Check a catalog candidate under combined load

A gearbox shaft analysis gives \(F_r=5.40\ \text{kN}\) and \(F_a=2.10\ \text{kN}\) at its locating ball bearing, with the inner ring rotating at \(720\ \text{rev/min}\). For \(30000\) hours at \(L_{10}\) and \(a_f=1.30\), assess a candidate rated \(C_{10}=92.0\ \text{kN}\), using supplied current-catalog factors \(e=0.22\), \(X=0.56\), and \(Y=1.55\).

Recognition strategy

Simultaneous radial and axial reactions require a catalog equivalent-load branch before the life calculation. The supplied factors are candidate-specific data, not universal constants.

Given information

\(F_r=5.40\ \text{kN}\), \(F_a=2.10\ \text{kN}\), \(V=1.00\), \(e=0.22\), \(X=0.56\), \(Y=1.55\), \(a_f=1.30\), and \(L_R=10^6\ \text{rev}\).

Model and interpretation

Positive axial direction follows the applied thrust. Shaft equilibrium has already supplied bearing reactions; the candidate is assumed correctly aligned and lubricated, and only basic fatigue is assessed.

Step-by-step solution

1. Select the catalog branch. \(F_a/(VF_r)=2.10/(1.00\times5.40)=0.389>e\), so both \(X\) and \(Y\) apply.

2. Compute equivalent load. \(F_e=XVF_r+YF_a=0.56(5.40)+1.55(2.10)=6.279\ \text{kN}\).

3. Establish required life. \(L_D=60(30000)(720)=1.296\times10^9\ \text{rev}=1296L_R\).

4. Compare capacity. \(C_{10,\mathrm{req}}=1.30(6.279)(1296)^{1/3}=89.0\ \text{kN}\), so \(92.0\ \text{kN}\) passes.

5. Back-check achieved life. \(L_{10}=10^6[92.0/(1.30\times6.279)]^3=1.431\times10^9\ \text{rev}=33130\ \text{h}\).

Verification

Units: Catalog coefficients and life ratio are dimensionless, leaving equivalent load and required rating in kilonewtons and converted life in hours.

Signs and direction: The positive axial reaction selects this locating bearing's thrust case; reversing thrust could transfer load to the opposed bearing and requires a new FBD.

Plausibility: The candidate exceeds the required rating by only 3.4 percent, so the back-calculated 33130 hours being modestly above 30000 hours is consistent.

Final result: \(\boxed{C_{10,\mathrm{req}}=89.0\ \text{kN};\ C_{10}=92.0\ \text{kN candidate passes},\ L_{10}\approx33100\ \text{h}}\).

Engineering interpretation

The fatigue margin is narrow and does not certify the assembly. The released selection must revalidate \(e,X,Y\), static rating, limiting speed, fit, clearance, preload, lubrication, contamination class, and reliability modifiers from the exact vendor record.

Chapter recap

Start with equilibrium and distinguish applied reactions from dynamic and static catalog ratings.
Match ball or roller exponents, reliability basis, combined-load branch, and revolution-weighted duty cycle.
Round a calculated requirement up to a real candidate, then iterate candidate-dependent factors.
Before moving on, be able to compute life or capacity and audit units, direction, plausibility, mounting, lubrication, and provenance.