Rolling Bearing Analysis, 4th Edition

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ROLLING BEARING ANAZYSIS

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This book is printed on acid-free paper. @ Copyright 0 2001 by John Wiley & Sons, Inc. All rights reserved. Published simultaneously in Canada,

No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 750-4744. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 605 Third Avenue, New York, NY 10158-0012,(212) 850-6011, fax(212) 850-6008, E-Mail: PERMREQ~W1LEY.COM. This publication is designed to provide accurate and authoritative information in regard to the subject matter covered. It is sold with the understanding that the publisher is not engaged in rendering professional services. If professional advice or other expert assistance is required, the services of a competent professional person should be sought.

Harris, Tedric A. Rolling bearing analysis / Tedric A. Harris. - 4th ed. p. cm. Includes index. ISBN 0-471-35457-0 (cloth : alk. paper) 1. Roller bearings. 2. Ball-bearings. TJ1071.H35 2001 621.8'22-dc21 Printed in the United States of h e r i c a . 1 0 9 8 7 6 5 4 3 2 1

00-038171

Linear Motion Bearings earings for Special Applications Closure

1 11 23 40 41 44

Tapered Roller Bearings Closure

47 48 49 66 73 77 79 V

81

123

1 Closure

130

134 135

s t ~ ~ u t Load e d Systems

3 0 153 153

157 157

Closure

181

General Load-~e~ection Relationships earings under Radial Load earings under Thrust Load gs under Combined Radial and Thrust Load arings under Combined Radial, Thrust, and Moment Load isalignment of Radial Roller Bearings Thrust Loading of Radial Cylindrical Roller Bearings Radial, Thrust, and Moment Loading of Radial oller Bearings Flexibly Supported Rolling Bearings Closure

233 34 35 45 256

List of Symbols General

307 308 309 313 317 330 335

olling and Sliding Orbital, Pivotal, and Spinning Motions in Ball Bearings oller End-Flange Sliding in Roller Bearings Closure

266 272 80 2 ~ 9 291 302

355 Closure

Closure

e

400 410 412 e

418 419 424 440 441 444 446

t a ~ a t i o nof Lu~ricant urface T o p o ~ a p ~ y ~ f f e c t s rease L u ~ r i c a t i o ~

456

Closure e

461 463 464 472 riction in the EHL 476 Contact Closure

479 483 485

o

~ of Friction ~ c

~

~

486 olling Element-

Cont~cts and Cage Forces ons and Forces

496 515 529 E”

c

losure

Closure e

Closure

Types of Lubricants Liquid Lubricants Grease Lubricants Polymeric Lubricants Solid Lubricants En~ironmentallyAcceptable Lubricants Seals Closure

643.

645 645 646 648 654 662 668 670 671 672 682

e

4

4

e

820 820

tatie Equivalent Load

i

earing Com~onents Permissible Static Load Closure

831 831 832

e

crostructures of Rolling Bearing Steels icrostructural Alterations Due to Rolling Contact esidual Stresses in Rolling Bearing Components of Bulk Stresses on Material Response to

835 835 836 37 843 853 857

Closure e

List of Symbols General Effect of Bearing Internal Load Distributi~non Fatigue Life Effect of Variable Loadingon Fatigue Life Fatigue Life of Oscillating Bearings and Fatigue Life brication on Fatigue Life aterial Processing on Fatigue Life Combining Fatigue Life Factors Li~itationsof the Lundber~-PalmgrenTheory The Stress-Life Factor Closure

861 863 864 874 879 886 890 894 89 903 904 906 909 931

e

List of Symbols

of a Lubricated ont tact

teracting Tribological Processes and Failure Modes ecommendations for ear Protection Closure

936 937 939

955 958

i e

Closure

963 963 964 968 980 997 1003 100~ 1010

1013 1014 arnped Forced ~ibrations

C~aracteristicsof Bearing ~tiffness ynamics Analysis Closure

~limi~ary ~nvesti~ation isassern~lyof Bearings xamination and Evaluationof Specific Conditions Fracto~aphy Closure

1024 1028 1033 1039 104

1043 1043 1044 1044 1049 1063 1068

and roller bearings, generically called r o Z Z ~~ ~e ~~ r i are ~ ~ coms , used machine elements. They are emplo d to permit rotary motion of, or about, shafts in simplecommercialvices such as bicycles, roller skates, and electric motors. They are also used in neering mechanisms such as aircraft gas turbines, rollin drills, ~ o s c o p e s and , power transmissions. Until appro the design and a~plicationof these bearings could be considered more art than science. Little was understood about the physical phenomena that occur during their operation. Since 1945, a date which marks the r I1 and the b e ~ n n i nof~the atomic age, scientific occurred at an e~ponentialpace. Since 1958, the date w marks the commencement of manned space travel, continually increasing demands are being madeof engineering equipment. To ascertain the effectiveness of rolling bearings in modern engineering applications, it is necessary to obtain a firm understanding of how these bearings ~erform r varied and often extremely demanding conditions of operation. ost information and data pe~tainingto the performance of rolling bearings are presented in manufacturers’ catalogs. These data are almost entirely empirical in nature, being either obtained from the testing of products by the larger bearing manufacturing companies or,more iii

likely for smaller rnanufacturing companies, based on information contained in the American National Standards Institute (ANSI) or Interrganization for Standards (ISO) publications or similar . These data pertain only to applications involvingslow speed, simple loading, and nominal operating temperatures. If an engineer wishes to evaluate the performance of bearing applications operating beyond these bounds, it is necessary to return to the basics of rolling and sliding motions over the concentrated contacts that occur in st books written on this subject was ing by Arvid Palmgren, Technical Di for many years. It explained, more completelythan had been done previously, the concept of rolling bearing fatigue life. Palmgren, together with Gustav Lundberg, Professor of Mechanical Engineering at Chalg, was the originator of rners Institute of techno lo^ in ~ o t e ~ o rSweden, the theory and formulas on which the current ANSI and IS0 standards for the calculation of rolling bearing fatigue life are based. Also, A. s’s book in two volumes,A n ~ l y s i of s Stresses and ~ e ~ ~ ~ t i o n s , xplanation of the staticloading of ball bearings. Jones, who worked in various technical capacities for New Departure sion of Motors ~orporation,Marlin-Rockwell Cor~oration, Fafnir ring company, and also as a consulting engineer,pioneered the use of digital computers to analyze the performance of ball and roller bearing shaft-bearing-housing systems. The remainder of other early and subsequent texts on rolling bearings were, and are, largely empirical in their approaches to applications analysis. Particularly since 1 9 ~ 0much , research has been conductedinto rolling bearings and rolling contact phenomena. The use of modern laboratory equipment and transmission electron microscopes, x-ray diffrachigh speed digital co~putershas shed much light on the mechanical,hydrodynamic, metal cal, and chemicalphenomena involved in rolling bearing operation. significanttechnical papers been published by various engineering societies-for ican Societyof Mechanical Engineers, the Institution e Society of Tribologists and Lubrication E ociety of Mechanical Engineers-analy~ing the performa~ce earings in exceptional a~plicationsinvolvinghighspeed, d e~traor~inary internal design and materials. Since 1960, substantial attention has been given to the mechanisms of rolling bearing lubrication and the rheology of lubricants, Notwithstanding istenco of the aforementioned literature, there remains a nee ce that presents a unified, up-to-date approach to the analysis of bearing performance. That is my intention in presenting this To acco~plishthis goal, I have attempted to review the most significant technical papers and texts covering the performance of rolling bear-

ings, their constituent materials, and lubrication. mathematical presentations contained in the reviewed technical literature have been condensed and simplified in this book for r ease of understanding. It should not be constru~d,howev book supplies a complete bib on rolling bearings tical analysis have bee atathat I foundmostuseful everal of the references cite own works, since in some eases these are theoriginal or are among the most s i ~ i f i c a navail t

~ ~ Z y sisi saimed at deve lling bearing operation, concepts of rolling bea basic bearing types, loading, plied loading and rollers, and con deformations. The analysis of load distribution among the rollingeleme speeds, and velocities, elastohydrodynamic lubrication, friction, temperstatistics of bearing endurance, and fatigue life are consi 1 topics depend almost entirely on the prece ossible, an attempt has been made to m iscussion, numerical stance, numerical ex ylindrical roller bear 7 spherical roller be earing are accum examples are carried out in metrl

tions presented in SI or metric system units ar system units as well. The material covered herein spans many metrx elasticity, statics, dynamics,hyd at transfer. Thus, many mathemathical sy In some cases, the same symbol has been cho Lain symbols have been lways ball or roller di ecause of the several scientific disciplines that this b treatment of each topic mayvary somewhat in scope and feasible, analytical solutions to problems have been pres other hand, empirical approaches to problems have been seemed more practical. The wedding of analytical and em~iricaltechniques is particularly evident in the chapters covering lubrication, friction, and fatigue life.

Chapter

ic

d so~etimes increase lso like to express my ylvania State University for his critical review of 1 s~ggestionsfor its modi~cation. c t to my editing, was he ater rial in thefollowing chap s, s ~ ~ j eonly Contri~utor(s)

Chapter

Topic

20

24

ar

rs remain unchange~from ofessor in theDepartment o Engineering at Penn State Great Valley; sociates, Inc., a tribological research and testingcom~any company and comm hanical Engineering at th ark, Pennsylvania, where in machine design and rolling contact tribology and conduct research,

of' my l o n ~ i m eassoci

manu~acturers aswell. I would like to express my a~~reciation to the

tion, Torrin~on, ~onnectic~t. The c o n t r i ~ ~ toofreach such.illustration is identi~ed.

TEDRICA. Professor of ~ e ~ h a n i~ngineering ~al The Pennsylvania State University Uniu~rsityPark, Pennsylvania

RIS

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After the invention of the wheel, it was learned that less effort was required to move an object on rollers than to slide the object overthe same surface. Even after lubrication was discovered to reduce the work required in sliding, rolling motionstill required less work when it could be used. For example, archeological evidence showsthat the E ~ p t i a n sca. , 2400 BC, employed lubrication, most likely water, to reduce the manpower required to drag sledges carrying huge stones and statues. The Assyrians, ca. 1100 BC, however, employed rollers under the sledges to achieve a similar result with less manpower. It was therefore inevitable that bearings using rolling motion would be developed use for in complex ~ a c h i n e r yand mechanisms. Figure 1.1depicts, in a sim~listicmanner, the evolution of rolling bearings, Dowson [l.11 provides a comprehensive presentation on the history of bearings and lubrication in general; his coverage on ball and roller bearings is extensive. though the concept of rolling motion was known and used forthousands of years, and simple forms of rolling bearings were in use ca. 50 AD during the lization, the general use of rolling bearings did not occur until the in[1.2], however, shows that Leonard0 da Vinci dustrial revolution.

d by the A ~ y r i a n sto move massive sto~esin 1100

and later, with crude cart wheals, man strived to overcome friction's drag.

r 19th century bicycles marked man's first i m ~ ~ a victov. n t

olution of rolling bearings (courtesy of SKI?).

( 1 4 ~ ~ - 1 ~AD) 1 9 in his Codex ~ ~ conceived d various ~ ~ forms d of pivot bearings having rolling elements and even a ball bearing with a device to space the balls. In fact, Leonardo, who among his prolific accomplishments studied friction, stated: I affirm, that if a weight of Rat surface moveson a similar plane their move~entwill be facilitated by interposing between them balls or rollers; and I do not see any difference betweenballs and rollers save the fact that balls have universal motion while rollers can move in one direction alone. But if balls or rollers touch each other in their motion, they will make the movement more difficult than if there were no contact between them, be-

cause their touching is by contrary motions and this friction causes contrariwise movements. But if the balls or the rollers are kept at a distance from each other, they will touch at one point only between the load and its resistance . . . and consequently it will be easy to generate this movement. Thus did Leonard0 conceive the basic construction of the modern rolling bearing; his ball bearingdesign is shown by Fig. 1.2. The universal acceptance of rolling bearings by design engineers was initially impededby the inabilityof manufacturers to supply rolling bearings that could compete in endurance with hydrodynamic sliding bearings.Thissituation, however, has beenfavorably altered during the twentieth century, and particularly since 1960, by development of supe-

1.2. ( a )Thrust ball bearing design (circa 1500) in Codex Madrid by Leonardo da Vinci [1.2];( b )Da Vincibearing with Plexiglas upper plate fabricated at Institut National des Sciences AppliquBes de Lyon (INSA) as a present for Docteur en M6canique Daniel NBlias on the occasion of his passing the requirements for “DirigerdesRecherches,” December 16, 1999.

rior rolling bearing steels and constant improvement in manufacturing, providing extremely accurate geometry, long-lived rolling bearing assemblies. Initially this development was triggered by the bearing requirements for high speedaircraft gas turbines; however, competition between ball and roller bearing manufacturers for worldwide markets increased served to provide consumers substantially during the 1970s, and this has with low-cost, standard design bearings of outstanding endurance. The ~ g s all forms of bearings that utilize the rollterm roZZing ~ e ~ r i includes u ~ constrained ing action of balls or rollers to permit m i n i ~ friction, motion of onebody relative to another. ostrolling bearings are employed to permit rotation of a shaft relative to some fixedstructure. Some rolling bearings, however, permit translation, that is, relative linear motion, of a fixture in the direction provided by a stationary shaft, and a few rolling bearing designs permit a combination of relative linear and rotary motion between two bodies. This book is concerned primarily with the standardized forms of ball and roller bearings that permit rotary motion between two machine elements. These bearings will always include a complement of balls or rollers that maintain the shaft anda usually stationary supporting structure, frequently called a ~ o ~ s i n in g ?a radially or axially spaced-apart relationship. Usually, a bearing may be obtained as a unit, which includes two steel rings each of which has a hardened raceway on which hardened balls or rollers roll. The balls or rollers, also called roZZing elements, are usually held in an angularly spaced relationship by a cage? whose function wasanticipated by Leonardo. The cage may also be called a se arator or retainer. , rollers, and rings of good quality, rolling bearings are normally ctured from steels that have the capability of being hardened to a high degree, at least on the surface. In universal use by the ball bearing industry is AIS1 52100, a steel moderately rich in chromium and easily hardened thro~ghout( t ~ r o ~ g ~ - ~ a the r ~ mass e n e of ~ )most bearing components to 61-65 Rockwell C scale hardness. This steel is also used in roller bearings by some manufacturers. ~ i n i a t u r eball bearing manufacturers, whose bearings are used in sensitive instruments such as to fabricate components from stainless steels such as r bearing manufacturers ~requently from sase-hardening steels such as 310. For some specialized applicati motive wheel hub bearings, the rolling components are manufactured from induction-hardening steels. In all cases, at least the surfaces of the rolling components are extremely hard. In some high speedapplications, to m i n i ~ i z einertial loading of the balls or rollers, these components are ~abricatedfrom lightweight, high compressive s t r e n ~ h ceramic materials such as silicon nitride. Also, these ceramic rolling elements tend to

endure longer than steel at ultrahigh temperatures and in applications with dry film or minimal fluid lubrication. Cage materials, as compared to materials for balls, rollers, and rings, rally required to be relatively soft. Theymust also -to-weight ratio; therefore, materials as widely div ical properties as mild steel, brass, bronze, aluminum, pol~amide(nylon), polytetra~uoroethylene(teflon or PTFE), fiberglass, and plastics filled with carbon fibers finduse as cage material. n this modern age of deep-space exploration and cybersp bearings have come into use, such as gas film etic bearings, and externally pressurized (hy ese bearing types excels in some s~ecializedfield of ' ple, hydrostatic bearings are excellentfor no problem, an ample supply of pressuri me r i ~ d i t yunder heavy loading is requi be used forapplications in which loadsare light, are high, a gaseous atmosphere exists, and friction must be minowever, are not quite so li all bearings such as shown such as inertial guidance e roller bearings, such as shown in Fig, 1.4, are ill applications9and even larger slewing bear1.5, were used in tunneling machines for the el tunneling) project. s find use in diverse precision machiner~ophigh load, high temperature, dusty environ1.Q the dirty environments of earthmoving

E 1.3. Miniature ball bearing (courtesy of SKI?).

of

sm).

1.4. A large spherical roller bearing for a steel rolling mill application (courtesy

and farming (Figs. 1.7 and 1.8), the life-critical applications in aircraft power transmissions (Fig. 1.9), and the extreme low-high temperature and vacuum environments of deep space (Fig. 1.10). They perform well in all of these applications. Specifically, rollingbearings have the following advantages compared to other bearing types: They operate with much less friction torque than hydrodynamic bearings and therefore considerably less power loss and friction heat generation. Starting friction torque is only slightly greater than moving friction torque. Bearing deflection is less sensitive to load fluctuation than in hydrodynamic bearings.

(b)

Large slewingbearing used in an English Channel tunneling machine. ( a ) Photograph; ( b ) schematic drawingof the assembly (courtesyof SKI?).

0

They require only small quantities of lubricant for satisfactory operation and have the potential for operation with a self-contained, life-long supply of' lu~ricant. y shorter axial length than conventionalhydro bearings.

1.6. Spherical roller bearings are typically used to support the ladle in a steelmaking facility (courtesy of SKI?).

.

Many ball and roller bearings must function in the high contamination environment of earthmoving vehicle operations.

.

Agricultural applications employ many bearings with special seals to assure long bearing life.

.

CH-53E Sikorsky Super Stallion heavy-lift helicopters employ ball, cylindrical roller, and spherical roller bearings t~ansmissions in which power the main andtail rotors (cou~esyof Sikorsky Aircraft, UnitedTechnolo~esCorp.).

1.10. The ball bearings in the Lunar Excursion Module and Lunar Rover operated well in the extreme temperatures and hard vacuum on the lunar surface.

omb bin at ions of radial and thrust loads can be supported simultaneously. ~ndividualdesigns yield excellent performance overa wide load-speed range. atisfactory performance is relatively insensitive to fluctuations in load, speed, and operating temperature.

~otwithstanding the foregoing advantages, rolling bearings have been considered to have a single disadvantage compared to hydrod~amic bearings. Tallian [1.3] defined three eras of modern rolling bearing development: an “empirical” era extending through the l~ZOs,a “classical” era lasting through the 1950s, and the“modern”era occurring thereafter. Through the empirical, classical, and even into the modern era, it was said that evenifrolling bearings are properly lubricated, properly mounted, protected from dirt and moisture, and otherwise properly operated, they will e v ~ ~ t u a lfail l y because of fatigue of the surfaces in rolling contact. Historically, as shown in Fig. 1.11,rolling bearings have been considered to have a life distribution statistically similar to that of light bulbs and human beings. esearch in the1960s [1.4]demonstrated that rolling bearings exhibit a mini mu^ fatigue life; that is, “crib deaths” due to rollingcontact ot occur when the foregoing criteria for good operation are oreover, modern manufacturing techniques enable production of bearings with extremely accurate component internal and exter-

Tungsten lamps

bearings

Total number of bearings failed

t

Total number of lamps failed

Total number of deaths

1.11. Comparison of rolling bearing fatigue life distribution with those of humans and light bulbs.

nal geometries and extremely smooth rolling contact surfaces, modern steel-making processes can provide rollingbearing steels of outstanding homogeneity with few impurities, and modern sealing and lubricant filtration methods act to minimize the incursion of harmful contaminants into the rolling contact zones. These methods, which are now being used in combination in many applications, can virtually eliminate the occurrence of rolling contact fatigue, even in some applications involving very heavy applied loading.In many lightly loaded applications, for example, most electric motors, fatigue life need not be a major design consideration. There are many different kinds of rolling bearings, and before embarking on a discussion of the theory and analysis of their operation, it is necessary to become somewhat familiar with each type. In the succeeding pages a description is given for each of the most popular ball and roller bearings in current use.

i n g l e - ~ o ~ ~ e e Conrad ~ - ~ r ossembly o ~ e Ball Bearing. This ball bearing is shown in Fig. 1.12, and it is the most popular rollingbearing. The

1.12. A single-row, deep-groove, Conrad-assembly,radial ball bearing.

inner and outer raceway grooves have curvature radii between 51.5 and 53% of the ball diameter for most commercialbearings. assemble these bearings, the balls are inserted between the inner and outer sings as shown by Figs. 1.13 and 1.14. The a s ~ e ~ arzgZe b ~ y4 is given as follows:

1.13. Diagram illustrating the method of assembly of a Conrad-type,deepgroove ball bearing.

~ ~ o t o showing ~ a p Conrad-type ~ ball bearing components just prior to snapping the inner ring to the position concentric with the outer ring.

stand m i s a l i ~ i n gloads (moment loads) of small the bearing outside surface a portion of a spher 1.15, however, the bearing can be made exter incapable of supporting a moment load. all bearing can be readily adapted with seals as shields as shown by Fig. 1.17 or both as illustrated function to keep l~bricant s and shields come in ma selective ap~lications; only as examples. form well a t high speeds provi available. Speed limits shown in mantain to bearing operation without the benefit of external cooling capability or special cooling t e c h n i ~ ~ e s . s can be obtained in different dimens Conrad assem ISO* standards. Figure 1.19 shows t ries according to ative dimensions of various ball bearing series.

.I&. A single-row deep-groove ball bearing assembly having a sphered outer surface to make it externally aligning.

*American National Standards Institute and International Organization for Standards.

1.16. A single-row deep-groove ball bearing having two seals to retain lubricant (grease) and prevent ingress of dirt into the bearing.

~ingle- ow Deep-Groove ~ i l l i n ~ - ~~l so st e ~ b lBall y Bearings. This bearing as illustrated in Fig. 1.20 has a slot machined in the side wall of each of the inner and outer ring grooves to permit the assembly of more balls than the Conrad type does, and thus it has more radial loadcarrying capacity. Because the slot disrupts the groove continuity, the bearing is not recommended for thrust load applications. Otherwise, the bearing has characteristics similar to those of the Conrad type.

Dou ble-Row Deep-Grooue Ball earings. This ball bearing as shown in Fig. 1.21 has greater radial load-carrying capacity than the single-row types. Proper load sharing between the rows is a function of the geometrical accuracy of the grooves. Otherwise, these bearings behave similarly to single-row ball bearings. ~ n s t r u ~ e nall t earings. In metric design, the standardized form of these bearings ranges in size from 1.5-mm (0.0~~06-in.) bore and 4-mm 1574~-in.) 0.d. to (0.35433-in.)bore and 26-mm (1.02~~2-in.) o.d. e reference [1.51. ailed in reference [1.6],standa~dizedform, inch design instrument ball bearings range from 0.635-mm (0.0250-i~.) bore

.

A single-row deep-grooveball bearing assembly having shields and seals, The shields are used to exclude large particles of foreign matter.

~iameter series

3 1

2

"

.

I

Size comparison of popular deep-groove ball bearing dimension series.

.

Cutaway view of a singlerow, deep-groove, filling slot-type ball bearing assembly.

. A double-row, deep-groove, radial ball bearing.

.

A delicate final assembly operation on an instrumentball bearing assembly is performed under ma~ificationin a “white room.”

earings, ~ ~ l a r - c o n t aball c t bear3 are designed to support combin ings as shown in Fig. ust loads depending on the conta thrust loads or heavy nitude. The bearing aving large contact angles can support heavier shows bearings having small and large contact thrust loads, Figure erally have groove curvature radii in therange angles, The bearing % of the ball diameter. The contact angle does not usually exceed bearings are usually mounted in pairs with the free end as shown in Fig. 11.25. These sets may be preloadedagainst other to stiffen the assembly in the axial direction. The bearin also be ~ o ~ n t in e dtandem as illustrated in Fig. 1.26 to achieve thrust-carrying capacity. ow ~ n g u l a r - ~ o n ~ a c ~earings. These bearings as depicted 27 can carry thrust either direction or a combination of radial and thrust load. earings of the rigid type are able to withstand

e

An angular-contact ball bearing.

\

(a) Small angle

. An~ular~contact ball bearings.

ALL

(a) Back-to-back mounted

(b) Face-to-face mount@d

Duplex pairs of an~lar-contactball bearings.

I-

”

(a) Nonrigid type

(b) Rlgid type

1.27. Double-row angular-contact ball bearings.

moment loading effectively. Essentially, the bearings perform similarly to duplex pairs of single-row angular-contact ball bearings.

~ e l ~ ~ l i g n i ~ g ~ Ball o u ~Bearings. l e - ~ o wAs illustrated in Fig. 1.28, the outer raceway of this bearing is a portion of a sphere. Thus, the bearings are internally self-aligning and cannot support a moment load. Because the balls do not conform wellto the outer raceway (it is not grooved)jthe outer raceway has reduced load-carrying capacity. This is compensated somewhat by use of a very large ball complement that minimizes the load carried by each ball. The bearings are particularly useful in applications in which it is difficult to obtain exact parallelism between the shaft and housing bores. Figure 1.29 shows this bearing with a tapered sleeve and locknut adapter. With this arrangement the bearing does not require a locating shoulder on the shaft. ing Ball Bearings. These bearings are illustrated in Fig. 1.30. As can be seen, the inner ring consists of two axial halves such that a heavy thrust load can be supported in either direction. They may also support, simultaneously, moderate radial loading. Thebearings have found extensiveuse in supporting the thrustloads acting on high speed, gas turbine engine mainshafts. Figure 1.31 shows the compressor and turbine shaft ball bearing locations in a high-performance aircraft gas turbine engine. Obviously, both the inner andouter rings must be locked

1.28. A double-row internally selfaligning ball bearingassembly.

up on both axial sides to support a reversing thrust load. It is possible with accurate ~~s~ grinding at the factory to utilize these bearings in tandem as shown in Fig. 1.32 to share a thrust load in a given direction.

The thrust ball bearing illustrated in Fig. 1.33 has a 90" contact angle; however, ball bearings whose contactangles exceed 45"are also classified as thrust bearings. As for radial ball bearings, thrust ball bearings are suitable for operation at high speeds. To achieve a degree of externally aligning ability, thrust ball bearings are sometimes mountedon spherical seats. This arrangement is demonstrated by Fig. 1.34. A thrust ball bearing whose contact angle is 90" cannot support any radial load.

Roller bearings are usually used for applications requiring exceptionally large load-supportingcapability, whichcannot be feasibly obtained using

.

A double-row internally self-aligning ball bearing assembly with a tapered sleeve and locknut adapter for simplified mounting on a shaft of uniform diameter.

1.30. A split inner ring ball bearing assembly,

1. Cutaway view of turbofan gas turbine engine showingmainshaft bearing locations (courtesyof Pratt and Whitney, United Technologies Corp.).

_..

1.32. A tandem-mounted pair of split inner ring ballbearings.

a ball bearing assembly. oller bearings are usually much stiffer structures (less deflection per unit loading) and provide greater fatigue endurance than do ball bearings of a comparable size.In general, they also cost more to manufacture, and hence purchase, than comparable ball bearing assemblies. Theyusually require greater care in mounting than do ball bearing assemblies. Accuracyof alignment of shafts andhousings can be a problem in all but self-aligning roller bearings.

earings. Cylindricalroller bearings asillustrated in Fig. 1.35 have exceptionally low friction torque characteristics that

.

A 90" contactangle ball bearing assembly.

thrust

make them suitable for high speedoperation. They also have high radialload-carrying capacity, Theusual cylindrical rollerbearing is free to float t has two rollering flanges onone ring and none on the shown. in Fig. 1.3 equipping the bearing with a guide flange on the opposing ring (illustrated by Fig. 1.37), the bearing can be made to support some thrust load. revent high stresses at the edges of the rollers the rollers are usually crowned as shown in Fig. 1.38.This crowning of rollers also g the ~ e a r i n gprotection against the effects of slight is alignment. crown is ideally designed for only one condition of loading. Crownedraceways may be used in lieu of crowned rollers. To achieve greater radial-load-carrying capacity,cylindricalroller bearings are frequently constructed of two or more rows of rollers rather than of longer rollers. This is done to reduce the tendency of the rollers to skew. Figure 1.39 shows a small double-row cylindrical rollerbearing designed for use in precision applications. Figure 1.40 illustrates a large multirow cylindrical rollerbearing for a steel rolling mill a~plication.

earings. A needle roller bearing is a Cylindrical roller bearing having rollers of considerably greater length than diameter. This

s

.

A 90" contact angle thrust ball bearinghaving a spherical seat to make it externally aligning.

bearing is illustrated in Fig. 1.41. ecause of the geometry of the rollers, they cannot be manu~actured asaccurately as other cylindrical rollers, nor canthey be guided as well. ~onsequently,needle rollerbearin eater friction than.other cylindrical roller bearings. Needle roller bearings are designed to fit in applications in which radial space is at a premium. Sometimes to conserve space the needles bear directly on a hardened shaft. They are useful for applications in which oscillatory motion occurs or in which continuous rotation occurs but loading is light and intermittent. The bearings may be assembled without a cage, as shown in Fig. 1.42. In this full-co~plement-type bearing, the rollers are fre~uently retained by turned-under flanges that are integral with the outer shell. The raceways are frequently hardened but not ground.

s The single-row tapered roller bearing shown in Fig. 1.43 has the ability inations of large radial and thrust loads or to carry thrust use of the difference betweenthe innerand outer raceway contact angles, there is a force componentthat drives the tapered rollers

:ylindrical roller bearing.

, Cylindrical roller bearings withoutthrust flanges.

against the guide flange. ecause of the relatively large sliding friction generated at this flange, the bearing is not suitable for high speed operation without special attention to cooling 'and/or lubrication. Tapered roller bearing terminolo~differs somewhat from that pertaining to other roller bearings, the inner ringbeing called the cone and

Cylindrical roller bearings havingthrust flanges.

"

(b)

.

(a)Spherical roller (fully crowned); ( b )partially crowned cylindrical roller (crown radius is greatly exaggeratedfor clarity).

the outer ring the cup. epe~dingon the magnitude of the thrust load to be supported, the bearing may have a small or steep contact angle, as shown in Fig. 31.44. Since tapered roller bearing rings are s e p a ~ ~ b lthe e, bearings are mounted in pairs as indicated in Fig. 1.45, and one bearing is adjusted against the other. To achieve greater radial load-carryingcapacity and eliminate problems of axial adjustment due to distance between bearings, tapered roller bearings may be combined as shown in Fig. 1.46 into two-row bearings. Fig. 1.47 shows a typicaldouble-row tapered roller bearing assembly for a railroad car wheel application. Double-row bearings may also be combined into four-row or quad bearings for exceptionally heavy radial load applications such as rolling mills. Figure 1.48 shows a quad bearing having integral seals.

. A double-row,cylindrical roller bearing for precision machine tool spindle applications.

As with cylindrical roller bearings, tapered rollers or raceways are usually crowned to relieve heavy stresses on the axial extremities of the rolling contact members. y e~uipping thebearing with specially contoured flanges, a special cage, and lubrication holes as shown by Fig. 1.49, a tapered roller bearing can be designed to operate satisfactorily under high load-high speed conditions. In this case, the cage is guided by lands on both the cone rib and the cup, and oil is delivered directly by centrifugal flow to the roller end-flange contacts and cage rail-cone land contact.

ost spherical roller bearings have an outer raceway that is a portion of a sphere; hence, the bearings, as illustratedby Fig. 1.50, are internally self-ali~ing.Each roller has a curved generatrix in the direction transverse to rotation that conforms relatively closely to the inner and outer

1.40. A multirow cylindrical roller bearing for a steel rolling mill application.

1.41. Needle roller bearing, nonseparable outer ring, cage, and roller assembly (courtesy of Torrington Company, Division of Ingersoll Rand Corp.).

.

(a)

(b)

Full-complement needle roller bearings. (a)Drawn cup assembly with trunnion-end rollers and innerring; ( 6 )drawn cup assembly with rollers retained by grease pack (courtesy of Torrington Company, Division of Ingersoll Rand Corp.).

1.43. Single-row tapered roller bearing showing separable cup and nonseparable cone, cage, and roller assembly (courtesy of the Timken Company). Ir

raceways. This gives the bearing high load-carrying capacity. Various executions of double-row, spherical roller bearings are shown in Fig. 1.51. Fig. 1.51a shows a bearing with asymmetrical rollers. This bearing, similar to tapered roller bearings, has force compone~ts thatdrive the rollers against the fixed central guide flange. earings such as illustrated in Fig. 1.51b and 1.5IChave symmetrical (barrel- or hourglass-shape) rollers, and these force components tend to be absent except under high

Small angle

,Steep angle

1.44. Small and steep contact angle tapered roller bearings.

E 1.45. Typical mounting of tapered roller bearings.

0

speed operation. ouble-row bearings having barrel-shape, symmetrical rollers frequently use an axially floating central flange as illustrated by Fig. 1.51d. This eliminates undercuts in the innerraceways and permits use of longer rollers, thus increasing the load-carrying capacity of the bearing. Roller guiding in such bearings tends to be acco~plishedby

r

e

Double-row tapered rolling bearings. ( a )Double coneassembly; (6) double

cup assembly.

F

1.47. Sealed, greased, and preadjusted double-row, tapered roller bearing for railroad wheel bearings (courtesy of the Timken Company).

the raceways in conjunction with the cage. In a well-designed bearing, the roller-cage loads due to roller skewing maybe m i n i m i ~ e(see ~ ChapBecause of the closeosculationbetween rollers and raceways and curved generatrices, spherical roller bearings have inherently greater

GS

LLE

.

A four-row tapered roller bearing with integral seals for a hot strip mill application (courtesyof SKI?).

High speed tapered roller bearing with radial oil holes and manifold. The '"z"-type cage is guided on the cone rib and cup lands (courtesy of the Timken Company).

friction than cylindrical rollerbearings. This is due to the degree of sliding that occurs in the roller-raceway contacts. Spherical roller bearings are therefore not readily suited for use in high speed applications. They perform wellin heavy duty applications such as rolling mills,paper mills, and power transmissions and in marine applications. ~ouble-rowbearings can c a m combined radial andthrust load; they cannot support moment loading. Radial, single-row, spherical roller bearings have a basic contact angle of 0".Under thrust loading, this angle does not increase appreciably;consequently, any amount of thrust loadingmagnifies roller-raceway loading substantially. Therefore, these bearings should

.

Cutaway view of a double-row, spherical roller bearing with symmetrical rollers and a floating guide fiange (courtesy of SKI?).

I/ I / I

(d)

I

(4

(f)

1.51. Various executions of double-row spherical roller bearings.

not be used to carry c o ~ b i n eradial and thrust loading whenthe thrust ent of the load is relatively large compared to the radial compospecial type of' single-row bearing has a toroidal outer raceway, this is ill~stratedby Fig. 1.5 ; it can a c c o ~ ~ o d aradial te load together with some moment load, however,little thrust load.

earings. The spherical roller th very high load-carrying capacity

IG Single-row, radial, toroidal roller bearing (courtesy of SKY).

1-53. Cutaway view of a spherical roller thrust bearing assembly (courtesy of

SKY).

osculation between the rollers and raceways. It can carry a combination thrust and radial load and is internally self-aligning. Becausethe rollers are asymmetrical, force components occur that drive the sphere ends of the rollers against a concave spherical guide flange. Thus, the bearings experience sliding friction at thisflange and do not lend themselves readily to high speed operation. oller Thrust Bearings. Because of its geometry, the cylindrical roller thrust bearing of Fig. 1.54 experiences a large amount of sliding between the rollers and raceways, also calledwashers. Thus, the bearings are limited to slow speed operation. Sliding is reduced somewhat by using multiple short rollers in each pocket rather than a single integral roller. This is illustrated by Fig. 1.55. oller Thrust Bearings. This bearing, illustrated in Fig. 1.56 has aninherent force componentthat drives each roller against theoutboard flange. The sliding frictional forces generated at the contacts between the rollers and flange limit the bearing to relatively slow speed applications. Needle Roller Thrust Bearings. These bearings, as illustrated byFig. 1.5'7, are similar to cylindrical roller thrust bearings except that needle

1.54. IG bearing.

.

Cylindricalroller

thrust

Cylindrical roller thrust bearing having two rollers in each cage pocket; the bearing has a spherical seat for external alignment capability.

1.56. Tapered roller thrust bearing. ( a ) Both washers tapered; ( b ) one washer tapered.

1.57. Needle roller thrust bearing (courtesy of Torrington Company, Divisionof Ingersoll Rand Gorp.).

rollers are used in lieu of normal size rollers. Consequently, roller-washer sliding is prevalent to a greater degree and loading must be light. The principal advantage of the needle rollerthrust bearing is that it requires only a narrow axial space. Figure 1.58 illustrates a needle roller-cage assembly that may be purchased in lieu of an entire bearing assembly.

1.58. Thrust needle roller-cage assembly (courtesy of Torrington Company, Division of Ingersoll Rand Gorp.).

Bearings for linear motion, such as those used in machine tool “ways”for example, V-ways-generally employ only lubricated sliding action. These sliding actions are subject to relatively high stick-slip friction, wear, and subsequent loss of locational accuracy. Ballbushing operating on hardened steel shafts, illustrated schematically by Fig. 1.59, provide many of the low friction, minimal characteristics o f radial rolling bearings. The ball bushing, which provides linear travelalong the shaft,limited y built-in motion. stoppers, contains three or more oblon~circuits of recirculating balls. As illustrated in Fig. 1.60, one portionof the oblong ball complement supports load on the rolling balls while the remain in^ balls operate with clearance in the return track. The ball retainer units can be fabricated relatively inexpensively of pressed steel or nylon (polyamide)material. Figure 1.61 is a p h o t o ~ a p h showing an actual unit with its components. Ballbushings of instrument quality are made to operate on shaft ~iameters assmall as 3.18 mm (0.125 in.).

FI

.

Schematic illustration of a ball bushing.

Load carryingballs I-

” I

~ecircui~ring balls in ~ t e a r ~ n c e

.

Schematic diagram of a ball bushing recirculating ball set.

1. Linear ball bushing showing various components.

all bushings can belubricated with medium-heavy weight oil or with a light grease to prevent wear and corrosion. For highlinear speeds, light oils are recommended. Seals can beprovided;however,friction is increased s i ~ i ~ c a n t l y . As with radial ball bearings, life can belimited by subsurface-initiated fatigue of the rolling contact surfaces. A unit is usually designed to perform satisfactorily for several million units of linear travel. Since the hardened shaft is subject to surface fatigue and/or wear, provision can be made for rotating the bushing or shaft to bring new bearing surface into play.

ings were grease-lubricated and, owing to the grease deterioration in this difficult application?needed to be regreased periodically. If this was not accomplished with care, inevitabl~contamination was introduced into the bearings an bearing longevity was substantia11 come thissitu ion,many bearings wereprovi greased and sealed-for-life, d plex sets as shown units needed to be press-fitte into the wheel hub semblyfor the automobile m cturerand to m was made integral with the bearing outer ring as thus, the unitcould be boltedto studs on the wheel, contained unit with a flange integral with each r o use for nondriven wheels; the unit can be bolted to the nd the wheel for simple assembly. uty vehicles such as trucks, tapered roller stead of ball bearings, to the bearing unit as shown by the taselfig. 1.64. This compact, preadjuste~? unit is equipped with an inte a1 s eed sensor to the anti-lock braking system arings to measure applied lo

To reduce friction associated with the follower contact on cams, rolling motion maybe employed. The needle rollerbearing is particularly suited to this application because it is radially compact. Figure 1.65 shows a needle roller bearing, cam follower assembly.

.

Modernautomobilewheel preadjusted, greased and sealed-€or-life,ball bearing units. ( a )Without flanges; unit is press-fitted into wheel hub and slip-fitted onto the axle; ( b ) with a single flange integral with the bearing outer ring; (e) with flanges integral with bearing outer and inner rings (photographs courtesy of SKI?).

.

Modern truck wheel preadjusted, greased, and sealed-for-life, Lapered roller bearing unit (courtesy of SKI?).

4. Self-contained?tapered roller bearing with an integral speed sensor to provide signal to the anti-lock braking system (courtesy of the Timken Company).

rplane and helicopterpower transmission bearing a~plications are nerally characterized by the necessity to carry heavy loads at h i g ~ speed while mini~izingbearing size. The bearings are generally man~facturedfromspecialhigh s t ~ e n ~high h , quality steels e weight and outof a steel bearing itself is signi~cant,minimizing bearing side diameter aids com~actnessin engine design, allowing surro~nding engine co~ponentsto be smaller and weigh less. us, aircraft power

.

Needlerollercamfollowerassembly(courtesy Ingersoll Rand Gorp.).

of ~ o r r i n ~ o Company, n

for the turbine e n ~ n application. e

ages have i l l ~ ~ t r a t eand d described various types and e~ecutionsof ball and roller bearings. It is not to be c o n s t ~ e d that every

.

Aircraft power transmissionbearings:(left)cylindricalrollerbearing; of ~ T N ) . (right) spherical planet gear bearing (courtesy

1.67. Aircraft gas turbine engine, cylindrical roller bearing (courtesy of SKI?).

F

1.68. Aircraft gas turbine engine mainshaft bearing components: lower leftsplit inner ring ball bearing; center and upper right-cylindrical roller bearing inner and outer ring units (courtesy of FAG OEM und Handel AG).

type of rolling bearing has been described; discussion has been limited to the most popular and basic forms. For example, there are cylindrical roller bearing designs that use snap rings, instead of machined and and IS0 standards on terminology [1.7] ore common bearing designs. It is also ings are specially designed for applicadiscussed herein only to indicate that imes warranted by the application. In ail additional cost for the bearing or bearing unit; however, such cost increase is usually offset by overall ef-

ficiency and cost reduction brought to mechanism and machinery design, manufacture, and operation.

1.1 D. Dowson, History of Tribology, 2nd ed., Longman, New York (1999).

1.3 T. Tallian, “Progressin Rolling Contact Technology,”SKF Report AL690007 (1969). 1.4 T. Tallian, “Weibull Distribution of Rolling Contact FatigueLife and Deviations Therefrom,”ASLE Trans. 5(1),183-196 (1962). 1.5 American National Standards Institute, American National Standard ( ~ S I I ~ ~ A ) Std, 12.2-1992, “Instrument Ball Bearings-Metric Design” (April 6, 1992). 1.6 American National Standards Institute, American National Standard ( ~ S I / ~ ~ A ) Std. 12.2-1992,“Instrument Ball Bearings-Inch Design” (April 6, 1992).

Symbol

A l3 d

D

Description

Distance between raceway groove curvature centers AID Raceway diameter Bearing pitch diameter Ball or roller nominal diameter Mean diameter of tapered roller Diameter of tapered roller at largeend Diameter of tapered roller at smallend rID effective Roller length Distance between cylindrical roller guide flanges Roller length end-to-end diametral Bearing clearance Bearing free endplay

Units mm (in.) mm (in.) mm (in.) rnm (in.) mm (in.) mm (in.) mm (in.) mrn (in.) mm (in.) mm (in.) mm (in.) mm (in.)

48

ROLLING EEARING ~ C R O G ~ O ~ T ~ S

Symbol

a

C 0

i

r

Units Raceway groove curvature radius Roller corner radius Roller contour radius Spherical roller bearing diametral play Number of rolling elements Free contact angle Contact angle Tapered roller bearing flange angle Roller angle Shim angle I) cos ald,, ~ i s a l i ~ m eangle nt Curvature Curvature difference Curvature sum Osculation Angular velocity

mm (in.) mm (in.) mm (in.) mm (in.) 0 0

radlsec

S U ~ S ~ R I ~ T ~ Refers to cage Refers to outer raceway Refers to inner raceway Refers to roller

Although balland roller bearings appear to be simple mechanisms,their internal geometries are quite complex. For example,a radial ball bearing subjected to thrust loading assumes angles of contact between the balls and raceways in accordance with the relative conformities of the balls to the raceways and the diametral clearance. On the other hand, theability of the same bearing to support the thrustloading dependson the contact angles f o r ~ e dThe . same diametral clearance or play produces an axial endplay that may or may not be tolerable to the bearing user. In later chapters it will bedemonstrated how diametral clearance affects not only contact angles and endplay but also stresses, deflections, load distributions, and fatigue life, Stresses, deflections, load distribution, and life in roller ~earingsare also affected by clearance. In the determination of stresses and deflections the relative conformities of balls and rollers to their contacting raceways are of vital interest. n this chapter the principal macrogeometric relationships govern in^ the operation of ball and roller bearings shall be developed and examined.

The ball bearing can be illustrated inits most simple formas inFig. 2.1. From Fig. 2.1 one can easily see that the bearing pitch diameter is approximately equal to the mean of the bore and 0.d. or

dm = +(bore + 0.d.)

(2.1)

More precisely, however, the bearing pitch diameter is the mean of the inner and outer ring raceway contact diameters. Therefore,

Generally, ball bearings and other radial rolling bearings such as cylindrical roller bearings are designed with clearance. From Fig. 2.1, the diametral* clearance is as follows:

2.1. Radial ball bearing showing diametral clearance.

* Clearance is always measuredon a diameter; however, becausemeasurement takes place in a radial plane, it is commonly called radial clearance. Thistext will use diametral and radial clearance interchangeably.

5

R O L L ~ GB

E

~ ~ ~C RGO G E O

Table 2.1 taken from Reference 2.1 givesvalues of radial internal clearance for radial contact ball bearings under no load.

.

A 209 single-row radial deep-groove ball bearing has the following dimensions: Inner raceway diameter, di Outer raceway diameter, do Ball diameter, D Number of balls, is Inner groove radius, ri Outer groove radius, ro

52.291 mm (2.0587 in.) 77.706 mm (3.0593 in.) 12.7 mm (0.5 in.) 9 6.6 mm (0.26 in.) 6.6 mm (0.26 in.)

Determine the bearing pitch diameter d m and diametral clearance Ipd.

P,

=

Q(52.3 i77.7)

=

65 mm (2.559 in.)

=

do - di - 2D

=

77.706 - 52.291 - 2

=

0.015 mm (0.0006 in.)

X

12.7

se The ability of a ball bearing to carry load depends in large measure on the osculation of the rolling elements and raceways. Osculation is the ratio of the radius of curvature of the rolling element to that of the raceway in a direction transverse to the direction of rolling. From Fig. 2.1it can be seen that for a ball mating with a raceway, osculationis given by

+ = -D

2r

Letting r

= fD,

osculation is

# = - 1. 2f

(2.5)

It is to be noted that the osculation is not necessarily identical for inner and outer contacts.

s

c - m m m w m o u3 m w w mwc-m

v?v?v?u?v?v?

0000000000ddddd

Determine the osculation in the 209 radial ball bearing of Example 2.1. f,,ri=-

L)

6.6 = 0.52 12.7 0.962

f O

=5=" 6.6 L)

12.7

-

0.52 0.962

Because a radial ball bearing is generally designed to have a diametral clearance in the no-load state, the bearing also can experience an axial play. Removal of this axial freedom causes the ball-raceway contact to assume an oblique angle with the radial plane; hence, a contact angle different from zero degrees will occur. Angular-contactball bearings are specifically designedto operate under thrust load, and the clearance built into the unloaded bearing along with the raceway groove curvatures determines the bearing free contact angle. Figure 2.2 shows the geometry

~-

.

Axis of rotation

Radial ball bearing showing ball-raceway contact dueto axial shift of inner and outer rings.

4

ROLLING BEARING ~ ~ R O G E O ~ T R ~

of a radial ball bearing with axial play removed. From Fig.2.2 it can be seen that the distance between the centers of curvature 0' and 0" of the inner and outer ring grooves is

A=r,+ri-D ~ubstitutingr = f D yields

in which B = f, + fi - I is defined as the totalcurvature of the bearing. Also from Fig. 2.2, it can be seen that the free contact angle is the angle made by the line passing through the points of contact of the ball and both raceways and a plane perpendicular to the bearing axis of rotation. The magnitude of the free contact angle can be described as follows:

or ao = cos-1 ( 1 -

$1

3. A 218 an~lar-contactball bearing has dimensions as follows: Inner raceway diameter, di Outer raceway diameter, do Ball diameter, D Inner groove radius, ri Outer groove radius, ro

102.79 mm (4.047 in.) 147.73 mm (5.816 in.) 22.23 mm (0.875 in.> 11.63 mm (0.4578 in.) 11.63 mm (0.4578 in.)

Determine the free contact angle of this bearing.

"= 11*63

22.23

0.5232

"-11*63 -

22.23

=

A

0.5232

0.5232

+ 0.5232 - 1 = 0.0464

= BD =

0.0464

(2.7) X

22.23

=

1.031 mm (0.0406 in.)

do - di - 2D

Pa =

147.73 - 102.79 - 2

a" = cos-l(l =

(2.3)

(

X

22.23

=

0.48 mm (0.019 in.)

2)

cos-1 1 -

=

2

X

40"

1.031

If in mounting the bearing an interference fit is used, then the diametral clearance must be diminished by the change in ring diameter to obtain the free contact angle. Hence

(

a" = cos-1 1 -

Pa + APd 2A

)

(2.10)

Because of diametral clearance a radial bearing is free to float axially under the condition of no load. This free endplay may be defined as the maximum relative axial movement of the inner ring with respect to the outer ring under zero load. From Fig. 2.2,

gP,

P,

=A =

sin a"

(2.11)

2A sin a'

(2.12)

Figure 2.3 shows the free contact angle and endplay versus P,/D for single-row ball bearings. Rouble-row angular-contact ball bearings are generally assembled with a certain amount of diametral play (smaller than diametral clearance). It can be determined that the free endplay for a double-row bearing is

0 -

0.12

U

0.11

0.10

0.09 0.08

0.07

3

0.06

0.05

0.04 0.03 0.02

0.01

Split inner ringball bearings as illustrated in Fig. 2.4 have inner rings that are ground with a shim between the ring halves. The width of this shim is associated with a shim angle that is obtained by removing the shim and abutting thering halves. From Fig. 2.5 it can be determined that the shim width is given by

-

F

~

bearing ball

ws

~Inner rings ~ of Esplit inner ring grinding. showing for shim

2.4.

2.6. Split inner ring ball bearingassembly showing shim angle.

w , = (2ri -

Since fi

=

D)sin a,

(2.14)

ri/D, equation (2.14) becomes w , = (2fi - l)D sin a,

(2.15)

The shim angle a,, and the assembled diametral play S, of the bearing accordingly dictate the free contact angle. The effective clearance P, of the bearing may be determined from Fig. 2.5 to be

P,

=

s, + (2fi - 1)(1- cos a,)D

(2.16)

Thus, the bearing contact angle which is shown in Fig. 2.2 is given by

ree ~urthermore? diametral clearance can allow a ball bearing to misalign slightly under no load. The free angle of misalignment is defined as the maximum angle through which the axis of the inner ringcan be rotated with respect to the axis of the outer ring before stressing bearing components. From Fig. 2.6, using the law of cosines it can be determined that (2.18)

(2.19) Therefore 8, the free angle of misalignment?is

e = 0i + eo

(2.20)

Since the following trigonometric identity is true, cos and since

ei + cos e* = 2 cos $(ei + eo>cos +(Oi

-

eo)

(2.21)

ei - eo approaches zero, therefore, e = 2 COS-^

(,,s

4

;

cos 00

(2.22)

or 0 = 2 cos-1

Determine the free contact angle a', free endplay P,, and free angle of misalignment of the 209 radial ball bearing in Example 2.1.

BALL B E ~ I N G S

2.6. (a) Free misalignment of inner ring of single-row ball bearing, (b) free misalignment of outer ring of single-row ball bearing.

ROLL~G ~~~G

fi = f,

~

~

R

~

Ex. 2.2

0.52

=

O

dm = 65 mm (2.559 in.)

Ex. 2.1

Pd = 0.015 mm (0.0006 in.)

Ex. 2.1

B=fi+f*-l

A

=

0.52

=

BD

=

0.04

+ 0.52 - 1 = 0.04 X

12.7 = 0.508

A ! ' = cos-1 (1 =

(

cos-1 1 -

2) 2

X

0.508

Pe = 2A sin ' a =

'Os-'

2

X

['

0.508

(2.12) X

0.015

-

sin (9'52')

=

0.174 mm (0.0069 in.)

(2 X 0.52 - 1) X 12.7 - (0.015/4) 0.52 - 1)X 12.7 - (0.01512)

E (65 + (2 x

- 1) X 12.7 - (0.015/4) + 65 (2- X(2 0.52 X 0.52 - 1) X 12.7 + (0.015/2)

Note "howsmall the free angle of misalignment is.

T w o bodies of revolution having different radii of curvature in a pair of

principal planes passing through the contact between the bodies may contact each other at a single point under the condition ofno applied load. Such a condition is called point contact. Figure 2.7 demonstrates this condition. In Fig. 2.7 the upper body is denoted by I and the ,lower body by 11. The principal planes are denoted by 1 and 2. Therefore, the radius of

~

O

~

2.7. Geometry of contacting bodies.

curvature of body I in plane 2 is denoted by r12.Since 7- denotes radius of curvature, curvature is defined as P";

1

(2.24)

Although radius of curvature is always of positive sign, curvature may be positive or negative, convex surfaces being positive and concave surfaces negative. To describe the contact between mating surfaces of revolution, the following definitions are used. 1. Curvature sum: 1

1

1

~ p = - + - + - + 7-11

r12

7-111

.

1

(2.25)

7-112

. Curvature difference: (2.26)

2

~ C ROLLING R O GB E~ OI N ~ ~G R ~

In equations (2.25)and (2.26)the sign convention for convex and concave surfaces is used. Furthermore, care must be exercised to see that F(p)is positive. By way of example, F(p) is determined for a ball-inner raceway contact as follows (see Fig. 2.8):

("-

rIIl = i d i = 1 - D) 2 cos a

Let Y=-

D cos a

(2.27)

dm

Then PI1 = PI2

2

=

5

For the ball-outer raceway contact pI1 = p12 = 2 / D as above; however, rIIl = 21

Therefore,

(-

COS dma

+ D)

3

~

PI11 =

I 2.8. Ball ~ bearing ~ geometry E

-(; +J 1

PI12 =

F(P)o =

"

f0D

fo

1

l + Y

gy

4"" fo

(2.30)

l + Y

F ( p ) is always a number between 0 and 1.

le 2.5. Determine the values of curvature sum and curvature difference for the 209 radial ball bearingof Example 2.1, subjected to radial load only.

6 = fo

=

0.52

dm = 65 mm (2.559 in.)

Ex. 2.2 Ex. 2.1

4

ROLLING BEARING ~ C R O G E O ~ ~ R S

D cos a (2.27)

dm

12.7

12.7

'

0.52

cos (0") = 0.1954 65

X

+ 1 -x 0*1954) = 0.202 mm-' 0.1954

(5.126 in.-1) (2.28)

1 2 X 0.1954 0.52 ' 1 - 0.1954 1 2 X 0.1954 0.52 "I- 1 - 0.1954

-."--A

-

=

0.9399

,

" 4

(2.30)

12.7 0.52

x

1 + 0.1954 >,,,*'

=

0.1378 mm-' (3.500 in.-1)

(2.31)

1 2 X 0.1954 0.52 1 + 0.1954 . 1 2 X 0.1954 4"0.52 1 + 0.1954 "

=

0.9120

~ F(p),. This condition will be used to Note the when fi = f,, F ( P ) > demonstrate in a later example that an elliptical area of contact of

greater ellipticity generally exists at the inner raceway contact as opposed to the outer raceway contact. Determine the magnitude of curvature sum and curvature difference for the 218 angular-contact ball bearingof Example 2.3 subjected to light axial loading.

fi = f, = 0.5232 (102.79 = 400*

=

Ex. 2.3

+ 147.73) = 125.26 mm (4.932 in.) Ex. 2.3

I) cos a

Y=-

(2.27)

dm

- 22.23

X cos (40") 125.26

=

o.1359

pi=$(4"+*) 1

fi 22.23 0.5232 =

(2.28) 1-Y

1 - 0.1359

0.108 mm-l (2.747 in.-')

(2.29)

c _

c _

2 X 0.11359 l + 0.5232 1 - 0.1359 2 X 0.1359 l + 4" 0.5232 1 - 0.1359

=

0.9260

*It will bedemonstratedin a later chapter that contact angle increases under thrust (axial) loading. This will not be consideredin this example.

x p o D=” (-4 ” ” =

fo

(2.30)

1 2+ yY )

1

” 22.23 .(4 - 0.5232 - 1 + 0.1359 0.0832 mm-’ (2.114 in.-’) 1

2Y

”~

(2.31) ’f

fo

l + Y

2 X 0.1359 0.5232 1 + 0.1359 1 2 X 0.1359 4”0.5232 1 + 0.1359 1

”

=

0.9038

Comparison of the F ( P ) and ~ F(p), values in Examples 2.6 and 2.7 indicates that magnitudes in the neighborhood of 0.9 are to be expected for ball bearings. Larger magnitudes of 6 and fo cause subsequently smaller valuesof F(p).

Equation (2.1) may also be used for spherical roller bearings to estimate pitch diameter. Radial internal clearance, also called diametral play, as illustrated by Fig. 2.9, is given by equation (2.32).

where ri and ro are theraceway contour radii. The diametralplay s d can be measured with a feeler gage. Table 2.2, excerpted Erom Reference 2.1, gives standard values of radial internal clearance, diametral play, under no load.

Radial spherical roller bearings are normally assembled with free diametral play and hence exhibit free endplay Pe. From Fig. 2.9, it can be seen that

7

""

"

Axis of rotation

F~~UR 2.9.E Schematic diagram of spherical roller bearing showing nominal contact angle a, diametral play S,, and endplay P,.

ro cos p

(

ro -

=

2)

cos a

(2.34)

or

p

=

[

cos-1 (1 -

2)

cos a

Therefore, it can be determined that

P,

=

4rJsin

-

sin a) + 2S, sin a

(2.35)

7. Estimate the magnitude of the free endplay P, for a 22317 spherical roller bearing having the following dimensions:

m m 0 m 0 0 m 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 P-Q,Ommrx,meoOm"MP-m"M"mmOooQ,o P?-i l-4 ?-i du m M M M d d m m eo eoP- rx, Q, 0_?4/-iemeqm_

omooomoooooooooooooooooooo eoP-rx,Omdrx,?-idrx,ddrx,dueoOmO~du~mm?-imdu ?-id*-(

d?-iddmmmMMMddmmeoeo~"rx,Q,oe?-i_me

d d d

omooomoooooooooooooooooooo eo"rx,Odudrx,?-idrx,ddrx,dueoOmoeomrx,mm?-idum ?-id?-idmdum~MmddmmeoeoP-P-rx,Q,oe?-i~me

T+?-id?-iNmmmMMMddmmeoeo""rx,Q,

mmomoomooooooooooooooooooo dmeo"Q,?-iMeoQ,mdeoQ,mmP-?-imomomoP-eoM

m o m m m o o o m o o o o o o o o o o o o o o o o o

~ddmeorx,omd"rx,Omdeorx,?-idP-?-idrx,Mrx,m?-i ?-idd?-i?-idummmduMMMdddmmeo"

d?-i?-i?-idmmmmmmMMdddmmeoP-

m o m m m o o o m o o o o o o o o o o o o o o o o o ~ddmeorx,OmdP-rx,Omdeorx,?-id"ddrx,Mrx,md

?-idd?-id?-iT+dudummmMMmdd

3momooommooooooooooooooooo NmMMdmeoP-Q,dmMdmP-Q,Odudeorx,dm"Mrx,

B

3

B

3

H

8

l-i

7

Roller diameter, D Number of rollers per row Roller effective length, Z* Roller contour radius, R Inner raceway contour radius, ri Outer raceway contour radius, ro Bearing pitch diameter, d, Nominal contact angle, a Diametral play

[ [

p

=

cos-1

p

=

cos-1 (1 -

(1 -

25 mm 14 20.762 mm 79.959 mm 81.585 mm 81.585 mm 135.07'7 mm 12" 0.102 mm

(0.9843 in.) (0.8154 in.) (3.148 in.) (3.212 in.) (3.212 in.) (5.318 in.) (0.004 in.)

5) .] cos

)

'"O2 cos (12")] = 12.167" 2 81.585

PC= 4r,(sin p - sin a) + 28, sin a

P,

(2.34)

=

4 * 85.585(0.2108 - 0.2079) + 2 0.102 * 0.2079

=

0.956 mm (0.0376 in.)

(2.35)

i ~ ~ applies to spherical roller bearings in that, as The term o s c ~ Z a t also illustrated by Figs. 2.9 and 2.10, the rollers and raceways have curvature in the direction transverse to rolling. In this case, osculation is defined as follows: R pr

(2.36)

in which R is the roller contour radius.

.8. Determine the osculation at each raceway contact for the 22317 two-row spherical roller bearing of Example 2.7.

*Roller effective length is the length presumed to be in contact with the raceway under loading. Generally, I = I, - Zr,, in which I",is the roller corner radius or the grinding undercut, whichever is larger. A combination of the corner radius and grinding undercut may also be required to estimate effective length.

71

S P ~ R I ROLLER C ~ B E ~ ~ G S

FIGURE 2.10. Spherical roller bearing geometry.

+.=R - = - =79.959 ri 81.585

R ro

+i="=-

79.959 81.585

o.98

=

0.98

(2.36) (2.36)

mature For spherical roller bearings with point contact between rollers and raceways, the equations for curvature sum and difference are as follows (see Fig. 2.IO):

72

(2.40)

9. Calculate the magnitudes of curvature sum and difference for the 22317 spherical roller bearing of Example 2.7.

Y=-

D cos a

(2.27)

dm

=

1

11 " " ri 79.959

25 cos (12")/135.077 == 0.1810

" "

_R

=

81.585

-

0.00025mm-'(0.0064in.-')

(2.37)

0.09793 mm-' (2.487 in.-')

(2.38)

-

1

1 ro

2/(1 - 0.1810) 2/(1 - 0.1810) 1 79.959

-

25

X

+ 25 X

0.00025 0.00025

1 81.585

" " " "

=

0.00025 m~-'(0.0064 in.-')

=

0.9951.

3

GS

(2.39)

+ 25 X 0.00025)

1 =

0.068 mm-' (1.726 in.-')

+D l + Y

;(

:>

(2.40)

---

- 2/(1 + 0.1810) - 25 X 0.00025 = o.9929 2/(l + 0.1810) + 25 X 0.00025

Note that the magnitude of F ( p ) approaches unity for spherical roller bearings.

ite Equations (2.1)-(2.3) are valid for radial, cylindrical roller bearings as well as ball bearings. Table 2.3 gives standard values of internal clearance forradial cylindrical rollerbearings. These data areexcerpted from Reference 2.1.

A 209 cylindrical rollerbearing has thefollowing dimensions: Inner raceway diameter, di Outer r~cewaydiameter, d, Roller diameter, D oller effective length, Z oller total l e n ~ hZ,, Number of rollers, i5

54.991 mm 75.032 mm 10 mm

(2.1 (2.9 (0.39~7in.)

14

ine the bearing pitch diameter dm and diametral clearance

5

E

tll

B

3 2

r-i

H

ROLLING BEARD46 ~ C R O G E O ~ ~ Y

Pd

=

$(54.991 + 75.032)

=

65.011 mm (2.559 in.)

=

do - di - 2D

=

75.032 - 54.991 - 2 10

=

0.041 mm (0.0016 in.)

0

Figure 2.11 illustrates a roller in a radial cylindrical roller bearing having two roller guide flanges on both inner and outer rings. In this case the roller is shown in contact with both inner and outer raceways, which would occur in the bearing load zone when simple radial loading is applied to the bearing (see Fig. 7.3). It is to be noted that clearance exists in theaxial direction betweenthe roller ends and the roller guide flanges. It can be seen from Fig. 2.11that the bearing will experience an endplay defined by

P,

=

2(2f - I,)

(2.41)

where Zf is the distance between the guide flanges of a ring and I , is the total length of the roller. As mentioned in Chapter 1 and discussed in Chapter 7, radial cylindrical roller bearings with guide flanges on both inner and outer rings can support small amounts of applied thrust load

1 2

__.

.11. Schematic drawingof a radial cylindrical roller bearing havingtwo integral roller guide flangeson the inner ring and one integral and one separable guide flange on the outer ring.

RINGS

77

ERED ROLLER

in addition to the applied radial load. Thebearing endplay influences the number of radially loaded rollers that will participate in supporting the thrust load. The endplay also influences the degree of roller skewing which can occur during bearing operation. See Chapter 14.

Most cylindricalroller bearings employcrowned rollers to avoid the stress-increasing effects of edge-loading (see Chapter 6 ) . For these rollers, even if fully crowned as illustrated by Fig. 1.31a, the contour or crown radius R is very large. Moreover, even if the raceways are crowned, R = ri = r, * a.Therefore, consideringequations (2.37) and (2.39),which describe the curvature sum for inner and outer raceway contact respectively, the difference of the reciprocals of these radii is essentially nil, and (2.42) (2.43)

be Examining equations (2.38) and (2.40), it can

seen that

=

F(p), = 1.

The nomenclature associated with tapered roller bearings is different than thatfor other types of roller bearings. For example, as indicated by Fig. 2.12, the bearing inner ringis called the cone and theouter ring the cap. It can be seen that the operation of the bearing is associated with a pitch cone; equation (2.1) can be used to describe the mean diameter of that cone. For manycalculations, this mean cone diameter will be used as the bearing pitch diameter dm.Figure 2.13 indicates dimensions and angles necessary for the p ~ r f o r ~ a n analysis ce of tapered roller bearings. From Fig. 2.13, it can be seen that ai,the inner raceway-roller contact e = 3 cone angle, a,, the outer raceway-roller contact angle = cup e, af,the roller large end-flange contact angle = 4 cone back face d aR= roller angle. Dm= = the large end diameter of the in = the small end diameter of the roller, whichhas anendto-end length of I,.

+

78

ROLLING BEARING ~ C R O G E O ~ T R ~

1 CUP FRONT FACE CAGE

CONE BACK FACE RIB

CONE FRONT FACE RIB,

CONE BACK FACE

CONE FRONT FACE

W

a UI

I I I I

I I

CAGE CLEARANCE

2.12. Schematic drawing of tapered roller bearing indicating nomenclature.

FI

2.13. Internal dimensions for tapered roller bearing performance analysis.

CLOSURE

Tapered roller bearings are usually mounted in pairs. In general, the clearance is removed so that a Z ~ ~ e - t o - Zfiti ~ise achieved under no load. It is possible, however, fora bearing set support in^ substantially applied radial loading, that a small amount of endplay is set at room temperature mounting to achieve desired distribution of load amongthe tapered rollers under higher temperature operating conditions. Endplay in tapered roller bearings is therefore associated with the bearing pair.

From Fig. 2.13, it is seen that theouter raceway contact angle is greater than the inner raceway contact angle. Therefore, considering equations (2.37) and (2.39), the curvature sums for the inner and outer raceway contacts are given by (2.44) (2.45)

where (2.46) (2.47) .Rm

cos a,

Y o ,=

(2.48)

dm

These equations give approximate answers in the respective calculations of curvature sum since the mean radius of the roller lies in a plane slightly angled to that in which the raceway rolling radius lies. As far c~lindricalroller bearings, F ( P ) = ~ F(p), = 1.

The relationships developed in thischapter are based onlyon the macroshapes of the rolling components of the bearing. When loadis applied to the bearing, these contours may be somewhat altered; however, the undeformed geometry must be used to determine the distorted shape.

ROLLING BEARING ~ ~ R O G E O ~ T R Y

Numerical examples developedin this chapter were of necessity very simple in format. The quantity of these simple examples is justified since the results from the calculations will subsequently be used as starting points in more complex numerical examples involving stresses, deflections, friction torques, and fatigue lives.

2.1. American National Standards Institute,American National Standard ( ~ S I / ~ ~ A ) Std. 20-1987,“Radial Bearings of Ball, Cylindrical Roller, and Spherical Roller Types, Metric Design” (October 28, 1987). 2.2 A. Jones, Analysis of Stresses and Deflections, v01. 1, New Departure Division, General Motors Corp., Bristol, Conn., 12 (1946).

T Symbol

Description Basic inner ring width Single width of an inner ring Basic outer ring width Single width of an outer ring Basic bore diameter Bearing inner raceway diameter g outer raceway diameter diameter of a bore Single plane mean bore diameter diameter er of an outside surface ngle plane mean outside ameter ~ o m m o ndiameter asic housing bore

Units mm (in.) mm (in.) mm (in.) mm (in.) mm (in.) mm (in.) mm (in.) mm (in.) mm (in.) mm (in.) mm (in.) mm (in.) mm (in.) mm (in.)

~ E ~ E R E ~1~~~ ~ C E

Symbol

Description Outside ring 0.d. Inside ring i.d. Basic shaft diameter Modulus of elasticity Interference Radial runout of assembled bearing inner ring Radial runout of assembled bearing outer ring Length Bearing clearance Pressure Ring radius Inside radius of ring Outside radius of ring Inner ring reference face runout with bore Outside cylindrical surface runout with outer ring reference face Axial runout of assembled bearing inner ring Axial runout of assembled bearing outer ring Radial deflection Bore diameter variation in a single radial plane Mean bore diameter variation Mean outside diameter variation Outside diameter variation in single radial plane Single inner ring widthdeviation from basic ingle outer ring width deviation from basic Single bore diameter deviation from basic Single plane mean bore diameter deviation from basic for a tapered bore small end Single plane mean bore diameter deviation at large end of tapered bore Single outside diameter deviation from basic

CE

Units mm (in.) mm (in.) mm (in.) N/mm2 (psi) mm (in.) pm (in.) pm (in.) mm (in.) mm (in.) N/mm2 (psi) mm (in.) mm (in.) mm (in.) pm (in.) pm (in.) pm (in.) pm (in.) mm (in.) pm (in.) pm (in.) pm (in.) pm (in.) pm (in.) pm (in.) pm (in.) p m (in.)

pm (in.) pm (in.)

83

GENE

Symbol

Description Single plane mean outside diameter deviation from basic Clearance reduction due to press fitting of bearing in housing Clearance reduction due to press fitting of bearing on shaft Clearance increase due to thermal expansion Temperature Strain in radialdirection Strain in tangentialdirection Coefficient of linear expansion

s 0-r

0-t

Poisson's ratio Normal stress in radialdirection Normal stress in tangential direction

Units pm (in.) mm (in.) mm (in.) mm (in.) "C("F) mm/mm (in./in.) mm/mm (in./in.) m m / ~ m / " C(in./ in./ O F ) N/mm2 (psi) N/mm2 (psi)

all and roller bearings are usually mounted on shafks or in housings with interference fits. This is usually done to prevent fretting corrosion that could be produced byrelative movement between the bearing inner ring bore and the shaft 0.d. and/or the bearing outer ring 0.d. and the housing bore. Theinterference fit of the bearing inner ringwith the shaft is usually accomplished by pressing the former member over the latter. In some cams, however, the inner ringis heated to a controlled temperature in an oven or in an oil bath. Then the inner ring is slipped over the shaft and allowed to cool, thus accomplishing a shrink fit. Press or shrink fitting of the inner ringon the shaft causes the inner ring to expand slightly. Similarly, press fitting of the outer ring in the housing causes the former memberto shrink slightly. Thus, the bearing's diametral clearance will tend to decrease. Large amounts of interference in fitting practice can cause bearing clearance to vanish and even produce negative clearance or interference in the bearing. Thermal conditions of bearing operation can also affectthe diametral clearance. Heat generated by friction causes internal temperatures to rise. This in turn causes expansion of the shaft, housing, and bearing components. Depending on the shaft and housing materials and on the magnitude of thermal gradients across the bearing and these supporting structures, clearance can tend to increase or decrease. It is also apparent

~

E

~

~ FI!CTING ~ N

C

E

CE

that the thermal environment in which a bearing operates may have a significant effect on clearance. In Chapter 2 it was demonstrated that clearance significantly affects ball bearing contact angle. Subsequently,the effect of clearance on bearing internalload distribution and life will beinvestigated. It is therefore clear that the mechanics of bearing fitting practice is an important part of this book.

~ t a n d a ~ ddefining s recommended practices for ball and roller bearing usage were first developed in the United States by the ti-Friction Bearing ~anufacturersAssociation (AE’BU), which has nowbecome the American Bearing ~anufacturersAssociation (AI3M.A). tinues the process of revising the current standards andproposing and preparin~new standards as deemed necessary by its bearing industry member companies. AI3M.A-generated standards are subsequently proposed to the American National Standards Institute (ANSI) as United States national standards. ANSI has a committee dedicated to rolling bearing standard activities; this committee has representatives of bearing user organizations such as major industrial manufacturers and the U.S. government. Other countries have national standards organizations similar to ANSI; for example, DIN in Germany and JNS in Japan. Currently, 26 documents, so tric and English unit system parts, havepublished been as standards. Any national stand quentlyproposed be to the International Organization for Standards?andafter extendednegotiation ~ublished asinternational Standard (ISO)”with an identifying number. In this chapter, various bearing, shaft, and housing tolerance data are pted from the American National Standards. ference [3.1] defines recommended practice in fitting bearing inner d outer rings in housings. These fits are recommended ormal, and heavy loadingas defined by Fig. 3. I. Figure ations and relative m a ~ i t u d e of s the shaft-bearing bore and ho~sing-bearing0.d. tolerance ranges. Each sha~-bearingfit tolerance range is d e s i ~ a t e by d a lower cas for example, g6,h5 and so on. Similarly, ea ists of an upper cas housing-bearing fit or Table 3.1 gives the forexample,6.7 r rings on shaft practicefor fitting ameter tolerance limits correspond in^ to milar data for fitting of bearing outer rings in housing in references [3. eraiice ranges on bearing bore and o.d. for various types of radial bear-

ST€?,W,N A T I O ~AND ~ , ~ E ~ A T I SOT A N N~D ~ D S

PfC,

3.1. Classification of loading for ball, cylindrical roller, and spherical roller bearings.

ings. Several of these bearing types, for example,tapered roller bearings, needle roller bearings, and instrument ball bearings, exist in too many variations to include all of the appropriate tolerance tables herein. On the other hand, reference E3.101 covers a wide range of standard radial ball and rollerbearings;Tables 3.6-3.10 are taken from reference[3.10]. For radial ball bearings these tolerances are grouped in ABEC* classes 1, 3, 5, 7, and 9 according to accuracy of manufacturing. Accuracy improves and tolerance ranges narrow as the class number increases. Tables 3.6-3.10 give tolerance ranges for all ABEC classifications. Additionally, Tables 3.6-3.8 provide the tolerances or bore and 0.d. for radial roller bearings as well as for ballbearings. The ABEC and tolerance classes correspondin every respect to the precision cla dorsed b the ISO. Table 3.5 shows the correspondencebetween the A.NSI/ and IS0 classifications. It is further noted that inch tolerances given in Part I1 of Tables 3.6-3.10 are calculated from primary metric tolerances given in Part I of those tables. To define the range of interference or looseness in themounting of an inner ring on a shaft or an outer ring in a housing, it is necessary to consider combinationof the shaft, housing, and bearing tolerances. *Annular BearingEngineers’ Committee of ABMA. t Roller Bearing Engineers’ Committee of ABMA.

8

~ T E ~ R E N C F E

I AND CL ~

Interference

Shafts

Clearance

he

Interference

F

I

~ 3.2 ~. ~Graphic E representation of fits.

~CE

~

I

a

8

I4

oooooF4

*222%2

*322%

000000

I

ri

000 d0

a,

u3 M

ba

W

rn

9

8 iii

M

W

rn

8

;ii

9

M

W

rn

8 iii

9

z

9

z

9

I

i

7

8

00

d

c:

l l

c:

00

ho

W

2

2

a

I

I

or(

I

I

ori

I

I

orc

I

wdr

+ + w+ m+ w+ m+ w+ m+ w+ m+

NcUSN

n

I

3

2%

SE

0

0 0 Nrl

Cvm ++

0 m

Nc-

4s

m

8

-

W W

om s

N N

I I -

m

0

O3

0

3

W

m 0 0

I ~

0

3 0

%

+

3% 3%

ljjj 0

3

25 0 0

++ -

s m

+" -

~

++

I

dim + I

J': dim + I

W E

++

mdi

-3 -

N m + I

oc-

I "TI,':

+ ---"-

I

--"--

I

7-4

7-47-4

98 07-4

---"--"-1 $24: $24: m r ' r+' +

m c -cm- m

I

I

+

1

2

%

w

---"-

:"r" m

1

I

0 4

I

I

I

0 %

I I

I

I

I

m I

0 0

I

m I

0 0

'

m I

0 0

2%

e5 *I++ ++

cow

I-0')

cu

n W

cu

n W

1-

a

7

LE 3.5. ANSI/AE3M.A vs IS0 Tolerance Classi~cations ABEC 1or RBEC 1 ABEC 3 or RBEC 3 ABEC 5 or RBEC 5 ABEC 7 BEC9

Normal class Class 6 Class 5 Class 4 Class 2

LE 3.6. PART 1. Tolerance Class ABEC-1, RBEC-1. Metric Ball and Roller Bearings [except tapered roller bearings"] of Dimensions Conforming to the Basic Plan for Boundary Dimensions of Metric Radial Bearings Given in Table 1 of [3.10].

Inner Ring (Tolerance values in micrometers) diameter series d rnm

over

incl. ~

a

0.6 2.5 10 18 30 50 80 120 180 250 315 400 500 630 800 1,000 1,250 1,600

7, 8, 9

Ah, 2.5 10 18 30 50 80 120 180 250 315 400 500 630 800 1,000 1,250 1,600 2,000

high

low

0, 1

-

ABS

2, 3, 4

-

max.

VhP

Kia

max.

max.

all high

6 6 6 8 9 11 15 19 23 26 30 34 38

10 10 10 13 15 20 25 30 40 50 60 65 70 80 90 100 120 140

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

-

-

normal -

modifiedd low

VBs Max.

~

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

-8 -8 -8 - 10 - 12 - 15 -20 -25 -30 -35 -40 -45 -50 -75 - 120 - 125 - 160 -200

10 10 10 13 15 19 25 31 38 44 50 56 63

8 8 8 10 12 19 25 31 38 44 50 56 63

6 6 6

8 9 11 15 19 23 26 30 34 38

-40

- 120 - 120 - 120 - 120 - 150

-200 -250 -300 -350 -400 -450 -500 -750 - 1,000 - 1,250 - 1,600 -2,000

-

-250 -250 -250 -250 -380 -380 -500 -500 -500 -630

12 15 20 20 20 25 25 30 30 35 40 50 60 70 80 100 120 140

ty7yyy

comomomomomooo l+cvmmd*mbocvwom I I I I I I I

0 0 0 0 0 0 0 0 0 0 0 0 0

1 *e

t;

Q)

-4J

o o o m o o o o o o o o o &i mcoml-loomoomooo cv m w m CD co o~cv~qo_q3 l+ l+

3.6. 2. Tolerance Class imensions ~ o n f o ~ i to n gthe Basic Plan for ~

o

etric Ball and Roller Bearings [except tapered roller bearings"] of ~ ~ id~ e an s ~ i of o Metric ~s Radial B ~ a r i n Given ~s in Table 1of 13.101.

Inner Ring (Tolerance values in .0001 in.) d'p

diameter series d mm

AdmP

over

incl.

0.6 2.5 10 18 30 50 80 120 180 250 315 400 500 630 800 1,000 1,250 1,600

2.5 10 18 30 50 80 120 180 250 315 400 500 630 800 1,000 1,250 1,600 2,000

a

high

low

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

-3 -3 -3 -4 -4.5 -6 -8 -10 -12 -14 -16 -18 -20 -30 -39 -49 -63 -79

7, 8, -

O, max.

2, 3>

4 4 4 5 6 7.5 10 12 15 17 20 22 25

3 3 3 4 4.5 7.5 10 12 15 17 20 22 25

2.5 2.5 2.5 3 3.5 4.5 6 7.5 9 10 12 13 15

43s

Vdmp max. 2.5 2.5 2.5 3 3.5 4.5 6 7.5 9 10 12 13 15

Kia mas. 4 4 4 5 6 8 10 12 16 20 24 26 28 31 35 39 47 55

all normal -~ high 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

- 16

-47 -47 -47 -47 -59 - 79 -98 - 118 - 138 - 157 - 177 - 197 -295 -394 -492 -630 - 787

modifiedd low -98

-98 -98 -98 - 150

- 150 - 197 - 197

- 197 -248 -

VBS

max. 4.5 6 8 8 8 10 10 12 12 14 16 20 24 28 31 39 47 55

TABLE 3.6. PART 2. (Continued) Outer Ring (Tolerance values in .0001 in.) Y

0 N

VDP" Capped Bearings

Open Bearings diameter series

D mm over

a

2.5 6 18 30 50 80 120 150 180 250 315 400 500 630 800 1,000 1,250 1,600 2,000

7, 8, 9

ADmp

incl. 6 18 30 50 80 120 150 18QL 250 315 400 500 630 800 1,000 1,250 1,600 2,000 2,500

high

low

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

-3 -3 -3.5 -4.5 -5 -6 -7 - 10 - 12 - 14 - 16 - 18 -20 -30 -39 -49 -63 -79 -98

0, 1

2, 3, 4 max.

4 4 4.5 5.5 6.5 7.5 9 12 15 17 20 22 25 37 49 -

3 3 3.5 4.5 5 7.5 9 12 15 17 20 22 25 37 49

-

-

2.5 2.5 3 3 4 4.5 5.5 7.5 9 10 12 13 15 22 30

-

-

"This diameter is included in the group. bNo values have been established for diameter series 7, 8, 9, 0 and 1. "Applies before mounting and after removal of internal or external snap ring. dThis refers to the rings of single bearings made for paired or stack mounting. "For tapered roller bearing tolerances see L3.8, 3.91.

2, 3, 4

vDm;

Kea

max.

max.

2.5 2.5 3 3 4 4.5 5.5 7.5 9 10 12 13 15 22 30

6 6 6 8 10 14 16 18 20 24 28 31 39 47 55 63 75 87 98

-

VCS

ACS

high

low

max.

Identical to A,, and V,, of inner ring of same bearing

mi

Q,

b"

23

!i E

E

4

3 0 0 0 0 0 0 0 0 0 0 0 0

o o o o o o o o o o o o c

3

1

l-li

d

m"

on

0"

l-ll

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Class ABEC-3, RBEC-3. Metric Ball and Roller Bearings [except tapered roller bearings"] of sic Plan for Boundary Dimensions of Metric Radial Bearings Given in Table 1 of [3.10]. hner Ring (Tolerance values in .0001in.)

diameter series

ABS

d

Ah,

7, 8, 9

over

incl.

high

low

0,6 2.5 10 18 30 50 80 120 180 250

2.5 10 18 30 50 80 120 180 250 315

400 500

500 630

0 0 0 0 0 0 0 0 0 0 0 0 0

-3 -3 -3 -3 -4 -4.5 -6 -7 -8.5 - 10 - 12 - 14 - 16

a

0, 1

2, 3, 4

VdmP max.

Kia max.

all high

normal

2 2 2 2.5 3 3.5 4.5 5.5 6.5 7.5 9 10 12

2 2 2 2.5 3 3.5 4.5 5.5 6.5 7.5 9 10 12

2 2.5 3 3 4 4 5 7 8 10 12 14 16

0 0 0 0 0 0 0 0 0 0 0 0 0

- 16 -47 -47 -47 -47 -59 -79 -98 -118 - 138 - 157 - 177 - 197

max, 3.5 3.5 3.5 4 5 6 7.5 9 11 12 15 17 20

3 3 3 3 4 6 7.5 9 11 12 15 17 20

modified" low

-98 -98

-98 -98 - 150 - 150 - 197 - 197 - 197

-248

VBEl max. 4.5 6 8 8 8 10 10 12 12 14 16 18 20

ce Class ABEC-5, RBEC-5. Metric Ball and Roller Bearings [except i n s t ~ e nbearings," t and ensions Conforming to the Basic Plan for B ~ ~ Dimensions d a ~ of Metric Radial Bearings Given. in Table 1of [3.10].

Inner Ring (Tolerance values in ~ i ~ r o ~ e ~ e r s ~ VdP

diameter series

ABS

d

mm

Ah*

7 7

over

incl.

high

low

a 0.6 2.5 10 18 30 50 80 120 180 250 315

2.5 10 18 30 50 80 120 180 250 315 400

0 0 0

-5 -5 -5 -6 -8 -9 - 10 - 13 - 15 - 18 -23

0 0 0

0 0 0 0

0

8, 9

5 5 5 6 8 9 10 13 15 18 23

0, 1, 2737 4 Max. 4 4

4 5 6 7 8 10 12 14 18

Vhp max.

Kia max.

s d max.

Si2 max.

dl high

3 3 3 3

4 4 4

4

5 5 6 8 10 13 15

7 7 7 8 8 8 9 10 11 13 15

7 7 7 8 8 8 9 10 13 15 20

0 0 0 0 0

5 5 7 8 9 12

4

0 0 0

0 0 0

normal modd VBS low max. -40 -40 -80 - 120 - 120 - 150 -200 -250 -300 -350 -400

-250 -250 -250 -250 -250 -250 -380 -380 -500 -500 -630

5 5 5 5 5 6 7 8 10 13 15

t.r

.

~~~ntinued) Outer Ring (Tolerance values in micrometers) diameter series ~

D mm over a

2.5

6 1 3 5 80 12 150 180 250 315 400 50 63 ~

7, 8, 9

ADmp

incl.

high

low

6 18 30 50 80 120 150 180 250 315 400 500 630 8

0 0 0 0 0 0 0 0 0 0 0 0 0

-5 -5 -6 -7 -9 - 10 -11 - 13 - 15 - 18 -20 -23 -28 -35

0,

2, 3,

max. 5 5 6 7 9 10 11 13 15 18 20 23 28 35

4 4 5 5 7 8 8 10 11 14 15 17 21 26

vDmp

Kea

SD

8,:

max.

max.

max.

max.

3 3 3 4 5 5 6 7 8 9 10 12 14 18

5 5

8 8 8 8 8 9 10 10 11 13 13 15 18 20

8 8 8 8 10 11 13 14 15 18 20 23

~

“This diameter is included in the group. bNo values have been estab~shedfor sealed or shielded bearings. “Applies to groove type ball bearings only. dThis refers to the rings of single bearings made for paired or stack mounting. “For instrument ball bearing tolerances see K3.2, 3.31. f For tapered roller bearing tolerances see 13.8, 3.91.

6 7 8 10 11 13 15 18 20 23 25 30

25

30

A,, high

vcs

low

Identical to ABsof inner ring of same bearing

max. 5 5 5 5 6 8 8 8 10 11 13 15 18 20

" a, ' E :

a,

&I

%

cr,

i

3

>oooooooooo

o o o o o o o o o o c

0 0 0 0 0 0 0 0 0 0 0 0 0 0

3 0 0 0 0 0 0 0 0

oooooc3c2oo

0 0 0 0 0 0 0 0 0

d

l

I I

0

+-,

4

5

~

B 0

t5

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

ooc>oooooo

~1

'I

43 U

8

(3

dE

B

0

r-4

6 3

!5

r-4

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

ad

ai "

E

w

6 3 B

0

w

6 3

0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0

3 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

1

0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0

11

EFFECT OF ~ E R F E R E N C EFI!t"I'mG CLEARANCE ON

NG The solution to this problem may beobtained by using elastic thick ring theory. Consider the ring of Fig. 3.3 subjected to an internal pressure p per unit length. The ring has a bore radius & and an outside radius 9j0. For the elemental area % d( % d+ the summation of forces in the radial direction is zero for static equilibrium:

(

)

crr%d# + 20, d % sin d+ - - cr + dcrr - % (%+ d@d+ 2 ' dmd Since d+ is small, sin higher order,

4d+

=

=

0

d+ and, neglecting small quantities of

direction, there is an elongation Corresponding to the stress in the radial u and the unit strain in the radialdirection is

F I ~ U 3.3. ~ E Thick ring loaded by internal pressure p .

1

~ ~ R F E ~ N C E

E,

F I T T ~ G

CE

du

d%

=

(3.3)

In the circumferential direction the unit strain is U

E, = -

(3.4)

9

According to plane strain theory,

Combining equations (3.3-3.6) yields

~ubstitutingequations (3.7) and (3.8) into (3.2) yields d2u -+"" dS2

1 du %d%

-u= gx2

0

(3.9)

The general solution to equation (3.9) is u = el%

+ c2S-l

(3.10)

~ubstitutingin equations (3.7) and (3.8) from (3.10) gives (3.11) (3.12) At the boundary defined by 9%= gl0, a, = p

=

0; therefore,

121

EFFECT OF ~ T E ~ E ~ FN C ~E ON CLEARANCE ~ G

(3.13)

At

=

q,ar= p

and, therefore, (3.14)

Substituting equations (3.13) and (3.14) into (3.11) and (3.12) yields (3.15) (3.16) Similarly, for a ring loaded by external pressure only, (3.17) (3.18) From equations (3.15), (3.16), and (3.5), the increase in the internal radius of a ring loaded by internal pressure p is given by (3.19) Similarly, the decrease in the external radius ternal pressure p is given by

soof a ring loaded by ex(3.20)

If a ring having elastic modulus E,, outside diameter g,,and bore I%, is mounted with a diametral interference I on a second ring having modulus E,, outside diameter 9, and bore g2, then a common pressure p develops between the rings. The radial interference is the sum of the radial deflection of each ring due to pressure p . Hence the diametral intederence is given by

~

I

= 2(u,

E

R

~ FITTING ~ ~

C

+ u,)

CE

E

(3.21)

In terms of the common diameter 9,therefore, I = P i n { -1 [

E,

(8,/9)2

+ 1+

8,/8)2- 1

~,] [

+1 (9/9,)2

+1

E, (8/9,), -1

It can be seen that equation (3.22) can be used known; thus,

to determine p if I is

9

p = +E]

+-[

E, 1

+ 1-

(9/9,)2 (9/g2)2 -

E,]

(3.23)

If the external ring is a bearing inner ring of diameter 9,and bore 8, as shown in Fig. 3.4, then the increase in 9,due to press fitting is

(3.24)

I

3.4. Schematic diagram of a bearing inner ring mounted on a shaft.

If the bearing inner ring and shaft are both fabricated from the same material, then

(3.25) For a bearing inner ring mounted on a solid shaft of the same material, diameter g2is zero and A, = I

(~)

(3.26)

By a similar process it is possible to determine the contraction of the bore of the internal ring of the assembly shown in Fig. 3.5. Thus,

For a bearing outer ring pressed into a housing of the same material,

3.5. Schematic diagram of a bearing outer ring mounted in a housing.

124

~

E

~

E

~ FITTING N C EAND C

L

~

C

(3.28)

If the housing is large compared to the ring dimensions, diameter gl approaches infinity and (3.29)

Considering a bearing having a clearance P d prior to mounting, the change in clearance after mounting is given by

The preceding equation takes no account of differential thermal expansion.

Since the pressure p between interfering surfaces is known, it is possible to estimate the amount of axial force necessary to accomplish or remove an interference fit. Because the area of shear is mgB, the axial force is given by

Fa = p7Ti3B p

(3.31)

in which ~ lis, the coefficient of friction. Accordingto Jones 13.111 the force required to press a steel ring on a solid steel shaft may be estimated by

Fa = 47100 BI

[ (~)"] 1

-

(3.32)

This is based on a kinetic coefficient of friction p = 0.15. Similarly; the axial force required to press a steel bearing into a steel housing is given by

[ (~)"]

Fa= 47100 CI 1 -

(3.33)

SI Rolling bearings are usually fabricated from hardened steel and are generally mounted with press fits on steel shafts. In many applications, such

E

as in aircraft, however, the bearing may be mounted in a housing of a dissimilar material. Bearings are usually mounted at room temperature; but they may operate at temperatures elevated AT above room temperature. The amount of temperature elevation may bedetermined by using the heat generation and heat transfer techniques indicated in Chapter 15. Under the influence of increased temperature, materialswill expand linearly to the following equation: U =

re(T - TJ

(3.34)

in which I' is the coefficient of linear expansion in mm per mm per "C and 2, is a characteristic length. Considering a bearing outer ring of outside diameter do at temperatures To - Taabove ambient, the increase in ring outside circumference is given approximately by (3.35)

Therefore the approximate increase in diameter is (3.36)

The inner ring will undergo a similar expansion: (3.37)

Thus the net diametral expansion of the fit is given by

When the housing is fabricated from a material other than steel, the interference I between the housing and outer ring may either increase or decrease at elevated temperatures. Equation (3.39) gives the change in I with temperature:

In which r b and rh are the coefficients of expansion of the bearing and housing, respectively. For dissimilar materials the housing is likely to expand more than the bearing, which tends to reduce any interference fit. Equation (3.30) therefore becomes

If the shaft is not fabricated from the same material (usually steel) as the bearing, then a similar analysis applies.

I ~ E ~ F E ~ NF C I E ~

~

CE

G

The interference 1' between a bore and 0.d. is somewhat less than the apparent dimensional value due to the smoothing of the minute peaks and valleys of the surface. The schedule of Table 3.11 for reduction of 1' may be used. It can be seen from Table 3.11 that for an accurately ground shaft mating witha similar bore, it may be expected that thereduction on the bore diameter would be 0.0020 mm (0.00008 in.) and on the shaft possibly 0.0041 mm (0.00016 in.) or a total reduction in I of 0.0061 mm (0.00024 in.). The 209 radial ball bearing of Example 2.4 is manuEC 5 specifications. The bearing is mounted on a solid steel shaft witha k5 fit and ina rigid steel housing with a K6 fit. The nominal bearing bore is 45 mm (1.7717 in.), and the nominal 0.d. is 85 mm (3.3465in.). Determine the bearingcontact angle and free endplay under light thrust loading. Shaft tolerance range from Table 3.2 is 0.0025 mm ( 0.0127 mm (0.0005 in.) or 0.0076 mm (0.0003 in.) mean. mean tolerance from Table 3.8 is 0.004 mm (0.00016 in.), a negative value; that is, -0.004 mm (-0.00016 in.) The mean interferenceon the shaft is

I

=

0.0076

+ 0.004

=

0.0116 mm (0.0005 in.)

Assuming the bearing is mounted on a ground surface, the reduction in 1' due to surface finish is approximately 0.0020 mm (0.00008 in.) (see Table 3.11) for the bearing bore and shaft. Therefore

I = di = 91 =

0.0116 - 2 X 0.0020 52.3 mm (2.0587 in.) di

=

0.0076mm(0.00030 in.) 2.4

Ex.

.ll. Reduction in Interference Due to Surface Condition

Finish Accurately ground surface Very smooth turned surface (24-56) Machine-reamed bores (40-94) Ordinary accurately turned surface (94-190)

Reduction 0.0001 mm

Reduction 0.0001 in.

20-51 61-142 102-239 239-483

(8-20)

127

EFFECT OF SURFACE FINISH

(3.26) 0.0076 - = 0.0065 mm (0.00026 in.) ( ~ 3 )

=

Housing tolerance range from Table 3.3 is -0.0178 mm (-0.0007 in.) to 0.0051 mm (0.0002 in.) or -0.0064 mm (-0.00025 in.) mean. is 0.005 mm Bearing 0.d. mean tolerance range fromTable3.8 (0.00020 in.), a negative value, that is, -0.005 mm (-0.00020 in.). The mean interference in the housing is

I

=

0.0064 - 0.005

=

0.0014 mm (0.00006 in.)

Assuming the bearing housing bore is accurately ground, the reduction i' due to surface finish is approximately 0.0020 mm (0.00008 in.) (see Table 3.11) for the housing bore and the bearing 0.d. Thus, the net interference in the housing is virtually zero.

& = 0.015 mm (0.0006 in) Apd

A

(3.30)

=

-As

=

-0.0065

=

0.508 mm (0.02 in.)

-

Ex. 2.4

Ah

+ 0 = -0.0065

mm (-0.00026 in.) Ex. 2.4 (2.10)

=

(

cos-1 1 -

0.015 - 0.0065 2 X 0.508

=

7'25'

Pe = 2A sin a' =

2

X

0.508

(2.12) X

sin (7'25')

=

0.1312 mm (0.0052 in.)

Comparison of these values of a' and Pe with those of Example 2.4 indicates the necessity of including the effect of interference fitting in the determination of clearance. e 3.2. The 218 angular-contact ball bearing of Example 2.3 has a 90 mm (3.5433 in.) bore, a 160 mm (6.2992 in.) 0.d. and is manufactured to ABEC 7 tolerance limits. The bearing is mounted on a hollow steel shaft of 63.5 mm (2.5 in.) bore with a k6 fit and in a

12

CE

titanium housing havingan effective 0.d. of 203.2 mm (8 in.) with an M6 fit. Determine the free contact angle of the bearing. Shaft tolerance range from Table 3.2 is t-0.0025 mm (0.0001 in.) to t-0.0254 mm (0.0010 in.) or a +0.0140 mm (0.00055 in.) mean. Bore mean tolerance range from Table 3.9 is 0.004 mm (0.00016 in.), a negative value, that is, -0.004 mm (-0.00016 in.) The mean interference on the shaft is

I

=

0.0140

+ 0.004 = 0.0180 mm

(0.00071 in.)

Assuming the bearing is mounted on a ground surface, the reduction in I due to surface finish is approximately 0.0020 mm (0.00008 in.) (see Table 3.11) for the bearing bore and shaft, therefore,

I

=

0.0180

d,

=

102.8 (4.047 mm

-

2

X

0.0020 = 0.0140 mm (0.00055 in.) in.)

g1 = d,

(2)[

I

A,

( 'f?'92)2

(91/92)2

-

:]

[

(3.25)

]

102.8 (90/63.5)2 - 1 90 (102.8/63,5)2 - 1

=

0.0140

=

0.00995 mm (0.00039 in.)

X

2.3 Ex.

Housing tolerance range from Table 3.3 is -0.033 mm (-0.0013 in.) to -0.0076 mm (-0.0003 in.), or a -0.0203 mm (-0.0008 in.) mean. Bearing mean 0.d. tolerance range from Table 3.9 is 0.005 mm (0.0002 in.), a negative value, that is, -0.005 mm (-0.OO02 in.). The mean interference is the housing is

I

=

0.0203

-

0.005 = 0.0153 mm (0.0006 in.)

Assuming the housing is mounted on a ground surface, the reduction in I due to surface finish is approximately 0.0020 mm (0.0008 in.) (see Table 3.11) for the bearing 0.d. and housing bore, therefore

I

=

0.0153 = 2

X

0.0020 = 0.0113 mm (0.00044 in.)

do = 147.7 mm (5.816 in.) g2 = do

For steel

2.3 Ex.

129

EFFECT OF SURFACE FINISH

E

206900 N/mm2 (30 X

=

lo6 psi)

5 = 0.3 For titanium

E

=

103500 N/mm2 (15 X

lo6 psi)

5 = 0.33

2

-

X

0.0113

X

(1601147.7) r

t,

(

[("")'

147.7

-

11 <

-

0.3

+

206900 103500

~

+1

)

2

+ 0.33

( ~ ) 2

-1

i

(3.27) =

0.0064 mm (0.00025 in.) (3.30)

A P d =,A8 - Ah =

-0.00995

-

0.0064 = -0.01635 mm (-00064 in.)

Ex.2.3

Pd = 0.483 mm (0.019 in.) (2.10) =

c0s-l

(

1-

0.483 - 0.01635) 2 X 1.031

=

39019,

~ ~ 3.3.~ TheZinnere ring of the 218 angular-contact ball bearing operates at a mean temperature of 148.9"C (300°F)and the outer ring is at 121.1"C (250°F). Consideringthat the bearing was assembled at 21.1'6 (70°F) and considering the press fits of Example 3.2, what free contact angle will occur? For steel

3130

I' = 11.7 X For titanium I' = 8.5 X

mm/mm/'C (6.5 mm/mm/'C (4.7

in./in./'F)

X

X

in./in./"F) Ex. 2.3

di = 102.8 mm (4.047) in.)

Ex. 2.3

do = 147.7 mm (5.816 in.) Because of differential expansion,

AT

- di(Ti -

= rb[do(Fo - Fa) =

11.7

=

0.0191 mm (0.00075 in.)

X

Ta)]

10"6(147.7 X 100 - 102.8

X

127.8)

(3.38)

The outer ring and housing havedifferent rates of expansion Ex. 3.2 $h = 160 mm (6.2992 in.)

AI

= (I'b

-

I'h)gh(To -

Ta)

(3.39)

=

(11.7 - 8.5) X

=

0.0508 mm (0.0020 in.)

Ah

=

0.0064 mm (0.000~5in.)

Ex. 3.2

I

=

0.0113 mm (0.00044 in.)

Ex. 3.2

X

160 X (121.1 - 21.1)

I=I+AI =

0.0113 + 0.0508

=

Ah = o*062 X 0.00635 0.0113

0.062 mm (0.00244 in.)

=

0.0348 mm (0.00137 in,)

A, = 0.00995 mm (0.00039 in.) APd = AT - A, - Ah =

0.0193. - 0.00995 =

Ex. 3.2

-

(3.40)

0.0348

-0.0257 mm (-0.00101 in.)

Pd = 0.483 mm (0.019 in.)

Ex. 2.3

A

Ex. 2.3

=

1.031 mm (0.0406 in.)

(2.10) =

(

cos-1 1 -

0.483 - 0.0257 2 X 1.031

=

38'54'

131

REFEREN~ES

4. The 209 ball bearing of Example 3.1 has a nominal width of 19 mm (0.7480 in.). What force is required to accomplish the shaft press fit in Example 3.1?

2-

=

0.0076 mm (0.00030 in.)

(Ex. 3.1

5B8

=

45 mm (1.7717 in.)

Ex. 3.1

=

52.3 mm (2.0587 in.)

Ex. 3.1 (3.32)

=

4.71

X

lo4 X

r

19 X 0.0076 1 L

=

1766 N (397 lb)

The important effect of bearing fitting practice on diametral clearance has been demonstrated for ball bearings in the numerical examples. Because the ball bearing contact angle determines its ability to carry thrust load and the contact angle is dependent on clearance, the analysis of the fit-up is important in many applications. The numerical examples herein were based on mean tolerance conditions. In many cases, however, it is necessary to examine the extremes of fit. Although onlythe effect of fit-up on contact angle has been examined, it is not to be construed that thisis the only effect of significance. Later, the sensitivity of other phases of rolling bearing operation to clearance will be investigated. The thermal conditions of operation have been shown to be of no less significance than thefit-up. In precision applications, the clearance must be evaluated under operating conditions. Tables 3.6-3.10 contain tolerance limits on radial and axial runout as well as the tolerance limits on mean diameters. Runout affects bearing performance in subtle ways such as through vibration as discussed in Chapter 25.

3.1. American National Standards Institute, American National Standard( ~ ~ ~ Std 7-1995, “Shaft and Housing Fits for Metric Ball and Roller Bearings (Except Tapered Roller Bearings) Conformingto Basic Boundary Plans” (October 27, 1995).

I

~

3.2. American National Standards Institute, American National Standard ~ S I / ~ ~ A ) Std 12.1-1992, “Instrument Ball Bearings Metric Design” (April6, 1992). 3.3. American NationalStandards Institute, American National Standard ( ~ S I I ~ ~ A ) Std 12.2-1992, “Instrument Ball Bearings Inch Design” (April6, 1992). 3.4. American NationalStandards Institute, American National Standard ~ S I f ~ ~ A ) Std 16.1-1988, “Airframe Ball, Roller,and Needle Bearings-Metric Design” (November 17, 1988). 3.5. h e r i c a n National Standards Institute, American National Standard ( ~ S ~ / ~ ~ A - Inch Design” (December Std 16.2-1990, “Airframe Ball, Roller, and Needle Bearings 20, 1990). 3.6. American NationalStandards Institute, American National Standard ( ~ S I l ~ ~ ) Std 18.1-1982,“Needle Roller Bearings Radial Metric Design” (December 2, 1982). 3.7. American NationalStandards Institute, American National Standard( ~ S I I ~ ~ A ) 14, 1982). Std 18.2-1982,“Needle Roller Bearings Radial Inch Design” (May 3.8. American NationalStandards Institute, American National Standard ~ S I / ~ ~ A ) 19, 1987). Std 19.1-1987, “Tapered Roller Bearings Radial Metric Design” (October 3.9. American NationalStandards Institute, American National Standard ( ~ S I f ~ ~ A ) Std 19.2-1994,“Tapered Roller Bearings-Radial Inch Design” (May 12, 1994). 3.10. American NationalStandards Institute, American National Standard ( ~ S I f ~ ~ A ) Std 20-1987, “Radial Bearingsof Ball, Cylindrical Roller, and Spherical Roller Types, Metric Design” (October28, 1987). s , Departure Division, GeneralMo3.11. A. Jones, Analysis of Stresses and ~ e ~ e c t i o nNew tors Corp., Bristol, Conn,, 161-170 (1946).

ST Synbol

a

S Description Distance to load point fromright-hand bearing center Gear train value Bearing radial load Gravitational constant Thread pitch of worm at thepitch radius Power Distance between bearing centers Length of connecting rod Speed Number of teeth on gear Applied radial direction load Force applied on piston pin Inertial force due to reciprocating masses Centrifugal force acting on connecting rod bearing due to rotating masses

Units mm (in.)

N (W mm/sec2 (in./sec2) mm (in.) watts (HI)) mm (in.) mm (in.) rPm

133

Symbol pcc

radiusr rP

T

X

is

Y h

- #+

Units

Centrifugal force acting on crankshaft bearing due to rotating masses Crank Gear pitch radius Applied moment load Applied per load unit length Weight of reciprocating parts Weight of connecting rod including bearing assemblies Weight of reciprocating portion of connecting rod Weight of rotating portion of connecting rod Weight of crank pin and crank webs with balance weights along Distance shaft Number of threads on worm, teeth on worm wheel Bevel gear cone angle Lead angle of worm at the pitch radius Gear pressure angle Gear helix angle

N (lb) mm (in.) mm (in.) N mm (lb in.) N/mm (lblin.) N (lb) N (lb) N (1b)

N (lb) N (lb) mm (in.)

O,

O, O, O,

rad rad rad rad

SUBSCRIPTS Refers to bearing location Refers to driving or driven gear Refers to axial direction Refers to radial direction Refers to tangential direction k

SUPERSCRIPT Refers to location of applied load or moment

The loading a rolling bearing supports is usually transmitted to the bearing through the shafton which the bearing is mounted. Sometimes,however, the loading is transmitted through the housing that encompasses the bearing outer ring; for example, a wheel bearing. In either case, and in most applications, it is sufficient to consider the bearing as simply resisting the applied load and not as an integral partof the loaded system. This condition will be covered in this chapter together with definition of the loads transferred to the shaft-bearing system by some common power transmission components.

135

The most elementary rolling bearing-shaft assembly is that shown by Fig. 4.1 in which a concentrated load is supported between two bearings. This load may be caused by a pulley, gear, piston and crank, electric motor, and so on. Generally; the shaft is relatively rigid, and bearing misalignment due to shaft bending is negligible. Thus, the system is statically determinate; that is, the bearing reaction loads F may be determined from the equations of static equilibrium. Hence

Z F = O Fl+F2-P=0

EM=O F,Z

-

F2(Z - a ) = 0

Solving equations (4.2) and (4.4) simultaneously yields

Fl=P(l-;)

(4.5)

a F2 = P -

(4.6)

Z

For an overhanging load as shown in Fig. 4.2, equations (4.5) and (4.6) remain valid if the distances measured to the left of the left-hand bearing support are considered negative.

4.1. ~ o - ~ ~ a ~ n system. ~ - s h ~ f t

BEARING LOADS ANL) SPEEDS

Ei: FI~URE 4.2. ~ o - ~ e a ~ n g -system, s h a ~ overhung load.

Equations (4.5) and (4.6)therefore become

F,

=

P (1

T

);

If the number of loads Pk act on the shaft as shown in Fig. 4.3, the magnitudes of the bearing reactions may be obtained by the principle of superposition. For these cases k=n

(4.9) k=l k=n

(4.10) k=l

Equations (4.9) and (4.10) are valid for loads applied in the same plane. If loading is applied in difl‘erent planes, then the loads must be resolved into orthogonal components; for example,Pt and P,”(assuming the shaft axis is aligned with the x-direction). Accordingly, the bearing radial reactions .Fly, Fk,F2y,and F2z will be determined. Then

Among the most common machine elements used in combi~ationwith rolling bearings in power transmissions are involute form spur gears.

F I ~ 4.3. ~ E Two-bearing-shaft system, multiple loading.

These gears are used to transmit powerbetween parallel shafts. As shown by machine design tests (for example, Spotts and Shoup 14.11, Juvinall and Marshek [4.2], Hamrock et al. [4.3], and several others), load is transmitted normal to the flanks of the gear teeth at anangle 4 to a tangent to the gear pitch circle at thepoint of contact. This normal load P can be resolvedinto a tangential load Ptand a separating or radial load Pr.Figure 4.4 illustrates the loads transmitted by spur gears at the gear pitch radius rp.The tangential load Pt can be determined from the power relationship (4.13) H = 2vnPtrP

The separating force is calculated using equation (4.12).

.4.

Loads transmitted by a spur gear.

138

Equations (4.13) and (4.14) are also valid for herringbone gears, which also transmit power between parallel shafts. These gears are illustrated by Fig. 4.5. Loading of planet gears in planetary gear transmissions is illustrated by Fig. 4.6. It can be seen that the total radial load acting on the shaft is 2 Pt. Also, the separating loads are self-equilibrating; that is, they cancel each other. In Chapter 7 the separating loads are shown to cause bearing outer ring bending, affectingdistribution of load among rollers.

Belts and pulley arrangements also produceradial loading, as illustrated by Fig.4.7. It can beseen from Fig.4.7 that theload appliedto the pulley

-.-

F3i 4.5. Loads t r a n s ~ i t ~bydherringbone a gear. J/ is the helix angle.

4.6.

Planet gear loading.

CONCE

EL) RARIAL LOADING

~

13

I 4.7. ~ Belt-and-pulley ~ arrangement. ~ E

shaft is a function of the sum of the tensions in the belt and the drive and driven pulley diameters. See references C4.11-[4.3] for means to determine the belt tension loads. Due to belt expansion and variation in the transmitted power, the belt is generally preloaded more than is theoretically necessary. The radial load on the shaft may be approxi~ated by (4.15)

I ) = flf2,'

where I)' is the theoretical pulley load. If the belt cross section is very large, then E" = f3A

(4.16)

where A is the cross-sectional area. Values of f l , f2, and f3 are given by Table 4.1. In Table 4.1 the higher f l values are appropriate when belt speed is low; the higher f i values should be used when the center distance is short and the operating conditions are not favorable.

Figure 4.8 schematically illustrates the loads in a friction wheel drive. In thiscase a coefficient of friction must be determined that is a function

LE 4.1. Factors for Belt and Chain Drive Calculations Type of Drive

fi

fi

f3

leather Flat belt with tension pulley Flat leather belt without tension pulley Fabric belt, rubberized canvas belt, nylon belt Balata belt TI-belt Steel belt Chain

1.75-2.5

1.0-1.1

550

2.25-3.5

1.0-1.2

800

1.5-2.0

1.0-1.2 1.0-1.2 1.1-1.5

275

4.0-6.0

I

-

BEAREVG LOADS ANI) SPEEDS

FIGURE 4.8. Schematic illustration of friction wheel loading.

of the application. The combinationsof materials and operating environments are too numerous for friction coefficient values to be included herein. It is suggested that a mechanical engineering handbook; for example, Avallone and Baumeister i4.41 be consulted.

ue to an Eccentric In some applications dynamic loadingis generated due to rotation of an eccentric mass. This is illustrated by Fig. 4.9.The force P d generated by the weight W located at distance e from the axis of rotation is given by (4.18)

where g is the gravitational acceleration and w is the speed of rotation in radlsec. The force P d is constant with regard to a rotating shaft angular position, and translates into a similar condition with regard to bearing loading. This condition must be considered in the evaluation of fatigue life (see Chapter 18).

FIG

.9. Rotor with eccentric mass.

141

C Q N C E N ~ T RADIAL E~ LQADING

ue to a Crankeehanism Reciprocating mechanisms, such as pistons in internal combustion engines, air compressors, and axial pumps, may employ rollipgbearings in several different locations. Each of these bearings experiences dynamic loading associated with the reciprocating motion. Figure 4.10 is a schematic representation of a crank mechanism. In Fig. 4.10 three bearing locations are indicated: (1) crankshaft support bearings, (2) connecting rod-crank bearing, and (3) the piston pin bearing. In this mechanism, usually 30-40% of the connecting rod is considered as reciprocating; therefore

w2,= (0.3 - (4.19) 0.4)w2 W21= (0.7 - 0.6)W2

(4.20)

All forces acting from left to right in Fig. 4.10 are considered positive; they are negative when acting in the opposite direction. The reciprocating masses comprise the piston and piston pin and the reciprocating portion of the connecting rod assembly. Therefore, the inertial force due to reciprocating masses is

Pil =

-(W1

r + W2') rco2(cos a + 7 cos (2a)

g

(4.21)

At top dead-center ( a = 0), Pilreaches a maximum of (4.22)

and at bottom dead-center ( a = 180')

4.10. Schematic diagram of crank mechanism.

(4.23)

The centrifugal force acting on the large-end bearing of the connecting rod is given by (4.24)

The centrifugal force transmitted to the crankshaft bearings is

In equation (4.25) rl is distance between the crankshaft axis and the avity (CG) of weight W.; the minus sign is used when the crank-pin and the CG are on opposite sides of the cranksha~t axis. The external force Ppand the inertial force Pixhave a c o m ~ o nline of action and may be combined such that the resultant is I") = Pp 4-

.Pi,

(4.26)

In some applications only a single bearing i s used to support the shafi as a cantilever subjected to load in a radial plane. As i n ~ i c ~ t by e d Fig. 4.11, in thiscase the b~aring must also support a moment or m i s a l i ~ i n ~ mome~tload. The equations for bearing radial and m o ~ e n load t are

143 k=n

(4.27)

F = Z P k k=l k=n

M

=

2 Pkak

(4.28)

k= 1

In a power transmission system involving helical, bevel, spiral bevel, hypoid, or worm gearing, in addition to the radial loads, thrust loads are transmitted between gears. These thrust loads occur at distances from the shaft axis and thereby create concentrated moments that act on the shaft. This is illustrated schematically by Fig. 4.12. For these cases, the equations defining radial loading of the bearings are

F,=&

ZPk k=n 1=1

(

41') 9"

1T"

"

(4.29) (4.30)

-

Helical gears also are mostly used to transmit power between parallel shafts; the radial andaxial loads transmitted are illustratedby Fig. 4.13. As before for spur gears, the tangential load component Pt is determined from equation (4.13). The separating load is given by

4.13. Two-bearing-shaft system with concentrated radial and moment loads.

144

BEARING LOADS AND SPEEDS

Loads transmitted by a helical gear.

tan Cp P,. = P, cos J/

(4.31)

where Cp is the pressure angle and J/ is the helix angle. The axial load component P, is given by

Pa = Pt tan IC/

(4.32)

Helical gears can also transmit power between crossed shafts. In this case, once P, is determined using equation (4.13), equations (4.31) and (4.32) can be used to determine Pal,Prl,Pa2,and P,.. for gears 1 and 2, which have helix angles +l and +2.

Bevel gears are used to transmit power between shafts whose axes of rotation intersect. Thus, the shaft axes lie in a common plane. As with helical gears, bevel gears, radial and axial loads are transmittedbetween the mating gear teeth. Because the axial load occurs at a distance rp from the shaft axis, a concentrated moment T = rpP , is created. Figure 4.14 shows the loads transmitted by a straight bevel gear. As for spur gears, the tangential load component P, is determined from equation (4.13); however,a mean pitch radius of the teeth-for example, r,-must be used. The radius to the pitch circle of the large end of the bevel gear is called the back cone radius rb. Using the half cone angle y and the gear tooth flank length Z, , it can be determined that

r,

=

I

rb - - sin y 2

(4.33)

The s e ~ a r a t i (radial) n~ andaxial loads are given by equations (4.34) and (4.35).

45

F

I

~ 4.14. ~ ELoads transmitted by a straight bevel gear.

Pr = Pt tan # cos y

(4.34)

Pa = Pt tan # sin y

(4.35)

+

Spiral bevel gears, similar to helical gears, have a helix angle as indicated by Fig. 4.15.In thiscase the axial and radialloads trans~itted between teeth are affected by the helix angle, the direction of the helix, and thedirection of rotation of the gear. Furthermore, a distinction must be made betweendriving and driven gears. Considering the ~ r ~ v i ~ g g e ~ r and for the followinghelix and rotational directions, the gear forces transmitted are given by equations (4.36) and (4.37).

Right-hand helix (RH) Left-hand helix (LH) Clockwiserotation(CW)Counterclockwiserotation(CCW)

Pal = t' (-sin cos

+

Prl = t' (sin cos

+

+ cos y1 + tan 4 sin yl)

(4.36)

+ sin y1 + tan ct) cos yl>

where subscri~t1refers to the driving gear, and yl is the hal~-con~ angle.

14

4.1Ei.

Loads transmitted by a spiral bevel

gear.

RH-CCW

* *

LH-CW

Pal = t‘ (sin cos

+ cos y1 + tan

sin

3/11

prl = pt (- sin t,b sin y1 + tan # cos yl) cos

(4.38) (4.39)

Considering the driven gear (subscript 2):

RH-CW

Pa2 =

~

t‘

cos

LH-CCW

*

(sin tc/ cos y2 + tan

sin y2)

Pt (-sin $ sin y2 + tan # COS Pr2= cos J,

y2)

(4.40) (4.41)

CORTCE

RH-CCW

LH-CW

pt (-sin sl/ cos Pa, = cos sl/

Pr2= t‘ (sin cos sl/

y2

+ tan 4 sin y2)

+ sin y2 + tan

cos y2)

(4.42) (4.43)

Hypoid gears are used to transmit power between shafts whose ases of rotation do not intersect; that is, the shaft axes lie in diEerent planes. Because of this arrangement, illustrated by Fig.4.16, a substantial amount of sliding occurs betweencontacting gear teeth, and a coefficient of sliding friction must be defined. For hypoidgears properly lubricated with a mineral or synthetic oil, the coefficient of friction p = 0.1 is representative. Similar to spiral bevel gears, the loading of the individual gears depends on the helix direction together with the direction of rotation. The equations for d r i u i gear ~ ~ loads are given as functions of P, the resultant gear tooth load, which is defined by equation (4.44).

4.16. Hypoid gear mesh.

14

P=

cos # cos

pt t,b1

RH-CW .pal

= P(--COS

4 sin t/t1 cos y1 + sin

LH-CCW

sin yl + p

Prl = P(cos # sin 11/1 sin y1 + sin # cos yl

"CCW

(4.44)

+ p sin t,b1

- p

COS $1 COS 71)

cos

+lsin

yl)

(4.46)

LH-CW

cos CJ, sin t/tl cos yl + sin sin yl - p COS +l COS 71) Prl = P ( - cos # sin +lsin y1 + sin # cos y1 + p COS q1 sin yl>

pal=

The driven gear tooth loads are given by equations (4.49)-(4.52).

RH-CW

(4.45)

LH-CCW

(4.47) (4.4

49

RH-CCW

Pa2= P(-cos (13 sin Pr2 = P(cos (13 sin

t,h2

cos y2 + sin (13 sin y2 +

t,h2 sin

y2

EA,

cos

t,h2

cos y2) (4.51)

+ sin (13 cos y2 - p cos t,h2 sin y 2 )

(4.52)

Worm gearing, which may be regarded as a case of 90" crossed helical gears, is used to effect substantial reduction in speed within a single reduction set. Even morethan in hypoid gears, substantial sliding occurs between the worm screw and the worm wheel teeth. Figure 4.17 illustrates the loads transmitted in a worm gear drive. The lead angle h of the worm thread at the pitch radius rplof the worm is defined in terms of the threadpitch h at thatpoint and thefriction coefficientby equation (4.53). tan A

=

h 2Wpl

(4.53)

Then for the worm

I

Schematic of tooth loading in a worm gear !

drive.

BEARING LOADS AND SPEEDS

"1

=

sin (b pt1 cos (b sin A + p cos h

(4.54)

Pal

=

cos (b - p tan h Ptl p + cos (b tan h

(4.55)

For the wormwheel, the following relationships are true: Pt2 = Pal, Pr2= Prl, and Pa, = Ptl.

In some applications only the input shaft speed is given; however, performance of both input and output shaftbearings must be evaluated. In general, simple kinematic relationships relying on instant center concepts are used to determine the input-output speed relationship. Using the friction wheel shownin Fig. 4.8, at thepoint of contact, the condition of no slippage is assumed, and therefore the surface velocity u of wheel 1equals that of wheel 2. Since u = or, (4.56)

The same relationship holds for pulleys and spur and helical gears. For straight and helical bevel gears, the following equation pertains: (4.57)

where rmland rm2 are the mean pitch radii of gears 1 and 2. For hypoid gears (4.58)

For worm gearing (4.59)

where Z, is the number of threads on the worm and is, is the number of teeth on the worm wheel. To achieve substantial speed reductions, gears are frequently combined in units called gear trains; Fig. 4.18 illustrates a simple train com-

S

SPEEDS

input

output

4.18. Simplefour-gear train.

posed of four gears. The train value is defined as the ratio of the output to input speeds; that is,

It can be further demonstrated (see Ref [l]etc.) that (4.61) More generally, for a train of several gears, the train value e is the ratio of the product of the pitch radii of the driving gears to the product of the pitch radii of the driven gears. (It is noted that thenumbers of teeth can be substitute^ for the pitch radii.) Hence, the output shaftspeed can be directly determined. Planetary gear or epicyclicpower transmissions are designed to achieve substantial speed reduction in a compact space. In its simplest form, the epicyclic transmission is shown schematically by Fig. 4.19, in which R refers to the ring gear, I> to the planet gear, and S to the sun gear. The sun gear typically is connected to the input shaft, and the output shaftis connected to the arm. In general, there are three or more planets; therefore, each planet gear shaR transmits one-third or less of the input power. A single planet is shown in Fig. 4.19 for the purpose of analysis of speeds. Using Fig. 4.19, it can be seen that the speed of the

1

4.19. Schematic diagram of a simple epicyclic power transmission.

sun gear relative to the arm is nsA = ns - nA. Furthermore, the speed of the ring gear relative to the arm is nRA= nR - nA. Therefore e = nSA

= nR nS

- nA - nA

(4.62)

Now, by holding the arm stationary, allowing the ring gear to rotate as in a simple gear train, and applying equation (4.61),

(4.63) The minus sign in equation (4.63) denotes that the ring gear (output) rotates in the opposite direction with regard to the sun gear (input). Equating (4.62) and (4,63), setting nR = 0, and rearranging yields the expression for the output shaft speed. Subsequently, the planet gear shaft speed can be determined.

(4.64)

Planetary gear transmission confi~rations are too numerous and varied to provide equations to calculate speeds for each case; however, the calculational method indicated herein is universally valid and may be applied.

Sometimes the load is distributed over a portion of the shaft as in a rolling millapplication. In general this type of loading maybe illustrated by Fig. 4.20. If the loading is irregular, then it may be considered as a series of loads Pk,each acting at its individual distance ab from the lefthand bearing support. Equations (4.9) and (4.10) may then be used to evaluate reactions F , and F2. For a distributed load for which the load per unit length w 'may be represented by a continuo~sfunction; for example, w = w, sin x or w = w,(l + bx), equations (4.9) and (4.10) become (see Fig. 4.21)

F,

=

la; ); (1 -

wdx

(4.65) (4.66)

In this chapter, methods and equations have been provided to calculate bearing loads in statically determinate shaft-bearing systems. These methods are adequate for performance analysis of ball and roller bearings in most applications. Also, methods and equations have been provided to estimate the loading transmitted through the shaft to its bearing supports resulting from various common power transmission, machine elements. Many applications are, however, moresophisticated than those covered in this chapter. These may involve statically indeterminate sys-

BEARING LOADS AND SPEEDS

7 a 2 -

1

tems, covered in Chapter11,andlor increased complexity of system loading, which can only be determined by detailed evaluation of the specific application.

4.1. M. Spotts and T. Shoup, Design of Machine Elements, 7th Ed., Prentice Hall, Englewood Cliffs, N.J. (1998). ls ~ o m ~ o n e Design, nt 2nd ed., 4.2. R. Juvinall and I(. Marshek, ~ u n ~ a m e ~ tofa Machine Wiley, New York (1991). 4.3. B. Harnrock, €3. Jacobson, and S. Schrnid, ~ u n ~ a m e n t aof z sM a c ~ i n eElements, McGraw-Hill, New York (1999). 4.4. E. Avalloneand T. Baumeister, Marks’ Standard ~ a n d ~ o for o kM e c ~ a ~ i c aEngineers, Z 9th ed., ~ c G r a ~ - H i New l l , York (1987).

Symbol

Units

Description

Ball or roller diameter diameter Pitch Force Centrifugal force Gravitational (in./sec2) mm/sec2 constant Mass moment of inertia Roller length Moment Gyratory moment Mass of ball or roller Rotational speed Orbital ball or roller speed, cage speed Ball or roller speed about its own axis Ball or roller normal load Radial direction load on ball or roller

mrn (in.) mm (in.) N (lb) N Ob) kg mm2 (in. lb * sec2) mm (in.) N mm (in. lb) N mm (in. * lb) kg (lb * sec2/in.) rPm rPm rpm N (W N (lb) 155

n

15

BALL AND R O L ~ R LOADS

Symbol &a

r

u

v

w

X

x X‘

Y Y

Y‘ z 2 x‘ a

P P’ Y YB

Y 6 P

# @ urn OR

a e

Units Axial direction load on ball or roller Radius Coordinate direction distance Coordinate direction distance Coordinate direction distance Coordinate direction distance Acceleration in x-direction Coordinate direction distance Coordinate direction distance Acceleration in y-direction Coordinate direction distance Coordinate direction distance Acceleration in x-direction Coordinate direction distance Contact angle Angle between W axis and z’ axis Angle between projection of the 7 7 axis on the x’y’ plane and the x’ axis D cos aid, Roller skewing angle Roller tilting angle Angle Mass density Angle in W plane Angle in yx plane Orbital angular velocity of ball or roller Angular velocity of ball or roller about its own axis SU~SCRIP~S Refers to axial direction Refers to rotation about an eccentric EEiS

f i j m 0

r

Refers to guide flange Refers to inner raceway Refers to rolling element at location j Refers to orbital rotation Refers to outer raceway Refers t o rolling element Refers t o radial direction

N (lb) rnm (in.) mm (in.) rnrn (in.) mm (in.) mm (in.) mm/sec2 (in./sec2) mm (in.) mm (in.) mm/sec2 (in./sec2) mm (in.) rnm (in.) mm/sec2 (in./sec2) mm (in.) rad rad O,

rad rad rad rad kg/mm3 (lb sec2 in.-*) rad rad

O, O,

rad/sec rad/sec

STATIC L

O

~

~

7

G

The loads carried by ball and roller bearings are transmitted through the rolling elements from one ring to the other. The magnitude of the loading carried by the individual ball or roller depends on the internal geometry of the bearing and on the type of load impressed on it. In addition to applied loading, rollingelements are subjected to dynamic loading due to speedeffects. Bearing geometryalsoaffects the dynamic loading. The objectof this chapter is to define the rolling element loading in ball and roller bearings under varied conditions of bearing operation.

A rolling element can support a normal load along the line of contact between the rolling element and the raceway (see Fig. 5.1). If' a radial lr is applied to the ball of Fig. 5.1, then the normal load supported by the ball is

Hence a thrust load of m a ~ i t u d e QP

I

,I

Radially loaded ball.

or Qa

=

Qr tan a

(5.3)

is induced in the assembly.

.

The 209 radial ball of Example 3.1 is subjected to a thrust (axial) load of 445 N (100 lb) per ball. What is the ~ a ~ i t u d e of the induced ball radial load assuming the contact angle* is not changed by the thrust load? a

=

Ex. 3.1

7'25'

Qa = Qr tan a

Qr

=

(5.3)

3419 N (768 lb)

This result indicates the degree of thrust load amplificationat small contact angles. The 218 an~lar-contactball bearing of Example 3.3 is subjected to a thrust load of 2225 N (500 lb) per ball. What is the magnitude of the normal ball load that is induced assuming the contact angle* is not changed by the thrust load? a = 38'54'

2225 = Q sin (38'54')

Q

=

3543

N (796.2 lb)

It can be seen that a bearing with a 40' contact angle can support a thrust load better than a bearing with an 7'25' contact angle. Equations (5.2) and (5.3) are also valid for spherical roller bearings using symmetrical rollers. For a double-row spherical roller bearing under an applied radial load the inducedroller thrust loads are selfequilibrating (see Fig. 5.2).

*This assumption is not accurate in this case and is made only to illustrate a point.

STATIC LO^^^

\

/

5.2. Radially loaded symmetrical rollers.

Qia =

Qir

tan ai

For static e~uilibrium the sum of forces in anydirection is equal tozero; therefore,

or

160

BALL AND ROLLER LOADS

5.3. Radially loaded asymmet-

rical roller.

Qir tan ai + Qf sin af-

Solving equations (5.8) and (5.9) for Q, and Qfyields

Qo

= Qir

Qf

=

Qir

(sin af+ tan ai COS af) sin (a, + cxf)

(5.10)

(sin a, - tan ai cos CY,) sin (a, + af)

(5.11)

The thrust load induced by the applied radial load is (5.12)

Under an applied thrust load Qia the tain, considering static equilibrium:

Qo

Qf

= Qia

=LI

Qia

follow in^ equations of load ob-

(cos af + ctn ai sin ai) sin (a, + af)

(5.13)

(ctn aisin a, - cos a,) sin (a, + af)

(5.14)

A 90000 series steep-angle tapered roller bearing has the following dimensions: ai = 22"

a, = 29"

D

=

22.86 mm (0.9000 in.) (mean)

I

=

30.48 mm (1.2 in.)

dm = 142.24 mm (5.600 in.) (mean) af =

64" (with bearing axis)

If the most heavily loaded roller supports a thrust of 22,250 N (5000 lb), what is the magnitude of the maximum load onthe guide flange?

Qf = Qia =

(ctn aisin a, - cos a,) sin (a, + af)

22250

(ctn 22" sin 29" - cos 29") sin (29" + 64")

(5.14) =

7245 N (1628 lb)

Compare this load with the maximum normal load on the cone. Qia Qi = -

sin ai

-

22250 sin 22"

Fig. 5.3 -

59,410 N (13,350 lb)

"

olling Element The development of equations in this section is based on the motions occurring in an angular-contact ball bearing because it is the most general form of rolling bearing. Subsequently, the equations developed can be so restricted as to apply to other ball bearings and also to roller bearings. Figure 5.4 illustrates the instantaneous position of a particle of mass in a ball of an angular-contact ball bearing operating at high rotational

w X

5.4.

Instantaneous position of ball mass element dm.

speed about an axis x. To simplify the analysis the following coordinate axes systems are introduced:

A fixed set of Cartesian coordinates with the x axis coincident with the bearing rotational axis. A set of Cartesian coordinates with the x' axis parallel to the x axis of the fixed set. This set of coordinates has its origin 0' at the ball center and rotates at orbital speed about the fixed x axis at radius dm. A set of Cartesian coordinates with origin at the ball center 0' and rotating at orbital speed am.The 77 axis is collinear with the axis of rotation of the ball about its own center. TheW axis is in theplane of the U axis and x' axis; the angle between the W axis and z' axis is ,6. A set of polar coordinates rotating with the ball.

+

In addition to the foregoing coordinate systems, the following symbols are introduced:

3

D

p’ The angle betweenthe projection of the U axis on the x’y’plane and the x’ axis. The angle betweenthe z axis andx’ axis, that is, the angular position of the ball on the pitch circle.

tl,

Consider that an element of mass d m in the ball has the following instantaneous location in the system of rotating coordinates: U, r, #. Since

u=u V = r sin #

W

=

(5.15)

r cos 4

and

x’ = U cos p cos p’

-

V sin p’

-

W sin /3 cos p‘

y‘ = U cos p sin p‘

+ V cos p‘

-

W sin /3 sin p‘

z‘ =

U sin p

(5.16)

+ W cos p

and

x

=

x‘

y =

$ dm sin tl, + y‘ cos tl, + z‘ sin tl,

z =

$ dm cos tl, - y‘ sin .IC/ + z’ cos sl/

(5.17)

therefore, by substitution of equations (5.15) into (5.16) and thence into (5.17), the following expressions relating the instantaneous position of the element of mass d m to the fixed system of Cartesian coordinates can be formulated.

x

=

y

=

U cos p cos p’ - r(sin p’ sin #

2

sin tl, +

+ sin p cos p‘

cos #)

(5.18)

cos p sin p’ cos tl, + sin p sin $1

+ r (cos p sin # cos tl, + cos /3 cos # sin tl, -

sin p sin p‘ cos # cos tl,)

(5.19)

x = - d m cos tl, + U(-cos p sin p’ sin IC/ + sin p cos tl,) 2

+ r(--cos p’ sin # sin tl, + cos p cos # cos tl, + sin p sin p’ cos # cos tl,)

(5.20)

In accordance with Newton’s second law of motion, the following re-

lationships can be determined if the rolling element position angle @ is arbitrarily set equal to 0”:

The net moment about the x axis must be zero forconstant speed motion. At each ball location (@,p), % (rotational speed d # / d t of the ball about its own axis U - 0‘) and wm (orbital speed ~~/~~of the ball about the bearing axis x) are constant; therefore, at @ = 0:

(5.26)

d “Y Y = - -d t - -2%wmr cos /3 sin #

+ uL[-U cos p sin p’ + r(-cos p’ sin (b + sin p cos # sin p’)] + wkr( -cos p’ cos # + sin p sin p’ sin 4) (5.2’7) d 2z z=”- -2%wmr(cos p’ cos 41) + sin p sin p’ sin #I dt

-

w;r cos p cos Ip

(5.28)

~ubstitutionof equations (5.26) to (5.28) into equations (5.21) to (5.25) and placing the latter into integral format yields

65

(5.29) (5.30) (5.31)

+ r (cos p’ sin 6, - sin p sin p‘ cos st))] + y [ U cos p cos p’ - r(sin p’ sin st) + sin p cos p’ cos #)I )rdrdU d6,

(5.32)

Io lo2“ {a[u + (r;-U2)1/2

My, =

sin p

-p -rR

-

r cos p cos st)]

%[Ucos p cos p’ - r(sin p’ sin 6,

+ sin

cos p’ cos 6,)l)rd6, dU dr

(5.33)

In equations (5.29)-(5.33) p is the mass density of the ball material and rR is the ball radius. Performing the integrations indicated by equations (5.29)-(5.33) establishes that the net forces in thex’ and y‘ directions are zero and that

(5.34) (5.35)

(5.36) in which m is the mass of the ball and J is the mass moment of inertia. rn and J are defined as follows:

Rotation abouttheBearing (5.34) and recognizing that

m

=

J

=

Q pvD3 pvD5

(5.37) (5.38)

Axis. ~ubstitutingequation (5.37) into

BALL AND ROLLER LOADS

Wm

=

2rnm 60

(5.39)

equation (5.40) yielding ball centrifugal force is obtained:

F,

=

r 3 p D3nidm 10800g

(5.40)

For steel balls,

F,

=

2.26

X

10-11D2nidm

(5.41)

For an applied thrust load per ball Qia and a ball centrifugal load F, directed radially outward, the ball loading is as shown in Fig. 5.5. For conditions of equilibrium, assuming the bearing rings are not flexible, (5.42) (5.43)

or

r

~

~ 5.5. GBall under ~ thrust R load ~ and centrifugal load.

D

Qia - Q, sin a, = 0 Qia ctn ai + Fc - Q, cos a,

=

0

(5.44) (5.45)

In equations (5.44) and (5.45), Q, and a, are unknown; therefore, these equations must be solved simultaneously for Q, and a,. Thus, a,

=

ctn-1 (ctn

ai

+

+ (ctn

ai

+

Q, = [1

2) 2)] 2

(5.46) 112 Qia

(5.47)

Further, Qi

=

Qia ~

sin ai

(5.48)

It is apparent from equation (5.46) that because of centrifugal force Fc, a, < ai. The quantity ai is the contact angle under thrust load and it is greater than the free contact angle a'. This condition will be discussed in Chapter 7.

4. If the 218 angular-contact ball bearing of Example 5.2 has a cage speed of 5000 rpm, what outer ring contact angle will ob2225 N (500 lb)

Ex. 5.2

38'54'

Ex. 3.3

22.23 mm (0.875 in.)

Ex. 2.3

125.3 mm (4.932 in.)

Ex. 2.6 (5.41)

2.26

X

10-193n24,

2.26

X

10-11(22.23)3(5000)2 X 125.3

775.2 N (174.2 lb) ctn-1 (ctn a,

+

2)

ctn 38"54'

+2225 775.2)

(5.46) 32012'

Compare the normal ball loads at the inner and outer raceways.

BALL ANI) ROLLER LOADS

Qi = 3543 N (796.2 lb)

Ex. 5.2

2)] 2

Q0= [ I + (ctn = =

ai

+

[

1 + (ctn 38'54'

1/2

(5.47)

&ia 2

+775*2) 2225

1

112

X

2225

4176 N (938 lb)

Ball thrust bearings with nominal contact angle a = 90" operating at high speed and light load tend to permit the balls to override the land on both rings. The contact angle thus deviates from 90" in the same direction on both raceways (see Fig. 5.6). From Fig. 5.6,

& = - PC 2 cos a a = tan-1

( ~ )

(5.49)

(5.50)

Ze 5.5. A 90" nominal contact angle ball thrust bearing has a cage speed of 5000 rpm. The bearing that has a 127 mm (5 in.) pitch diameter and 12.7 mm (0.5 in.) balls, supports a thrust load of 445 N (100 lb) per ball. What contact angle obtains?

/ 5.6. Ball loading in a ball thrust bearing.

F~ = 2.26

X

10-193n3,

=

2.26

X

10-11(12.7)3(5000)2X 127

=

146.7 N (32.97 lb)

(5.41)

(5.50)

=

2 x 445 tan-1 146.7

=

80’38’

(

)

Equation (5.34) is not restrictive as to geometry and since the mass of a cylindrical (or nearly cylindrical) roller is given by

the centrifugal force for a steel roller rotating about a bearing axis is given by

Fc = 3.39

X

10-11R2Z,d,n~

(5.52)

For a tapered roller bearing, however, rollercentrifugal force alters the distribution of load between the outer raceway and central guide flange. Figure 5.7 demonstrates this condition for an applied thrust ia. For equilibrium to exist,

(5.53) (5.54) or ia

+ Qf sin af - Q, sin a, = 0 COS

af+ Fc -

Q,

COS

a, = 0

Solving equations (5.55) and (5.56) simultaneously yields

f =

ia(ctnaisin a, + cos a,) sin (a, + af)

+ Fc sin a,

(5.55) (5.56)

17

5.7. Asymmetrical roller under thrust load and centrifugal load.

If the tapered roller bearing of Example 5.3 has a cage speed of 1700 rpm, what is the guide flange load due to the most heavily loaded roller? Qia = 22,250 N (5000 lb) I) = 22.86 mm (0.9000

dm= 142.2 (5.600 mm

f =

in.)

x. 5.3

lo-11~2zt~m~~

(5.52)

3.39

10-11(22.86)2 X 30.48

=

221.9 IN (49.86 lb)

&,(ctn a, sin a, - cos a,) sin (a,+ af)

X

142.2(170~)2

+ .Fc sin a,

- 2225O(ctn 22" sin 29" - cos 29") + 221.9 sin 29" sin (29" + 64") =

Ex. 5.3

Ex. 5.3

=

=

X

in.)

30.48 mm (1.2 in.)

I,

PC= 3.39 x

Err. 5.3

7351 N (1652 lb)

What is the maximum normal load at theouter raceway?

(5.58)

171

0

-

,(ctn aisin af+ cos (xcuf) + Fc sin af sin (a, + af)

- 22250(ctn 22" sin 64" + cos 64") sin (29" + 64") =

(5.57)

+ 221.9 sin 64"

59,500 N (13,370 lb)

Care must be exercised in operating a tapered roller bearing at very high speed. At some critical speed related to the magnitude of the applied load, the force at theinner ringraceway contact approaches zero and the entire axial load is carried at the flange-roller end contact. Since this contact has only sliding motion, veryhigh friction results with a~tendant high heat generation. Most modern radial spherical roller bearings have complements of barrel-shaped rollers and relatively small contact angles; for example, (x 5 15". When the bearings are operated at high speed, roller loading is as illustrated in Fig. 5.8. Equilibri~mof forces in the radial and axial directions gives (5.58) (5.59) Solving these equations simultaneously yields

5.8. Loading of barrle-shaped (spherical)rollersubjected to applied and centrifbgal forces.

172

QO

=

Fc sin ai sin (ai- a,)

(5.60) (5.61)

Therefore, it appears that roller-raceway loading is uniquely determined by roller centrifugal loading. Clearly,in thisinstance the innerand outer raceway contact angles are functions of the bearing applied radial and thrust loading, and these must be determined from the equilibrium of loading on the bearing. To do this requires the determination of the bearing contact deformations.This will be discussedin subsequent chapters. Another way to view the operation of a loaded spherical roller operting at high speed is to resolve the centrifugal force into components collinear with, and normal to, the roller axis of rotation. Hence

Fca = Fc sin ao

(5.62)

Fcr= Fc cos ao

(5.63)

where ao is the nominal contact angle. ~quilibriumof forces acting in the radial plane of the roller gives

&,

=

Qi

+ Fc cos ao

(5.64)

and thecomponent Fc sin aocauses the inner andouter raceway contact angles to shift; slightly from ao to accommodate the roller axial loading. In general, spherical roller bearings do not operate at speeds that will cause s i ~ i f i c a nchange t in the nominal contact angle. Also, consider a double-row spherical roller bearing having barrel-shaped rollers subjected to a radial load while rotating at high speed. The speed-induced roller axial loads are self-equilibrated within the bearing; however, the outer raceways carry larger thrust co~ponents than do the inner raceways.

~ o t a t ~ about o n a n ~ c c e n t Axis. ~ i c The foregoing section dealt with rolling element centrifugal loading when the bearing rotates about its own axis. This is the usual case. In planetary gear transmissions, however, the planet gear bearings rotate about the input and/or output shaft axes as well as about their own axes. nce an additional inertial or centrifrollin lement. F i ~ r 5.9 e shows a schematic ugal force is in~uced in the am of such a system. From Fig. 5.9 it can be seen that the instantaneous radius of rotation is by the law of cosines

F I ~ 5.9. ~ E Rolling element centrifugal forces due to bearing rotation about an eccentric axis.

r

=

(ri

+ rz

- 2r,r, cos

+)1/2

(5.65)

Therefore, the corresponding centrifugal force is

+

This force F,, is maximum at = 180" and at thatangle is algebraic~lly additive to F,. At = 0 the total centrifugal force is Fc - Fee. The angle between F, and F,, as derived from the law of cosines is

+

F,, can be resolved into a radial force and a tangential force as follows:

174

BALL AND ROLLER LOADS

Fcer= Fce cos 0

(5.68)

Feet = FCesin 8

(5.69)

Hence the total instantaneous radialcentrifugal force acting on the rolling element is

Fc. = rnw;r, + rnw:(r,

- re cos $)

(5.70)

or

(5.71) in which the positive direction is that taken by the constant component Fc. For steel ball and roller elements the following equations are respectively valid:

FCr= 2.26

X

FCr= 3.39

X

+ n:) - den: cos $1 10-11D2Zt[d,(n& + n:) - den:cos $1

10-11.Z)3[d,(n;

(5.72) (5.73)

The instantaneous tangentialcomponent of eccentric centrifugal force is

Fct= rnw :re sin $

(5.74)

For steel ball and roller elements, respectively, the following equations pertain:

Fct= 2.26

X

10-11D3den: sin $

(5.75)

Fct= 3.39

X

10-11D2Ztden~ sin $

(5.76)

This tangential force alternates direction and tends to produce sliding between the rolling element and raceway. It is therefore resisted by a frictional force between the contacting surfaces. The bearing cage'also undergoes this eccentric motionand if it is supported on the rolling elements, it will impose additional load on the individual rolling elements. This cage load may be reduced by using a material of smaller mass density.

It can usually be assumed with minimal loss of calculational accuracy that pivotal motiondue to gyroscopic moment is negligible; then the an-

gle p' is zero and equation (5.36) is ofno consequence. The gyroscopic moment as defined by equation (5.35) is therefore resisted successfully by friction forcesat the bearing raceways for ball bearings and by normal forces for rollerbearings. ~ubstitutingequation (5.39) into (5.35) the following expressionis obtained for ball bearings: (5.77) since (5.78)

and 0,

=

2vnm 60

(5.79)

The gyroscopic moment forsteel ball bearings is given by

Mg = 4.47

X

10-12D5nRnm sin p

(5.80)

Figure 5.10 showsthe direction of the gyroscopic moment in a ball bearing. Accordingly, Fig. 5.11 shows the ball loading due to the action of gyroscopic moment and centrifugal force ona thrust-loaded ball bearing.

2 IGURE 6.10. Gyroscopic moment due to s i ~ u l t a ~ e o rotation us about nonparallel axes.

176

BALL AJ9D ROLLER LOADS QW

F

I

~ 5.~11. EForces acting on a high speed ball under applied thrust load.

~ 5.7. In ~ the ball ~ thrust~ bearingZ of Example e 5.5, the balls rotate at a speed of 50,000 rpm about axes perpendicular to the bearing axis, that is, /3 = 90'. What is the gyroscopic moment that must be resisted by raceway friction forces and what coefficient of friction is required t o resist gyroscopic pivotal rotation?

~

D mm = 12.7 n,,

=

Ex. 5.5

(0.5 in.)

Ex. 5.5

5000 rpm

M g = 4.47

X

10-12D5nR72, sin /3

=

4.47

X

10-12(12.7)5X 50,000

=

0.369 N m (3.265 in lb)

(5.80) X

5000

Fc = 146.7 N (32.97 lb) a = $0'38'

&=-

li'C

2 cos a -

146*7 = 450.8 N (101.3 lb) 2 cos (80'38')

X

1

Ex. 5.5 Ex. 5.5 (5.49)

177

ROLLER AXIAL LONXNG IN l&WIAL BEARINGS

(Fig. 5.11

p(450.8

+ 450.8) X - 0.369 x 2

10-3 = p =

o 0.0645

Gyroscopic moments also act on roller bearings. The rollers, however, are geometrically constrained from rotating; therefore, a gyroscopic momentof significant magnitude tends to alter thedistribution of load across the roller contour. For steel rollers the gyroscopic moment is given by

nil, = 8.37 X

10-12D4Z,nRn,sin p

(5.81)

Axial LoadingEffects in Radial Roller Bearings Axial loading of the rollers in a radial roller bearing can significantly affect bearing performance. In cylindrical and tapered roller bearings a roller asial load is reacted between the roller end and a guide flange. The combination of load and sliding in thiscontact can cause the bearing to generate excessive heat and, under certain conditions, lead to wear and smearing of the contacting surfaces. Conversely, with proper management of roller end-flange design and lubrication, bearing heat generation and axial load-carrying capacity can be optimized. Axial roller loads interact with roller radial forces and ring deflections t o determine roller tilting and skewing motionsthat influence roller-raceway contact stresses and sliding velocity distributions. Roller tilting and skewing substantially affect roller bearing heat generation, friction torque, and rolling contactfatigue life. Many modernspherical roller bearing designs do not use flanges as theprimary source of roller guidance. Such designs rely on proper management of roller tilting and skewing motions via roller-raceway traction forces to provide roller guidance while minimizing bearing heat generation and friction torque.

Many radial roller bearing designs have the ability to carry applied axial or thrust load in addition to their predominant radial loading. Cylindri-

17

BALL AND ROLmR L

cal roller bearings can carry thrust loading by virtue of flanges fixed to inner andouter rings. The angular-contact roller arrangement of tapered roller bearings reacts applied thrust load overthe combination of rollerraceway contact surfaces and roller end-flange contacts. Spherical roller bearings react applied thrust loading through their roller-raceway contacts (and flanges if asymmetric rollers are employed). Axial loading of radial roller bearings causes alteration of internal load distribution and roller response that can significantly affectbearing performance. Cylindrical roller bearings with fixed inner and outer ring flanges, as shown in Fig. 5.12a, can carry axial load through contacts between the roller end facesand theflanges. The couple produced on the roller by the axial roller end-flange forcesQajresults in a tilting 4 about the center of the roller. As the roller tilts, the roller-raceway load distribution shifts asymmetrically, as represented in Fig. 5.13. The roller-raceway load dis-

(a)

(b)

(e)

5.12. Applied thrust loading on radial roller bearings. (a)Cylindrical roller. ( b ) Tapered roller. (e) Spherical roller.

5.13. Roller tilting displacement and contact load distribution due to combined axial and radial load on a cylindrical roller bearing.

ROLLER, AXIAL L

O

~ IN ~ RADIAL G BEARINGS

79

tribution of Fig. 5.13 can be compared with the ideal load distribution in Fig. 6.23b.

In roller bearings skewing is defined as anangular rotation of the roller axis (in a plane tangent to its orbital direction) with respect to the axis of the contacting ring. The magnitude of the skewing motion may be expressed as a skewing angle 3/s. The skewingangle of a roller at a given azimuth position might be different for inner and outer ring contacts if misalignment exists between inner andouter rings. In general, the skewing angle will vary with roller azimuth position +. Skewing is caused by forces acting on the roller, which result in a moment loading about an axis oriented in the bearing radial direction are and passing through the roller at its midpoint. These skewing forces often due to asymmetrical distribution of tangential friction forcesat the roller-raceway contacts arising from asyxtmetrical normal loading and/ or sliding velocity distributions along the contacts. Normal and frictional forces acting on the roller due to contact with guide flanges (including ial loads) and cage may contribute to, or serve to limit, otions. Operating conditions associated with asymmetrical load distributions include misalignment of inner andouter rings in cylindrical rollerbearings and tapered roller bearings and applied roller axial loading in cylindrical roller bearings. In high-speed roller bearing operation the dynamic effects of roller mass unbalance and impact loading between roller and flange or cage can become significant causative factors in roller skewing. Forces that give rise to roller skewing motions may also, by virtue of their points of application or radial force components, becoupled to the roller tilting action. ~uantificationof roller skewing and coupled tilting behavior require computer p r o ~ a m sdesigned for this purpose. As an illustration of roller skewing, consider the cylindrical roller bearing roller subjected to radial and axial applied loads shown in Fig, 5.14. The appliedaxial load QG causes the roller to tilt through the angle 4.The tilting motion gives rise to the asymmetrical contact load distribution in Fig. 4.14a.Assuming no gross sliding or skidding in the application, the tilting also causes sliding motion to occur alongthe inner and outer raceway contacts. Roller deformations cause the sliding velocity on the uncrowned roller to be zero at only one point along each raceway contact. The tangential sliding velocity distributions for the unskewed roller are represented by Fig. 5,14b. Positive and negative values obtain correspond in^ to sliding in the z direction. Tangential friction forces in both contacts are related to the magnitude of the normal contact load and magnitude and direction of tangential sliding velocity. The asymresults, metrical tangential friction forcedistribution shown in Fig. 5 . 1 4 ~ This distribution causes a skewing moment about the y axis. The skew-

180

BALL AND ROLLER LOADS

Outer ring X

Inner ring .-

out of page

(a)

X

i

Msi

(c)

Roller loadingand tangential sliding velocity in a cylindrical roller bearing with applied axial and radial load. (a)Applied roller loads and roller load distribution. ( b ) Tangential slidingvelocity distribution. (e) Tangential traction force distribution and skewing moments. (d) Roller-flange contact forcesand resultant axial friction force (view from outer ring). 5.14.

ing moment tends to cause the roller to skew and generate an axial friction force at both contacts, the resultant of which, Q,, is shown in Fig. 5.14d. Also contributing to the roller skewing moment are the fiictional forces generated at the flange contacts. In the illustration the skewing moment and axial friction force must be reacted by the flange contacts. In principle, a skewing angle y8 may

be achieved whereby roller skewing momentand axial force are in equilibrium with the flange contact forces. Note that, in general, the locations of the roller-flange contacts, the roller-raceway normal loaddistribution, sliding velocities, and tangential and axial friction forces are all functions of the roller skewing angle. The skewing angle at which this force balance is obtained is known as an equilibrium skewing angle. As bearing operational speed increases, dynamic effects become significant. Brown et al. [4.1] investigated the problem of roller axial load due to skewing in cylindrical rollerbearings under high-speed conditions. Their analytical and experimental work highlighted the detrimental effects of unbalance forces due to roller corner radius runout. Analytical models were developed for rollerimpact loading on flange and cage. The effect of bearing design parameters was empirically correlated with observed roller end wear.

To analyze rolling bearing performance in a given application, it is usually necessary to determine the load on individual balls or rollers. How well the balls or rollers accept the applied and induced loads willdetermine bearing endurance. For example, light radial load applied to a 90” contact angle thrust bearing can cause the bearing to fail rapidly. Similarly, thrust load applied to a 0” contact angle, radial ball bearing is largely magnified according to the final contact angle that obtains. In Chapters 7 and 9 this book will concern itself with the distribution of load among balls and rollers. It will be shown that themanner in which each rollingelement accepts its load willdetermine in large measure the loading of all others. Moreover, in an~lar-contactball bearings the ball loading can affectball and cage speeds significantly. In high-speed roller bearings if roller loading is too light, rolling motion may be preempted by skidding. The material in this chapter is therefore fundamental to even the most rudimentary analysis of a rolling bearing application.

5.1. I?. Brown, D. Robinson, L. Dobek, and J. Miner, “Mainshaft High-speed, Cylindrical Roller Bearings for Gas Turbine Engines,” U.S. Navy Contract NOOOl40-76-C-0383 Interim Report (1978).

This Page Intentionally Left Blank

LS Symbol

Description Semimajor axis of the projected contact Dimensionless semimajor axis of contact ellipse Semiminor axis of the projected contact ellipse Dimensionless semiminor axis of contact ellipse Modulus of elasticity Complete elliptic integral of the second kind Elliptic integral of the second kind Complete elliptic integral of the first kind Elliptic integral of the first kind Force Shear modulus of elasticity Roller effective length Normal force between rolling element and raceway

Units mm (in.)

mm (in.) N/mm2 (psi)

N (1b) N/mm2 (psi) mm (in.)

183

184

Symbol

CONTACT STBESS AND D E F O ~ T I O N

Description Radius of curvature Principal stress Deflection in x direction Arbitrary function Deflection in y direction Arbitrary function Deflection in x direction Principal direction distance Dimensionless parameter Principal direction distance Dimensionless parameter Principal direction distance Depth to maiimum shear stressat x = 0, y = o Depth to maximum reversing shear stress y z0 , x = 0 Dimensionless parameter Shear strain Deformation Dimensionless contact deformation Linear strain zlb, roller tilting angle Angle Auxiliary angle a/b Parameter Poisson’s ratio Normal stress Shear stress Auxiliary angle Auxiliary angle Curvature difference Curvature sum S U ~ S C ~ ~ ~ S Refers to inner raceway Refers to outer raceway Refers to radial direction Refers to x direction Refers to y direction Refers to X direction Refers to yz plane

Units mm (in.) N/mm2 (psi) mm (in.) mm (in.) mm (in.) mm (in.) mm (in.) mm (in.) mm (in.) mm (in.) mm (in.) *, rad rad rad

N/mm2 (psi) N/mm2 (psi) rad rad or *

is6

THEORS OF ELASTICITS

Spbol xz

I I1

1

Description

Units

Refers to xz plane Refers to contact body I Refers to contact body I1

Loads acting between the rolling elements and raceways in rolling bearings develop only small areas of contact between the mating members. Consequently, although the elemental loading may only be moderate, stresses induced on the surfaces of the rolling elements and raceways are usually large. It is not uncommon for rolling bearings to operate continuously with normal stresses exceeding 1380 N/mm2 (200,000 psi) compression on the rolling surfaces. In some applications and during endurance testing normal stresses on rolling surfaces may exceed 3449 N/mm2 (500,000 psi) compression. Since the effective area over which load is supported rapidly increases with depth below a rolling surface, the high compressive stress occurring at the surface does not permeate the entire rolling member. Therefore, bulk failure of rolling members is generally not a significant factor in rolling bearing design; however, destruction of the rolling surfaces is. This chapter is therefore concerned only with the determination of surface stresses and stresses occurring near the surface. Contact deformations are caused by contact stresses. Because of the rigid nature of the rolling members, these deformations are generally of a low order of magnitude, for example 0.025 mm (0,001 in.) or less in steel bearings. It is the purpose of this chapter to develop relationships permitting the determination of contact stresses and deformations in rolling bearings.

The classical solution forthe local stress and deformation of two elastic bodies apparently contacting at a single point was established by Hertz i6.11 in 1881. Today, contact stresses are frequently called Hertzian or simply Hertz stresses in recognition of his accomplishment. To develop the mathematics of contact stresses, one must have a firm foundation in principles of mechanical elasticity. It is, however, not a purpose of this text to teach theory of elasticity and therefore only a rudimentary discussion of that discipline is presented herein to demonstrate the complexity of contact stress problems. In that light consider an infinitesimal cube of an isotropic homogeneous elastic material sub-

186

CONTACT STRESS AND ~ E F O R ~ T I O N

jected to the stresses shown in Fig. 6.1. Considering the stresses acting in the x direction and in the absence of body forces, static equilibrium requires that axdx dy

+ rw dx dz + T= dx dy - (rw

+~

d

y d x)d z

( +2 ) ax

-dx dz dy

- (rm

+2

d x ) dxdy

=

0

(6.1)

Therefore,

Similarly, for the y and z directions, respectively,

2 I

.""_.

/ I

Y

X

F

~ 6.i. ~ Stresses~acting E on an infinitesimal cube of material under load.

QRY OF E ~ T I C I ~

Equations (6.2)--(6.4)are the equations of equilibrium in Cartesian coordinates. Hooke’s law for an elastic material states that within the proportional limit a

E‘==.-

E

in which E is strain and E is the modulus of elasticity of the strained material. If u, v, and w are the deflections in the x,y, and x directions, then

Ey -

av aY

If instead of an elongation or compression the sides of the cube undergo relative rotation such that the sides in the deformed conditions are no longer mutually perpendicular, then the rotational strains are given as

” au

Yxy =

yxz

-

du

av

+ ajt

aw

dz + -ax

When a tensile stress a, is applied to two faces of a cube, then in addition to an extension in the x direction, a contraction is produced in the y and x directions as follows:

CONTACT STRESS AND ~ E ~ O ~ T I O N E, =

-

Ey -

a; E

Sa, -E

In equation (6.8)’ 6 is Poisson’s ratio; for steel 6 0.3. Now the total strain in each principal direction due to the action of normal stresses a,, ay,and a, is the total of the individual strains. Hence

Equations (6.9) wereobtained by the method of superposition. In accordance with Hooke’s law, it can further be demonstrated that shear stress is related to shear strain as follows:

(6.10)

in which G is the modulus of elasticity in shear andit is defined (6.11) One further defines the volume expansion of the cube as follows: E = E,

+ E y + E,

(6.12)

189

S n F A C E STRESSES AND ~ E F O R ~ T I O N S

Combining equations (6.9),(6.11), and (6.12)one obtains for normal stresses

:(-+-+(~~

ffx=2G

ffy=2G

1-25 1-25

4 4

(6.13)

Finally, a set of “compatibility” conditions canbe developed by differentiation of the strainrelationships, both linear androtational, and substituting in the equilibrium equations (6.2)-(6.4): v2u

+--1 - 12 5 a a€ = 0 x

v2v

1 a€ +--0 1-25ay

v2w

1 at: +--0 1-2582

(6.14)

in which

a2

a2

a2

ax2

ay2

az2

v2=-+-+-

(6.15)

Equations (6.14) represent a set of conditions that by using the known stresses acting on a body must be solved to determine the subsequent strains and internal stresseso f that body. See Timoshenko and Goodier C6.21 for a detailed presentation of the foregoing.

STRESSES

DEFQ

Using polar coordinatesrather than Cartesian, Boussinesq C6.31 in 1892 solved the simple radial distribution of stress within a semiinfinite solid as shown in Fig. 6.2. With the boundary condition of a surface free of shear stress, the following solution was obtained for radial stress:

CONTACT STRESS AND ~ E F O ~ T I O ~

x FIGURE 6.2. Model for Boussinesq analysis.

2 F cos

Q"--"-

r

m

e

(6.16)

It is apparent from equation (6.16) that as r approaches 0, 0;. becomes infinitely large. It is further apparent that this condition cannot exist without causing gross yielding or failure of the material at thesurface. Hertz reasoned that instead of a point or line contact, a small contact area must form, causing the load to be distributed over a surface, and thus alleviating the condition of infinite stress. In performing his analysis, he made the following assumptions: The proportional limit of the material is not exceeded, that is, all defor~ationoccurs in the elastic range. Loading is perpendicular to the surface, that is, the effect of surface shear stresses is neglected. 3. The contact area dimensions are small compared to the radii of curvature of the bodies under load. The radii of curvature of the contact areas arevery large compared to the dimensions of these areas. e

The solution of theoretical problems in elasticity is based on the assumption of a stress function or functions that singly or in combination fit the com~atibilityequations and the boundary conditions. For stress

SURFACE STR.ESSES AND ~ E F O ~ T I O N S

distribution in a semiinfinite elastic solid, Hertz introduced the assumptions:

x = bX

y = Y-

(6.17)

b

in which b is an arbitrary fixed length and hence, X , Y , and Z are dimensionless parameters. Also,

(6.18)

in which c is an arbitrary length such that deformations u / c , u / c , and w l c are dimensionless. U and V are arbitrary functions of X and Y only such that

v2u= 0 v2v= 0

(6.19)

Furthermore, b and c are related to U as follows:

bE

-= C

a2U

-2 aZ2

(6.20)

Theforegoing assumptions, which are partly intuitive and partly based on experience, when combined with elasticity relationships (6.7), (6.10), (6.12), (6.13),and (6.14) yield the following expressions:

CONTACT STRESS AND D E F O ~ ~ O ~

z""a2v

-V = X

".=z"" a2V 0-0

2 -av

a2U

axz ax2

a 0

a2U

az

aV

2-

az

aY2aY2

(6.21)

in which

E'rom the preceding formulas, the stresses and deformations maybe determined for a semiinfinite body limited by the xy plane on which rm = ryZ= 0 and a; is finite on the surface, that is, at X = 0. Hertz's last assumption was that the shape of the deformed surface was that of an ellipsoid of revolution. The function V was expressed as follows:

in which so is the largest positive root of the equation

X2 IC2

+ s;

+ - +Y- 2= 1+

s;

z2 5;

(6.23)

and K

=

alb

(6.24)

Here, a and b are the semimajor and semiminor axes of the projected

193

S W A C E S’IYWSSES AND ~ E ~ O ~ T I O N S

elliptical area of contact. For an elliptical contact area, the stressat the geometrical center is 3Q 2vab

(6.25)

= --

00

The arbitrary length c is defined by c = - 3& 4vGa For the special case K

= m,

(6.26)

then 0-0

2Q vb

(6.27)

= --

c = -Q

(6.28)

TG

Since the contact surface is assumed to be relatively small compared to the dimensions of the bodies, the distance between the bodies may be expressed as z = - x2 +-

Y2

(6.29)

2ry 2r,

in which rx and ry are the principal radii of curvature. Introducing the auxiliary quantity F ( p ) as determined by equation (2.26),this is found to be a function of the elliptical parameters a and b as follows: (K2

F(p) =

+ l)G

(IC2

-

23

(6.30)

- l)G

in which 3 and G are thecomplete elliptic integrals of the first and second kind, respectively. (6.31) (6.32) By assuming values of the elliptical eccentricity parameter

K,

it is pos-

194

sible to calculate corresponding values of F(p) and thus create a table of K VS

F(p).

Brewe and Hamrock [6.4], using a least squares method of linear regression, obtained simplified approximations for K, 3,and 6. These equations are:

i~) 0.636

K

= 1.0339

(6.33) (6.34)

-

1.5277 + 0.6023 In

(2)

(6.35)

For 1 5 K 5 10, errors in calculation of K are less than 3%, errors on 6 are essentially nil except at K = I and vicinity where they are less than 2%,and errors on tF are essentially nil except at K = I and vicinity?where are defined they are less than 2.6%. The directional equivalent radii 9, by

K1 = PXI + PXII

(6.36)

where subscript x refers to the direction of the major axis of the contact and y refers to the minor axis direction. Recall that F(p) is a function of curvature of contacting bodies.

It was further determined that (6.38) (6.39) (6.40) (6.41)

STJRFACE STRESSES AND DEFO

(6.42) =

2.79 X 10-46*&2/3Ep1/3(for steel bodies)

(6.43)

in which S is the relative approach of remote points in the contacting bodies and

(6.44) (6.45) (6.46)

Values of the dimensionless quantities a*, b*, and 15" as functions of F ( p ) are given in Table 6.1. The values of Table 6.1 are also plotted in Figs. 6.3-6.5. For an elliptical contact area, themaximum compressivestress occurs at the geometrical center. Themagnitude of this stress is

pmax

-

3Q 2~ab

(6.47)

The normal stress at other points within the contact area is given by equation (6.48) in accordance with Fig. 6.6:

(6.48)

The foregoing equations (6.30)-(6.43) of surface stress and deformation apply to point contact. ~ x u ~ 6~.1.Z For e the 218 angular-contact ball bearing of Example

5.2 determine the maximum normal contact stresses and contact deformations at the inner and outer raceways.

196

CONTACT STRESS AND ~ E F O ~ T I O ~

LE 6.1. Dimensionless Contact Parameters 0 0.1075 0.3204 0.4795 0.5916

1 1.0760 1.2623 1.4556 1.6440

1 0.9318 0.8114 0.7278 0.6687

1 0.9974 0.9761 0.9429 0.9077

0.6716 0.7332 0.7948 0.83495 0,87366

1,8258 2.011 2.265 2.494 2.800

0.6245 0.5881 0.5480 0.5186 0.4863

0.8733 0.8394 0.7961 0.7602 0.7169

0.90999 0.93657 0.95738

3.233 3.738 4.395

0.4499 0.4166 0.3830

0.6636 0.6112 0.5551

0.97290

5.267

0.3490

0.4960

0.983797

6.448

0.3150

0.4352

0.990902 0.995112 0.997300 0.9981847

8.062 10.222 12.789 14.839

0.2814 0.2497 0.2232 0.2072

0.3745 0.3176 0.2705 0.2427

0.9989156 0.9994785 0.9998527 1

17.974 23.55 37.38

0.18822 0.16442 0.13050 0

0.2106 0.17167 0.11995 0

co

Q = 3543 N (796.2 lb)

Ex. 5.2

a = 38" 54'

Ex. 3.3

I) = 22.23 mm (0.875 in.)

Ex. 2.3

dm = 125.3 mm (4.932 in.)

Ex. 2.6 (2.27)

-

22.23 cos (38" 54') = 0.1381 125.3

Since this value is only slightly larger than the values of y obtained

197

660

860

L 60

zoo WO 900 800

01'0

ZI'O $1'0

91'0

81'0

ozo ZZO

PZO

920 820

200

CONTACT STRESS AND ~ E F O ~ T I O N

Y

x

F

I

~ 6.~ 6. E Ellipsoidal surface compressive stress distribution of point contact.

in Example 2.6, it is sufficient for the purpose of this example to use , and F(p), obtained in that example, Hence, values of 2pi, F ( P ) ~2po, Xpi = 0.108 mm-l (2.747 in?)

F ( P )= ~ 0.9260 Xpo = 0.0832 mm-l (2.114 in.”’) F(p), = 0.9038

From Fig. 6.4,

a? = 3.50, b j

=

0.430, 6;

=

0.630

( ~ ) 113

ai = 0.0236aj =

0.0236

=

0.0236bj

=

0.0236 “ X 0.430

=

0.324 mm (0.01277 in.)

X

3.50

(6.39) =

( ~ ) (

2.64 mm (0.1040

in.)

113

bi

X

-

(6.41)

201

SURFACE STRESSES AND D E F O ~ ~ O N S

c i r n a x

=

3Q 27ra,bi

-

3 x 3543 2 7 ~X 2.64 X 0.324

=

1976 N/mm2 (286,400 psi) (compression)

(6.47)

Si = 2.79

X

10-46fQ2f3Zp;’3

=

2.79

X

lom4X

=

0.0195 mm (0.000766 in.)

0.630 X (3543)2’3(0.108)1f3

(6.43)

At the outer raceway, from Fig. 6.4, a$

=

3.10, bz

=

0.455, 6: = 0.672

( ~ ) 113

a, = 0.0236aZ =

0.0236

X

3.10

(6.39) =

X

2.56 mm (0.1007 in.)

( ~ ) (~~~~2)1f3 1/3

bo = 0.0236bz =

0.0236

=

0.3754 mm (0.01478 in.)

X

cornax

0.455

(6.41)

X

3&

=

27raob,

-

3 x 3543 2 7 ~X 2.56 X 0.3754

=

1762 N/mm2 (255,500 psi) (~ompr~ssion)

(6.47)

6, = 2.79 X 10-46$Q2f3Zp~/3 =

2.79

=

0.01902 mm (0.000749 in.)

X

X

0.672

X

(3543)2’3(0.0832)1’3 (6.43)

Note that cbaxis greater than cornax. This is true for most ball and roller bearings in static loading.

202

CO~ACT STRESS AND DEFO

For ideal line contact to exist, the length of body I must equal that of body 11. Then K approaches infinity and the stress distrib~tion in the contact area degenerates to a sernicylindrical form as shoyn by Fig. 6.7. For this condition,

(6.49) (6.50) (6.51)

For steel roller bearings the semiwidth of the contact surface may be appro~imatedby

b = 3.35

X

(

~

~

2

(6.52)

The contact deformation for a line contact condition was determined by Lundberg and Sjovall E6.51 to be

i

6.7. Semic~lind~ical surface compressive stress distribution o f ideal line con-

tact.

SmFACE STRESSES ANI) D E F O ~ T I O ~ S

03

(6.53)

Equation (6.53) pertains to an ideal line contact. In practice, rollers are crowned as illustratedby Fig. 6.26b,c, and d. Based on laboratory testing of crowned rollers loaded against raceways, Palmgren I6.61 developed equation (6.54) for contact deformation.

In addition to Hertz [6.l] and Lundberg and Sjovall I6.51, Thomas and Hoersch E6.71 analyzed stresses and deformations associated with concentrated contacts. These references provide more complete information on the solution of the elasticity problems associated with concent~ated contacts. 2. Estimate the maximum normal contact stress anddeformation at theinner raceway of the 90000 series tapered roller bearing of Example 5.3:

Q,= 59,410 N (13,350 lb)

Ex. 5.3

CY; =

22"

Ex. 5.3

L) =

22.86 mm (0.9000 in.)

Ex. 5.3

1 = 30.48 mm (1.2 in.)

Ex. 5.3

dm = 142.2 mm (5.600 in.)

Ex. 5.3

L)

cos CY;

- 22.86 -

(2.27)

X cos(22") = 0.1490 142.2

(2.37)

=

Assuming that neither rollers nor raceways are crowned, that is, R ri = and ideal line contactoccurs,

204

CONTACT SmESS AND D E F O ~ T I O N

xpi =

bi

1 D(1 -

ri)

-

2 = 0.1028 mm-’ (2.611 in.-’) 22.86(1 - 0.1490)

=

3.35

X

10-3

=

3.35

X

10-3

=

0.461 mm (0.01815 in.)

T

(6.52)

( ~ ) 1 ’ 2

59410 30.48 X 0.1028 (6.49)

2 X 59410 = 2692 N/mm2 (390,200 psi) X 30.48 X 0.461

This roller is very heavily loaded and probably requires crowning to avoid edge loading (see Figs. 6.24 and 6.25):

(6.54) ==

3.84

X

X

(59410)0*9= 0.0491mrn(0.001932 (30.48)0*s

in.)

Hertz’s analysis applied onlyto surface stresses caused by a concentrated force applied perpendicular to the surface. Experimental evidence indicates that failure of rolling bearings in surface fatigue caused by this load emanates from points below the stressed surface. Therefore, it is of interest to determine the magnitude of the subsurface stresses. Since the fatigue failure of the surfaces in contact is a statistical phenomenon dependent on the volume of material stressed (see Chapter l8>, thedepths at which significantstresses occur below the sudace arealso of interest. Again, considering only stresses caused by a concentrated force normal to the surface, Jones [6.8] after Thomas and Hoersch E6.71 gives the following equations by which to calculate the principal stresses S,, Sy’ and 8, occurring alongthe i! axis at any depth below the contact surface.

205

SUBSURPACE STRESSES

Since surface stress is maximum at the Z axis7 therefore the principal stresses must attain maximum values there (see Fig. 6.8):

s, = -*A(;

(6.55)

.)

-

in which (6.56)

y

=

(

~

)

1

'

(6.57)

2

2 p-

(6.58)

b

a;= -1 + v + &F(c$) =

lo4[

1

-

(1 -

&(+)I

-

5)

(6.62) -112

sin261

d+

(6.63) (6.64)

The principal stresses indicated by the foregoing equations are graphically illustrated by Figs. 6.9-6.11. Since each of the maximurn principal stresses can be determined, it is further possible to evaluate the maximum shear stress on the z axis below the contact surface.By Mohr's circle (see reference[6.21),the maximum shear stress is found to be

As shownbyFig.

6.12 the maximum shear stress occurs at various

CONTACT STR,ESSAND ~ E F O ~ T I O N

St stresses occurring on element on Z axis below contact surface.

0

0.2 t

_.

b

0.4

0.6 0.8 1.o 0

0

-b 2

0.2

0.4

0.6 0.8 1.o

-ab 6.10. S,/am,,vs bla and zlb.

6.11. s,/am,, vs bla and zlb.

0.

0.7

0.6

0.4

0.3

0.2

0.4

6.12.

T

0.6

4 ~

~

~and ~

1.0

0.8

z,lb ~ /vs ubla. -

~

~

~

depths z, below the surface, being at 0.467b for simple point contactand b for line contact. uring the passage of a loaded rolling element over a point on the raceway surface, the maximum shear stresson the z axis varies between 0 and T ~ If the ~ ~ element . rolls in the direction of the y axis, then the shear stressesoccurring in the yz plane below the contact surface assume values from negative to positive for values of y less than and greater than zero, respectively. Thus, the maximum variation of shear stress in the y z plane at any point for a given depth is 2trYz. Lundberg and Palmgren E6.91 show that cos2 4 sin (I, sin 6 + b2 cos2 (;b

a2 tan2 6

wherein

~6.66)

y

=

(b2 + a2tan2

x

=

a tan 6 cos (I,

sin ct,

(6.67) (6.68)

Here, 6 and ct, are auxiliary angles such that

of the shear stress. Further, 6 and ct, are

which defines the amplitude related as follows:

tan2+ = t tan2 6

=

t-1

(6.69)

in which t is an auxiliary parameter such that b a

- = [(t2-

l)@t -

1)]1/2

(6.70)

Solving equations (6.66)-(6.70) simultaneously, it is shown in reference [5.8] that (6.71) and 1 =

(t + l)(Zt - 1)1/2

(6.72)

Figure 6.13 shows the resulting dist~butionof shear stress at depth x. in the direction of rolling for b l a = 0, that is, a line contact. Figure 6.14 shows the shear stress amplitude of equation (6.71) as a function of bla. Also shown is the depth below the surface at which this shear stress occurs. Since the shear stress amplitud 6.14 is greater than that of Fig. 6.12, Lundberg and sumed this shear stress,called the masimum orthogonal shear stress,to be significant in causing fatigue failure of the surfaces in rolling contact, As can be seen from Fig. 6.14, fora typical rolling bearing point contact of b l a = 0.1, the dep below the surface at which. this stress occurs is approximately 0.49b.oreover, as seen by Fig.6.13, this stress occurs

at any instant under the extremities of the contact ellipse with regard to the direction of motion, that is, at y = ir 0.9b. etallurgical research [6.10] based on plastic alterations detected in sub-surf'ace material by transmission electron microscopic investigation gives indications that thesubsurface depth at which significantamounts of material alteration occur is approximately 0.75b. Assuming such plastic alteration is the forerunner of material failure, then it would appear that the m ~ i m u mshear stress of Fig. 6.12 may be worthy of consideration as thesignificant stress causing failure. Figures 6.15 and 6.16 from reference [6.10] are photomicro~aphsshowing the subsurface changes caused by constant rolling on the surface. Many researchers consider the von ~ises-Hencky distortion energy theory [6.11] and thescalar von Mises stress a better criterion for rolling contact failure failure. The latter stress is given by

S

CE S T ~ S S E S

6.14.

~

T

~

/ and c Fz,/b ~ ~ vs

b / a (concentrated normal load).

(6.73)

As compared to the m ~ m u m orthogonal shear stress ro which occurs at depth x. appro~imately equal to 0.5b, at y approximately equal to t- 0.9b in the rolling direction, crw,max occurs at x between 0.7b and 0.8b and at y = 0. Octahedral shear stress,a vector quantity favored by some researchers, is directly proportional to aVM.

(6.74)

Figure 6.17 compares the magnitudes of ro,masimum shear stress, and roctvs depth.

.

Subsudace metallurgical structure (1300 times ma~ificationafter picral etch) showing change due to repeated rollingunder load. (a)Normal structure; ( b ) stresscycled structure-white deformation bands and lenticular carbide formations are in evidence.

6.16. Subsurface structure (300 times magnification after picral etch) showing orientation of carbides to direction of rolling. Carbidesare thought to be weakness locations at which fatigue failure is initiated.

.

Determine the amplitude of the maximum orthogonal shear stress at the inner and outer raceways of the 218 ,angularcontact ball bearing of Example 6.1. Estimate the depths below the rolling surfaces at which these stresses occurs. ai= 2.64 mm (0.1040 in.)

x. 6.1

b. = 0.324 mm (0.01277 in.)

x. 6.1

-

0.1227

"

2.64

CONTACT STRESS AND D E ~ O R ~ T I O N T~ T

- orthogonal shear stress ______________ ~ maximum ~ shear stress

__________

Tact - octahedral stress shear

.rkrrnax

0.6

0.4

0.2 1_1

0

;

zlb

I

I_

0

1.o

0.5

1.5

2.0

2.5

E 6.17. Comparison of shear stresses at depths beneath the contact surface (x = y = 0).

From Fig. 6.14, 0.498, zgi bi

=

0.493

1976 N/mm2 (286,400 psi) 0.498

X

1976

2

=

491.9 N/mm2 (71,310 psi)

2.558 mm (0.1007 in.)

Ex. 6.1

0.375 mm (0.01478 in.)

Ex. 6.1

0.375 2.558

Ex. 6.1

-

0.1468

=

0.497,

"

From Fig. 6.14,

(fOlllSiX

zoo -=

b0

0.491

aomax = 1762 N/mm2 (255,500 psi)

Ex. 6.1

EFFECT OF S ~ A C SHEAR E STRESS

700

=

0'497 2x 1762 = 438 N/mm2 (63,490 psi)

zoi = 0.493

X

0.324

=

0.160 mm (0.00630 in.)

zoo = 0.491 X 0.375 = 0.184 mm (0.00726 in.) For case-hardening bearings the value of zoi and zoo can be used to estimate therequired case depth. Note that themaximum shear stress at the center of contact occursat zli = 0.763, and zlo = 0.7553, for the inner and outer raceways, respectively (see Fig. 6.12). Hence zli = 0.246 mm (0.00867 in.) and zlo = 0.281 mm (0.01108 in.). It is more conservative to base case depth on these values. Case depth should exceed zo or z1by at least a factor of three.

In the determination of contact deformation vs load only the concentrated load applied normal to the surface need be considered for most applications. Moreover, in most rolling bearing applications, lubrication is at least adequate, and the sliding friction between rolling elements and raceways is negligible (see Chapter 14). This means that the shear stresses acting on the rolling elements and raceway surfaces in contact, that is, the elliptical areas of contact, are negligible compared to normal stresses. For the determination of bearing endurance with regard to fatigue of the contacting rolling surfaces, the surface shear stress cannot be neglected and in many cases is the most significant factor in determining endurance of a rolling bearing in a given application. Methodsof calculation of the surface shear stresses (traction stresses) will be discussed in Chapter 14. The means for determining the effect on the subsurface stresses of the combination of normal and tangential (traction) stresses applied at the surface are extremely complexrequiring the use of digital computation. Among others, Zwirlein and Schlicht [6.10] have calculated subsurface stress fieldsbasedupon assumed ratios of surface shear stress to applied normal stress. Reference E6.101 assumes that the von Mises stress is most significant with regard to fatigue failure and gives illustrations of this stress inFig. 6.18. Figure 6.19 also from reference E6.101 shows the depth at which the various stresses occur. Figure 6.19 shows that as the ratio of surface shear to normal stress increases, the maximum von Mises stress moves closer to the surface. At a ratio of ~ C J =. 0.3, the maximum von Mises stress occurs at thesurface. Various other investigators have found that if a shear stress is applied at the contact surface in addition to the normal stress, the maximum shear stresstends to increase and it is located

x/b-

-2.5 -2.0 -1.5

-1

.O -0.5

0

0.5

1.0

1.5

2.0

2

0.5 1 .o z lb

JI

X

2.0

z

p=0

2.5

3.0

-2.5

-2.0 -1.5

-1.0

-0.5

0

0.5

1.0

1.5

2.0

2.5

alb X

6.18. Lines of equal von Misea stress/normal applied stress for various surface shear stresses dnormal applied stress a.

E F F E C ~OF S

~

~ SC

E STRESS

6.19. Material stressing ( c q , J o - ) vs depth for different amountsof surface shear stress (do).

closer to the surface (see references [6.11-6.151. References [6.16-6.181 give indications of the effect of higher order surfaces on the contact stress solution. The references cited above are intended not to be extensive, but to give only a representation of the field of knowledge. Theforegoingdiscussion pertained to the subsurface stress field caused by a concentrated normal load appliedin combination with a uniform surface shear stress. The ratio of surface shear stress to normal stress is also called the coefficient of friction (see Chapter 14). Because of in~nitesimallysmall irregularities in the basic surface geometries of the rolling contact bodies,neither uniform normal stress fields as shown by Figs. 6.6 and 6.7 nor a uniform shear stress field are likely to occur in practice. Sayles et al. [6.19] use the model shown by Fig. 6.20 in developing an “elastic conformity factor.” Kalker [6.20] developed a mathematical model to calculate the subsurface stress distribution associated with an arbitrary distribution of shear and normal stresses over a surface in concentrated contact. Ahmadi et al. [6.21] developed a “patch” method that can be applied to determine the subsurface stresses for any concentrated contact surface subjected to arbitrarily distributed shear stresses. Using superposition, this method combined with that of Thomas and Hoersch [6.7], for example, for Hertzian surface loading, can be applied to determine the subsurface stress distributions occurring in rolling element-raceway contacts. Harris andYu E6.221, applying this method of analysis, determined that the range of maximum orthogonal shear stress, i.e., 2 ~ is~not , altered by the addition of surface shear stresses to the Hertzian stresses. Fig. 6.21 illustrates this condition.

C

I area = 71 ab I

O

~ STRESS A ~ AND ~ DEFO

I 1

s

( b)

6.20. Models for less-than-idealelastic conformity. (a)Hertzian contact model ( b )Elastic conformity envisagedwith real used in developing elastic conformity parameter. the figure roughness would bepreferentialto certain asperity wavelengths. For convenience of similar moddmws only one compliant rollingelement, whereas in practice materials if ulus were employed the deformation would be shared.

f

(a) y = -0.9b

location x

(b) y = +0.9b

6.21. Orthogonal shear stress T ~ ~ / (abscissa) c F ~ ~ vs ~ depth z / b at contact area = 0 for friction coefficients f = 0, 0.1, 0.2.

Since the Lundberg- lmg+ren fatigue life theory [6.9] is based on maximum orthogonalshear stress as the fatigue failure-initiating stress, the adequacy of using that method to predict rolling b durance is called into question. Conversely, for simple directly under i.e., f = 0, maximum octahedral shear stress the center of the contact. Fig. 6.22 further shows that the magnitude of T ~ and the ~ depth ~ at , which ~ it ~occurs ~ is substantially influenced by surface shear stress. The question of which stress should be used for fatigue failure life prediction will be revisited in Chapters 18 and 23.

occurs

Basically, two hypothetical types of contact can be defined under conditions of zero load. These are a

Point contact, that is, two surfaces touch at a single point.

. Line contact, that is, two surfaces touch along a straight or curved line of zero width.

Obviously, after load is applied to the contacting bodies the point expands to an ellipse and the line to a rectangle in ideal line contact, that is, the bodies have equal length. Figure 6.23 illustrates thesurface compressive stress distribution which occurs in each case. When a roller of finite length contacts a raceway of greater length, the axial stress distribution along the roller is altered from that of Fig. 6.23. Since the material in the raceway is in tension at theroller ends because of depression of the raceway outside of the roller ends, the roller end compressive stress tendsto be higher than that in the center of contact. Figure 6.24 demonstrates this condition of edge loading.

(a) Hertz loading only 6.22. Orthogonal shear stress

xla.

(b) Hertz plus f = 0.2 shear stress loading

la,, (y-direction)vs depth z / b and location

C O ~ ~ STRESS C T AND D E F O R ~ T I O N

8 2

X

2 ” -

Y

x

3. Surface compressive stress distribution. (a)Point contact; ( b ) ideal line contact.

To counteract this condition, cylindricalrollers (or the raceways) may becrowned as shown by Fig. 1.38. The stress distribution is thereby made more uniform depending upon applied load. If the applied load is increased si~ificantly,edge loading will occur once again. e d contact Lundberg et al. [6.9] have defined a condition of ~ o d i ~ line for roller-racewaycontact. Thus, when the major axis (2a)of the contact ellipse is greater than the effective roller length l but less than 1.51, ~ o d i line ~ econtact ~ is said to exist. If 2a c I , then point contact exists; if 2a > 1.51, then line contact exists with attendant edge Zoa~ing.This condition may be ascertained appro~imatelyby the methods presented in the section“Surface Stresses and Deformations,” using the roller crown radius for 22 in equations (2.37)-(2.40).

E OF C O ~ A C T

t ' l Tension

Tension (a)

Actual area 2b of contact\

Apparent area f of contact

Y

X

4 (c)

6.24. Line contact. (a)Roller contacting a surface of infinite length; ( b ) rollerraceway compressive stress distribution; (c) contact ellipse.

The analysis of contact stress and deformation presented in this section is based on the existence of an elliptical area of contact, except for the ideal roller under load, which has a rectangular contact. Since it is desirable to preclude edge Zoading and attendant high stress concentrations, roller bearing applications should be examined carefully according to the ~ o d i ~ Line e d contact criterion. Where that criterion is exceeded, redesign of roller and/or raceway curvatures may be necessitated. Rigorous mathematical/numerical methods have beendeveloped to calculate the distribution and magnitude of surfaces stresses in any "line" contact situation, that is, including the eEects of crowning of rollers, raceways, and combinations thereof'(see references L6.231 and [6.f214]. ~dditionall~, finite element methods (FEN) have been employed r6.251

2

C O ~ ~ STRESS C T AND DEFO

to perform the sameanalysis. In all cases, a rather great amountof time on a digital computer is required to solve a single contact situation. Figure 6.25 shows the resultof an FEM analysis of a heavily loaded typical spherical roller on a raceway. Note the slight “dogbone” shape of the contact surface. Note also the slight pressure increase where the roller crown blends into the roller end geometry. The 22317 spherical roller bearing of Example 2.9 experiences a peak roller load of 2225 N (500 lb). Estimate the type of raceway-roller contact at each raceway Z = 20.71 mm (0.8154 in.)

Ex. 2.7

x p i = 0.0979 mm-’ (2.487 in.-’)

Ex. 2.9

0.9951

Ex. 2.9

=

x p , = 0.068 mm-’

(1.726 in.”’)

F(p), = 0.9929

Ex. 2.9 Ex. 2.9

From Fig. 6.5, a:

=

10.2

a:

=

8.8

( ~ ) (113

ai = 0.0236a:

2ai

=

0.0236

=

13.64 mm (0.537 in.)

X

(6.38)

10.2 ~ ~ ~ ~ 9=) 6.828 ’ 1 3mm (0.2688 in.)

Since 2ai < Z point contact occurs at the inner raceway a,

2a,

=

0.0236a:

(

=

0.0236

8.8

=

13.3 mm (0.5236 in.)

X

113

=

6.65 mm (0.2618 in.)

Since 2a, < I , point contact occurs at the outer raceway also.

(6.38)

E OF C O ~ A C T

Distribution of maximum transverse pressure

I

I

Position along roller(mm)

Contact area plan view ~ . ~ 5Heavy . edge loaded roller bearing contact (example of non-Hertzian con-

tact).

5. Estimate the type of contact that occurs at each raceway of a 22317 spherical roller bearing if the peak roller load is 22,250 N (5000 lb.). At 2225 N (500 lb.),

ai = 6.828 mm (0.2688 in.)

6.4Ex.

a,

=

6.65 mm (0.2618 in.)

Ex. 6.4

I

=

20.71 mm (0.8154 in.)

Ex. 2.7

(

ai = 6.828 ~ ~ ~ ~

) 1='14.69 3

mm (0.5785 in.)

22

2ai = 29.39 mm (1.157 in.) 1.51 = 1.5 X 20.71 = 31.06 mm (1.223 in.)

Since 1 < 2ai < 1.51, modified line contact occursat the innerraceway. 113

=

14.31 mm (0.5632 in.)

2a0 = 28.6 mm (1.126 in.) Since 1 < 2a, < 1.51, modified line contact occursat the outer raceway. The circular crown shownin Fig. 1.38aresulted from the theory of Hertz E6.11 whereas the cylindrical/crowned profileof Fig. 1.3% resulted from the work of Lundberg et al. [6.5].As illustrated inFig. 6.26, eachof these surface profiles, while minimizing edge stresses, has its drawbacks. Under light loads, a circular crowned profile does not+enjoy full use of the roller length, somewhat negating the use of rollers in lieu of balls t o carry heavier loads with longer endurance (see Chapter 18). Under heavier loads,whileedge stresses are avoidedformost applications, contact stress in the center of the contact can greatly exceed that in a straight profile contact, again resulting in substantially reduced endurance characteristics. Under light loads, the partially crowned roller of Fig. 1.38b as illustrated by Fig. 6.26~experiences less contact stress than does a fully crowned roller under the same loading, Under heavy loading the partially crowned roller alsotends to outlast the fully crowned roller because of lower stress in the center of the contact; however, unless careful attention is paid to blending of the intersections of the “fiat,’ (straight portion of the profiles) and thecrown, stress concentrations can occurat the intersections with substantial reduction in endurance (see Chapter 18). When the roller axis is tilted relative to the bearing axis, both the fully crowned and partially crowned profiles tend to generate less edge stress under a given load as compared to the straight profile. After many years of investigation and with the assistance of mathematical tools such as finite difference and finite element methods as practiced using computers, a “logarithmic”profilewasdeveloped[6.261 yielding a substantially optimized stress distribution under most conditions of loading (see Fig. 6.262d). The profile is so named because it can be expressed mathematically as a special logarithmic function. Under all loading conditions,the logarithmic profile uses more of the roller length than either the fully crowned or partially crowned roller profiles.Under misalignment, edge loadingtends to be avoided under all but exceptionally heavy loads. Under specific loading (Q/ZD) from 20 to 100 N/mm2

25

(d)

6.26. Roller-raceway contact load vs length and applied load; a comparison of straight, fully crowned, partially crowned, and logarithmic profiles (from [5.241).

(2900-14500psi),Fig.6.2'7 taken from[6.26] illustrates the contact stress distributions attendant to the various surface profiles discussed herein. Figure 6.28, also from [6.26], compares the surface and subsurface stress characteristics for the various surface profiles.

The contact stresses between flange and roller ends may be estimated from the contact stress and deformation relationships previously presented. The roller ends are usually flat with corner radii blending into

mm'

-

2

1

1

6.27. Compressive stress vs length and specific roller load (&/ID)for various roller (or raceway) profiles (from i6.261).

logadthmic

B

0.1

0.2

0.3 (A) Compression stress for different pmfiles uz (B) Maximum von Mises' stress D IC) Depth at which it acts z

~

~ 6.28.~ Comparison U of R surface ~ compressive stress a;, maximum von Mises stress

am, and

depth z to the maximum von Mises stress for various roller (or raceway) profiles (from l6.261).

C O ~ ~ A STRESS C ~ AND DEFO

the crowned portionof the roller profile.The flange may also be a portion of a flat surface. This is the usual design in cylindrical roller bearings. When it is desired to have the rollers carry thrust loads between the roller ends and the flange, sometimes the flange surface is designed as a portion of a cone. In this case, the roller corners contact the flange. The angle between the flange and a radial plane is called the layback angle. Alternatively, the roller end may be designed as a portion of a sphere that contacts the flange. Thelatter arrangement, that is a sphereend roller contacting an angled flange, is conducive to improved lubrication while sacrificing some flange-roller guidance capability. In this case, some skewing control mayhave to be provided by the cage. For the case of sphere-end rollers and angled flange geometry, the individual contact may be modeled as a sphere contacting cylinder. For the purpose of calculation the sphere radius is set equal to the roller sphere end radius, and the cylinder radius can be approximated by the radius of curvature of the conical flange at the theoretical point of contact. By knowing the elastic contact load, roller-flange material properties, and contact geometries, the contact stress and deflection can be calculated. This approach is only approximate, because the roller end and flange do not meet the Hertzian half-space assumption. Also, the radius of curvature on the conical flange is not a constant but will vary across the contact width. This method applies only to contacts that are fully confinedto the spherical roller endand conical portionof the flange. It is possible that improper geometry or excessive skewing could cause the elastic contact ellipse to be truncated by the flange edge, undercut, or roller cornerradius. Such a situation is not properly modeledby Hertz stress theory and should be avoided in design because high edge stresses and poor lubrication can result. The caseof a flat end roller and angled flange contactis less amenable to simple contact stress evaluation. The nature of the contact surface on the roller, being at or near the intersection of the corner radius and end flat, is difficult to model adequately. The notion of an “effective”roller radius based on an assumed blend radius between roller cornerand end flat is suitable for approximate calculations. A moreprecisecontact stress distribution can be obtained by using finite element stress analysis technique if necessary.

he information presented in this chapter is sufficient to make a determination of the contact stress level and elastic deformations occurring in a statically loaded rolling bearing. The model of a statically loaded bearing is somewhat distorted by surface tangential stressesinduced by

rolling and lubricant action. However, under the effects of moderate to heavy loading, the contact stresses calculated herein are sufficiently accurate for the rotating bearing as well as the static bearing. The same is true with regard to the effect of “edge stresses” on roller load distribution and hence deformation. These stresses subtend a rather small area and therefore do not influence the overall elastic load-deformation characteristic. In any event, from the Simplified analytical methods presented in thischapter, a level of loading canbe calculated against which to check other bearings at the same or different loads. The methods for calculation of elastic contact deformation are also sufficiently accurate, and these can be used to compare rolling bearing stiffness against stiffness of other bearing types.

6.1. H, Hertz, “On the Contact of Rigid Elastic Solids and on Hardness,”in ~iscellaneous Papers, MacMillan, London, 163-183 (1896). 6.2. S. Timoshenko andJ. Goodier, Theory of Elasticity, 3rd ed., McGraw-Hill, New York (1970). 6.3. J. Boussinesq, Compt. Rend., 114, 1465 (1892). 6.4. D. Brewe and B. Hamrock, “Simplified Solution for Elliptical-Contact Deformation Between Two Elastic Solids,” ME Trans., J: Lub. Tech. 101(2), 231-239 (1977). 6.5. G. Lundberg and H.Sjovall, Stress and Deformation in Elastic Contacts, Pub. 4, Institute of Theory of Elasticity and Strength of Materials, Chalmers Inst. Tech., Gothenburg (1958). 6.6. A. Palmgren, Ball and Roller Bearing Engineering, 3rd ed., Burbank, Philadelphia (1959). 6.7. H. Thomas and V. Hoersch, “Stresses Due to the Pressure of One Elastic Solid upon Another,” Univ. Illinois Bull. 212: (July 15, 1930). 6.8. A. Jones, Analysis of Stresses and Dejlections, New Departure Engineering Data, Bristol, Conn., 12-22 (1946). 6.9. A. Palmgren and G. Lundberg, “Dynamic Capacity of Rolling Bearings,”Acta Polytech. Mech. Eng. Ser. 1, R.S.A.E.E., No. 3, 7 (1947). ~ n~al~beanspruchung-Ein~uss g 6.10. 0.Zwirlein and H. Schlicht,‘ ‘ ~ e r k s t o ~ a n s t r e nbei von Reibung und Eigensp~nungen,” 2. W e r ~ s t o ~ e c11, h . 1-14 (1980). 6.11. IC. Johnson, “The Effectsof an Oscillating Tangentialforce at the Interface Between Elastic Bodies in Contact,’’ (Ph.D. Thesis, Universityof Manchester, 1954). 6.12 J. Smith and C. Liu, “Stresses Due to Tangential and Normal Loads on an Elastic Solid with Application to Some Contact Stress Problems,”ASME Paper 52-A-13 (December 1952). 6.13 E. Radzimovsky, “Stress Distribution and Strength Condition of Two Rolling Cylinders Pressed Together,”Univ. Illinois Eng.Experiment Station Bull., Series 408 (February 1953). 6.14. C. Liu, “Stress and Deformations Due to Tangential and Normal Loads on an Elastic Solid with Application to ContactStress,” (Ph.D. Thesis, Universityof Illinois, June 1950).

23

CON’IACT STRESS AND DEFO

6.15. M. Bryant and L. Keer, “Rough Contact Between Elastically and Geometrically IdenE J. Applied Mech. 49, 345-352 (June 1982). tical Curved Bodies,”A S ~ Trans., 6.16. C. Cattaneo, “ A Theory of Second Order Elastic Contact,” Uniu. Roma Rend. Mat. Appl. 6,505-512 (1947). 6.17. T. Loo, “A Second Approximation Solution on the Elastic Contact Problem,”Sei. Sin‘ ,1235-1246 (1958). ica 7 6.18. H. Deresiewicz, “A Note on Second Order Hertz Contact,” A S M E Trans., J. AppL Mech. 28,141-142 (March 1961). 6.19. R. Sayles, G. desilva, J. Leather, J. Anderson, and P. MacPherson, “Elastic Conformity in Hertzian Contacts,”Tribology Intl. 14, 315-322 (1981). 6.20 J. filker, “Numerical Calculation of the Elastic Field in a Half-space Due to an Arbitrary Load Distributed over a Bounded Region of the Surface,” SKI? Eng. and Res. Center Report NL82D002, Appendix (June 1982). 6.21 N. Ahmadi, L. Keer, T. Mura, and V. Vithoontien, “TheInterior Stress Field Caused by Tangential Loading of a Rectangular Patch on an Elastic Half Space,”ASME Paper 86-Trib-15 (October 1986). 6.22 T. Harris and W. Yu, “Lundberg-Palmgren Fatigue Theory: Considerations of Failure Stress and Stressed Volume,”A S M E Trans., J. Tribology 121,85-90 (January 1999). 6.23 I(. Kunert, “Spannungsverteilung im Halbraum bei Elliptischer Flachenpressungsverteilung uber einer Rechteckigen Drucldiache,” Forsch. Geb. Ingenieur~es27(6), 165-174 (1961). 6.24. H. Reusner, “Druckflachenbelastung und Ove~achenverschiebungin Walzkontakt von Rotationkorpern (Dissertation, Schweinfurt, Germany, 1977). 6.25. B. Fredriksson, “Three-Dimensional Roller-Raceway Contact Stress Analysis,” Advanced Engineering Corp. Report, Linkoping, Sweden (1980). 6.26. H. Reusner, “The Logarithmic Roller Profile-the Key to Superior Performance of Cylindrical and Taper Roller Bearings,”Ball Bearing J. 230, SKI? (June 1987).

Symbol

Description Distance between raceway groove curvature centers + f, - 1,total curvature Crown drop Influence coefficient Ball or roller diameter Bearing pitch diameter Eccentricity of loading Modulus of elasticity Raceway groove radius -+ D Applied load Friction force due to roller end-ring flange sliding motions Roller thrust couple moment arm Number of rows of rolling elements Ring section moment of inertia

Units mm (in.) mm (in.) mm/N (in./lb) mm (in.) mm (in.) mm (in.) MPa (psi)

N (lb) mm (in.) mm4 (in.4) 31

Symbol

Units Axial load integral Radial load integral Moment load integral Number of laminae Load-deflection factor; axial load deflection factor Roller length Distance between rows Moment Moment applied to bearing Load-deflection exponent Diametral clearance Load per unit length Ball or roller-raceway normal load Roller end-ring flange load Raceway groove curvature radius Radius to contact in tapered roller bearing Tapered roller radius to flange contact at roller large end

N/mmn (lb/in.") mm (in.) mm (in.) N * mm (lb in.) N mm (lb in.) 9

mm (in.) N/mm (lb/in.) N(W N (1b) mm (in.) mm (in.) mm (in.) mm (in.)

of locus of raceway groove mm (in.) Distance between loci of inner and outer raceway groove curvature centers Ring radial deflection Strain energy Number of rolling elements Mounted contact angle Free contact angle tan-l l / ( d m - I)) D cos a / d m eflection or contact deformation istance between inner and outer rings ontact deformation due to ideal normal gular spacing between rolling elements oad distribution factor earing misalignment angle Lamin~mposition Coefficient of sliding friction between roller end and ring flange

mm (in.) mm (in.) N mm (lb * in.) rad, rad, rad,

O O O

mm (in.) mm (in.)

mm (in.) rad, O rad, O rad, O rad, O

~E~

Symbol

Units

Description

Curvature sum skewing Roller angle Position angle Contact deformation at laminum h due to skewing roller Azimuth angle

mm-l (in.-') rad, O rad, O mm (in.) rad, O

~ ~ ~ S ~ R I P T ~ efer to axial direction Refers to inner raceway Refers to ring angular position Refers to rolling element position Refers to rolling element position Refers to line contact Refers t o rolling element position Refers to raceway Refers to moment loading Refers to direction collinearwith normal load Refers to outer raceway Refers to point contact Refers to radial direction Refers to rolling element Refers to gear separating load Refers to gear tangential load Refers to bearing row Refers to outer raceway efers to inner raceway efers to tapered roller bearing, roller endflange contact Refers to angular location

aving deter~inedin Chapter 5 how each ball or roller in a bearing carries load, it is possible to determine how the bearing load is d i s t ~ b uted among the balls or rollers. To do this it is first necessary to devel load-deflection relationships for rollingelements contacting raceways. using Chapters 2 and 6 these load-deflection relationships can be de r any type of rolling element contacting any type of rac , the material presented in this chapter is completel~depe revious chapters, and a quick review might be advanta~e

ost rolling bearing applications involve steady-state rotation of either the inner or outer raceway or both; however, the speeds of rotation are usually not so great as to cause ball or roller centrifugal forces or copic moments of magnitude large enough to affect si ibution of applied load among the rolling elements. most applications the frictional forces and moments acting on the rolling elements also do not signi~cantlyinfluence this load distribution. Consequently, in analyzing the distribution of rolling element loads, it is usually satisfactory to ignore these eEects in most applications. In this chapter the load distribution of statically loaded ball and roller bearings will be investigated,

From equation (6.38) it can be seen that for a given ball-raceway contact (point loading) 6

N

Q213

(7.1)

Inverting equation (7.1) and expressing it in equation format yields

Similarly, fora given roller-raceway contact (line contact) Q

K1;510/Q

(7.3)

In general then

in which n = 1.5 for ball bearings and n = 1.I1 for roller bearings. The total normal approach between two raceways under load separated by a rolling element is the sum of the approaches between the rolling element and each. raceway. Hence 6n =

Therefore,

and

si + 6,

(7.5)

=

Knsn

(7.7)

For steel ball-steel raceway contact,

K~ = 2.15

X

105

x

p2(6*)-3/2

Similarly, for steel roller and raceway contact,

K~ = 7.86

X

104 ~9

For a rigidly supported bearing subjected to radial load, the radial deflection at any rolling element angular position is given by 6, = 6, cos if$ - +Pd

(7.10)

in which 6, is the ring radial shift, occurring at if$ = 0 and P, is the diametral clearance. Figure 7.1 illustrates a radial bearing with clearance. Equation (7.10) may be rearranged in terms of maximum deformation as follows: 6,

in which

=

[ 2:

Smax 1 - - (1 - cos if$)]

+-z) (z) 2

(7.11)

(7.12)

From equation (7.12) the angular extent of the load zone is determined by the diametral clearance such that

h = cos-1

I

(7.13)

For zero clearance, h = 90". From equation (7.41, (7.14) Therefore, from (7.11) and (7.14),

BUTION OF

~~~

LOADENG EN S T A T I C ~ L L S O ~ BE ~

Before displacement (a)

4 pd

(7.15) For static equili~rium, the applied radial load must equal the sum of the vertical components of the rolling element loads: (7.16)

or

[

*=Lt&

Fr = Qmax

It:

*=O

1-

1 g (1 - cos

+)In

cos

+

(7.17)

Equation (7.17) can also be written in integralform: Fr =

ZQm,

X

2T

1'" [ -&

1 1 - - (1 - cos +) 236

I"

cos t,b d+

(7.18)

(7.19) in which Jr(E) =

I+"[

22T -&

1 1 - - (1 - cos +) 2E

(7.20)

The radial integralof equation (7.20) has been evaluated numerically for various values of E. This is given in Table 7.1. From equation (7.7),

Therefore,

For a given bearing with agiven clearance under agiven load, equation (7.22) may be solved by trial and error. A value of 8, is first assumed and E is calculated from equation (7.12). This yields J,(E)from Table 7.1. If equation (7.22) does not then balance, the process is repeated. ,

Line E

Load Distribution Integral J,<E)

Point Contact Contact

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

l/Z 0.1156 0,1590 0.1892 0.2117 0.2288 0.2416 0.2505

E

1/z 0.1268 0.1737 0.2055 0.2286 0,2453 0.2568 0.2636

Contact Contact

0.8 0.9 1.0 1.25 1.67 2.5 5.0 G3

0.2559 0.2576 0.2546 0.2289 0.1871 0.1339 0.0711 0

0.2658 0.2628 0.2523 0.2078 0.1589 0.1075 0.0544 0

238

D I $ T R ~ ~ I QOF N ~E~~

LQADING

IN STATICALLY LOADED ~ ~ I N G $

t

F

I

~ 7.2. ~ E J,(E)vs E for radial bearings.

Figure 7.2 gives values of Jrvs E . Figure '7.3 shows radial load distribution for various values of E . Here, E is the ratio of the projected load zone on the bearing diameter. For ball bearings under pure radial load and zero clearance Stribeck C7.11 concluded that

Qrnax

=

4.37 Fr is cos a

(7.23)

Accounting for nominaldiametral clearance in thebearing, one may use the following approximation.

(7.24)

.

The209 radial ball bearing of Example 2.1 experiences a radial load of 8900 N (2000 lb). Determine the loading at each ball location.

(a) c = 0.5, $ 1 = t90", 0 clearance

(b) 0

(e) 0.5 < c < 1, 90" <$ r

< t < 0.5, 0 < $ 1


< 180", preload

~ I ~ 7U .3. RRolling ~ element load distribution for different amounts of clearance.

Z=9

Ex. 2.1

Pd = 0.0150 mm (0.0006 in.)

Ex. 2.1

dm = 65 mm (2.559 in.)

Ex. 2.1

I>

=

fi =

x x

12.7 mm (0.5 in.) fo

=

0.52

Ex. 2.2

y = 0.1954

Ex. 2.5

pi = 0.202 mm-' (5.126 in.-')

Ex. 2.5

I"'(P)~= 0.9399 p, = 0.138 mm-' (3.500 in.?)

F(p), = 0.9120

From Fig. 6.4,

Ex. 2.1

Ex. 2.5 Ex. 2.5 Ex. 2.5

240

DISTRIBUTION OF INTERNALLOADING IN STATICALLYLOADED ~~~~S

8; = 0.602,

6:

=

0.658

=

2.15

=

1.026 X IO6N/mm1.5(2.951 X IO7 l b / i ~ ~ . l . ~ )

K~~= 2.15

X

X

IO5 X (O.202)-li2 (0.602)-3/2

105

x p,-1i2a3-3i2

(7.8)

(7.22)

€+2) 2

="I--) 2

(a)

(7.12)

0.015

Equations (a) and(b) may be solved by trial and error or iteration. 6, = 0.06041 mm (0.002379 in.) E =

0.438

(7.19)

Q,,

=

4536 N (1019 1b)

(7.15) =

a*

4536 (1.142

360" 360" = --

-0.9397

__

4oo

"------"

z

0

0.7660

1c/ - 0.142)1.5

COS

9

1

t- 40 t- 80 t- 120 t- 160

4536 (1019 lb) 2846 (638.6 lb) 6 1 (13.7 lb) 0 0

0.1737 -0.5000

le 7.2. By using Stribeck's equation (7.23), determine the load distribution for the 209 radial ball bearing of the preceding example. 4.37 F,

&Inax = ZGE -

4.37 X 8900 9 x cos (0")

=

4321 N (971.2 lb)

Corresponding to equation (7.23),

E

(7.23)

= 0.5

(7.15) =

4321 COS^.^

* %

0" (650.7 2897 t- 40" k 80" t- 120" t- 160"

4321 (971.2 lb) lb) (70.1 313 lb) 0 0

+

For radial roller bearings with zero clearance under pure radial load, it can also be determined that 4.08 F,

_.

(7.25)

- Z-ZE

Equation (7.24) is also a valid approximation for radial roller bearings with nominal clearance.

3. The 209 cylindrical rollerbearing of Example 2.7 supports a radial load of 4450 N (1000 lb). Determine the loading at each roller location and the extent of the load zone.

(0.3780mm

14

Ex. 2.7

65 mm (2.559 in.)

Ex. 2.7

0.041 mm (0.0016 in.)

Ex. 2.7

9.6

Ex. 2.7

in.)

7.86

X

104 z8/9

7.86

X

104(9.6)8'9

5.869

Kn =

X

lo5 N/mml*ll(4.799 X lo6 lb/in.'.'')

r

):(

(7.9)

1 1/1.11

7 1.11 1/1.11

+

(~)

(0.5)1*11 X 5.869 2.720

ZKn(6, 4450

=

X

X

lo5

lo5 N/mrn1.l1 (2.222 x lo6 lb/in.'.")

- +pd)""C&(

14 X 2.720

X

(7.22)

E)

IO5

(6, - 0.0205)1.11~r(~) = 0.001169

(a)

€=!(1-2)

(7.12)

2

= i ( 1 - w0.041 )=0.5-2 Eq~ations (a) and (b) may be solved by trial and error.

0.01025

4

(b)

B

E

~ ~ ~ E~

S R

43

Sr = 0.0320 mm (0.00126 in.) E

=

0.1824

Jr(0.1824) = 0.165

Fr 4450

(7.19)

= 2Qrn~~r(€)

=

14 X QrnmX 0.165

Qrna = 1926 N (432.8 lb) (7.15)

['

-2

1 (1 - cos *) 0.1824

=

1926

=

1926 (2.741 COS @ - 1.741)1.11

0" 25.71' 51.42' 77.13" 102.84' 128.55" 154.26' 180"

X

1 0.9010 0.6237 0.2227 -0.2227 -0.6237 -0.9010 -1 *I

=

cos-'

= COS-'

( ~ ) (x

0.041 o , o ~ 2 0 )=

1926 (432.8 lb) 1355 (304.7 lb) 0 0 0 0 0 0

(7.13) rfi 50'10'

By using equation (7.24) to determine the load distribution which occurs for the 209 cylindrical roller bearing of the preceding example.

rnax

5Fr -2 cos a -

x 4450 = 1589 N (357.1 lb) 14 X cos (0')

(7.24)

24

DIST~IBUTIONOF INTERlWL LOADING IN S T A T I C ~ L LOADED ~ BE~INGS

From Fig. 7.2, E

=

0.28

(7.15)

=

1589 (1.786 COS

+ - 0.786)'*''

+ 0 25.71' 5 I .42" 77.13" 102.84" 128.55" 154.26" 180"

1589 (357.1 lb) 1280 (287.6 lb) 461 (103.6 lb) 0 0 0 0 0

=

cos-1

( ~ )

(7.13)

since

E=ql-g) 2

+I =

=

(7.12)

cos-l (1 - 2 E ) cosv1(1 - 2

X

0.28)

= rir 63'54'

For lightly loaded rolling bearings, the appro~imation of equation (7.24) is not adequate to determine maximum rolling element load and it should not be used in that instance.

45

GS

ST L

Thrust ball and roller bearings subjected to a centric thrust load have the load distributed equally among the rolling elements. Hence

&=-

F a

(7.26)

is sin a

In equation (7.26), a is the contact angle that occurs in the loaded bearing. For thrust ball bearings whose contact angles are nominally less than 90", the contact angle in the loaded bearing is greater than the initial contact angle a" that occurs in the nonloaded bearings. This phenomenon is discussed in detail in the following paragraphs.

In the absence of centrifugal loading, the contact angles at inner and outer raceways are identical; however, they are greater than those in the unloaded condition. In the unloaded condition, contact angle is defined bY -?d cos ao = 1 - 2191)

(7.27)

in which -?d is the mounted diametral clearance. A thrust load Fa applied to the inner ring as shown in Fig. 7.4 causes an axial deflection 6,. This axial deflection is a component of a normal deflection along the line of contact such that from Fig. 7.4, (7.28) Since G;!

=

Kn6k5, =

(-

Kn(B1))1-5

1.5

cos ao - 1) cos a

(7.29)

Substitution of equation (7.26) into (7.29) yields 1.5 Fa

isKn(191))1.5

(7.30)

Since Kn is a function of the final contact angle a, equation (7.30) must

-Shifted of inner ring

position

Q

"

7.4.

Angular-contact ball bearing under thrust load.

be solved by trial and errorto yield an exact solution for a. Jones f7.21, however, defines an axial deflection constant K as follows:

(7.31)

+

in which y = (D cos cx)/d, and g( y ) refers to the inner raceway and g( - y) refers to the outer raceway. Jones f7.21further indicates that the sum of g( y ) and g( - y) remains virtually constant for all contact angles

+

ing dependent only on total curvature is related to Kn as follows: I

ence,

. The mial deflectionconstant

6 5 UNDER THRUST LOAD 1.5 Fa

(7.33)

Taking K from Fig. 7.5, equation (7.33) may be solvednumerically by the ~ e w t o n - R a ~ h s omethod. n The equation to be satisfied iteratively is 1.5

Fa ~ZD2K a'=a+

cos ao cos a (cos a - 1) ~

1.5

+

("----

cos CYo 1.5 tan2 a - 1) cos a

0.5

cos a* (7.34)

Equation (7.34) is satisfied when a' - a is essentially zero. The asial deflection Sa corresponding to Sn may also be determined from Fig. 7.6 as follows:

700,000

.-

500,000

300,000 ~00,000 100,000 0

F B

I = f,

~ 7.5. ~ EAsia1deflection = rlD [7.21).

+ fi - 1,f

constant K vs total curvature B for ballbearings.

aa = (SD + an) sin a

-

SD sin a"

(7.35)

~ubstituting6n from equation (7.28) yields

a,

=

SD sin ( a - a")

(7.36)

cos a

Figure 7.6 presents a series of curves for the rapid calculation of the change in contact angle ( a - a"), and axial deflection as functions of initial contact angle and t = Fa/ZD2K. Ze 7.5. The 218 angular-contact ball bearing of Example 2.3 experiences a statically applied thrust load of 17,800 N (4000 lb). Determine the contact angle, normal ball load,and axial deflection of the bearing considering a ball complement of 16.

2.3

Ex. 3

=

0.0464

a" = 40"

D From Fig. 7.5, K

=

=

22.23 mm (0.875

2.3 in.)

896.7 N/mm2 (130,000 psi)

Ex. 2.3 Ex.

1.5

(7.33) 17800 16(22.23)2X 896.7

Fa ZD2K

sin a

"

a'=a+

1.5

cos a" cos a (cos a - 1)

(2

-- 1)1.5

+ 1.5 tan2 a

(-

cos a" cos a

0.5

-

1)

cos a" (7.34)

=a+

Equation (b) is solved byassuming values of a. The iteration is continued until the absolute value of a' - a approaches zero. a is determined to be 0.7260 radians (41.6"). This result is sufficiently accurate for the illustrative purpose intended here. A similar result could have been obtained by using Fig. 7.6 at t = 0.0025 and a" = 40". (7.26) -

17800 = 1676 N (376.6 lb) 16 X sin (41.6")

( a - a") sa= BD sin cos a -

0.0464

X 22.23 sin (1.6") (0.00152 mm = 0.0386 cos (41.6")

(7.36) in,)

SingleDirectionBearings. Figure 7.7 illustrates asingle-row thrust bearing subjected to an eccentric thrust load. Taking 1(/ = 0 as the position of the maximum loaded rollingelement, then

7.7. 90"ball thrust bearing under an eccentric load.

8, =

sa + $0 dmcos tr/

(7.37)

Also, Smax=

sa + $0 dm

(7.38)

From equations (7.37) and (?.38), one may develop the familiar relationship

s,

=

[ 2:

smax1 - - (1 - cos *)

1

(7.39)

in which

.=l(l+Z) 2

(7.40)

The extent of the zone of loading is defined by (7.41)

As before,

Static equilibrium requires that

Fa =

&+sin a

(7.43)

x4

(7.44)

*=0

%, =

eFa =

&*dmsin a cos II/

*=0

Equations (7.43) and (7.44) may also be written in terms of thrust and moment integrals as follows:

in which (1 - cos +) %=

eFa = 4& m ~ d m J m (sin E)a

I'

d+

(7.46) (7.47)

in which (7.48)

2E

Table 7.2 as shown by Rumbarger L7.31 gives values of Ja(e)and J,(E) as functions of 2eIdm,.Figures 7.8 and 7.9 yield identical data in graphical format. Assuming that the contact angle remains constant a t 41.6", determine what the magnitude would be of the maximum ball load in the 218 angular-contact ball bearing of Example 7.5 if the 17,800 N (4000 lb) thrust load was applied a t a point 50.8 mm (2 in.) distant from the bearing's axis of rotation. dm= 125.3 mm (4.932 in.)

2e - 2 X 50.8 dm 125.3

"

=

From Fig. 7.8, Ja= 0.285, Jm= 0.233,

0.8110

E =

0.525

2.6 Ex.

5%

~ ~ S T R ~ B U OF T ~INTERNAL ON LOADING IN STATICALLY LOADED B E ~ ~ G $ , J,(E) and Jm(e)for

Single-Row Thrust Bearings

Contact Line Contact Point 2e

2e

0 0.1 0.2 0.3

1.0000 0.9663 0.9318 0.8964

1/ z 0.1156 0.159 0.1892

1/x 0,1196 0,1707 0.2110

1.0000 0.9613 0.9215 0.8805

1l z 0.1268 0.1737 0.2055

1/x 0.1319 0.1885 0.2334

0.4 0.5 0.6 0.7

0.8601 0.8225 0.7835 0.7427

0.2117 0.2288 0.2416 0.2505

0.2462 0,2782 0.3084 0.3374

0.8380 0.7939 0.7488 0.6999

0.2286 0.2453 0.2568 0.2636

0.2728 0.3090 0.3433 0.3766

0.8 0.9 1.o 1.25

0.6995 0.6529 0.6000 0.4338

0.2559 0.2576 0.2546 0.2289

0.3658 0.3945 0,4244 0.5044

0.6486 0.5920 0.5238 0.3598

0.2658 0.2628 0.2523 0.2078

0.4098 0.4439 0.4817 0.5775

1.67 2.5 5.0

0.3088 0.1850 0.0831 0

0.1871 0.1339 0.0711 0

0,6060 0.7240 0,8558 1.0000

0.2340 0.1372 0.0611 0

0.1589 0.1075 0.0544 0

0.6790 0.7837 0.8909 1.0000

co

Ex. 7.5

is = 16 Fa = ZQmaJa(~) sin a 17800 = 16 X Qma

X

0.285

(7.45) X

sin (41.6")

QmaX= 5878 N (1321 lb)

An identical result is obtained by using J,(E) in equation (7.47),

Figure 7.10 demonstrates a typical distribution of load in a 90" thrust bearing subjected to eccentric load.

Double Direction Bearings. The following relationships are valid for a two-row double-direction thrust bearing:

3

(7.49) (7.50)

It can also be shown that €1

+ €2 = I

(7.51)

and (7.52) Considering equation (7.4), equation (7.52) becomes

DI$TRIBUTION OF I N T E ~ ~

7.9.

L O IN~$ T I ~NTG I C ~ LLYO ~ E D B E ~ I ~ G $

J,(E),J,(E),E vs 2eld, for line-contact"thrust bearings.

55

B E ~ I N UNDER ~ S THRUSTLOAD

(7.53)

In equation (7.53), n = 1.5 for ball bearings, and n = 1.11 forroller bearings. From conditions of equilibrium one may conclude that

Fa = Fal- Fa2 = ZQmm,Jasin a

(7.54)

in which (7.55) %=

+ % = -ijZQmm,d,Jm

(7.56)

sin a

in which (7.57)

Table 7.3 below gives values of Jaand Jmas functions of 2eldm for tworow bearings. Figures 7.11 and 7.12 give the same data in graphical format.

Line Contact

ContactPoint 2e €1

€2

dJam

&ma,, Jm

dm

Qmxl

2e Ja

Jm

&ma,

0.50 0.51 0.60 0.70

0.50 0.49 25.72 0.40 2.046 0.30 1.092

0.4577 0.4476 0.3568 0.3036

0 0.0174 0.1744 0.2782

1,000 0.941 28.50 0.544 2.389 0.281 1.210

0.4906 0.4818 0.4031 0.3445

0 0.0169 0.1687 0.2847

1.000 0.955 0.640 0.394

0.80 0.90 1.0 1.25

0.20 0.10 0 0

0.800 0.671 0.600 0.434

0.2758 0.2618 0.2546 0.2289

0.3445 0.3900 0.4244 0.5044

0.125 0.037 0 0

0.823 0.634 0.524 0.360

0.3036 0.2741 0.2523 0.2078

0.3688 0.4321 0.4817 0.5775

0.218 0.089 0 0

0.309 0.185 0.083 0

0.1871 0.1339 0.0711 0

0.6060 0.7240 0.8558 1.0000

0.234 0.137 0.061 0

0.1589 0.1075 0.0544 0

0.6790 0.7837 0.8909 1.0000

~0

1.67 0 2.5 0 5.0 0 0

~0

~~~

DISTR~UTIONOF INTERNALLOADING IN STATICALLYLOADED B

~

~

0.9

0.8

0.7 r)

E 0.6

2cu

z

2 0.5 w

M

w

4

0.4

t 3

0.3

0.2 0.1

0

Jm, J,, E

0.2 0.4 0.6 0.8 I

1.2 1.4 1.6 2e/drn

1.8 2 2.2 2.4

QmmJQmmlvs & / d m for double-row point-contact thrust

~ E , ~ ,

bearings.

ER CO

If a rolling bearing without diametral clearance is subjected simultaneously to a radial load in the central plane of the rollers and a centric thrust load, then the inner and outer rings of the bearing will remain parallel and will be relatively displaced a distance aa in the axial direction and 6, in the radial direction. At any regular position II/ measured from the most heavily loaded rollingelement, the approach of the rings is 8,

=

aa sin a +

6, cos a cos ~!,t

(7.58)

Figure 7.13 illustrates this condition. At II/ = 0 maximum deflection occurs and is given by

G

S

F I G ~7 E.12. J,, J,, el, e2, Qmax2/Qmaxl vs 2e/d, for double-row line-contactthrust bearings.

cos

F I G U ~ E7.13. Rolling bearing displacements due to combined radial and axial. loading.

25

D I S T ~ ~ OF ~ I O N ~T E LOADING ~ ~ IN S T A T I C ~ YLOADED ~

Smax=

sasin a + 8, cos a

E

(7.59)

~ombiningequations (7.58) and (7.59) yields

[

s,, = srnm1 - l (1 - cos $1 2E

1

(7.60)

This expression is identical in form to equation (7.11), however,

satan a

(7.61)

8r

It s h o ~also l ~ be apparent that

[

Q,, = Qmax

1, -

- (1 - cos $)

E;

(7.62)

~

As in equation (7.4), n = 1.5 for ball bearings and n = 1.11for roller bearings. For static equilibrium to exist, the summation of rolling element forces in each direction must equal the applied load in that direction.

x

*=+h

Fr =

Q,, cos a cos II/

(7.63)

+= -44

(7.64)

in which the limiting angle is defined by

$1

=

cos-l

(

-

satan

sr

a

(7.65)

Equations (7.63) and (7.64) may be rewritten in terms of a radial integral and thrust integral, respectively.

where

~

(7.67) and (7.68) where (7.69) The integrals of equations (7.67) and (7.69) wereintroduced by Sjovall 117.41. Table 7.4 gives values of these integrals for point and line contact as functions of Fr tan a/Fa. Note that the contact angle a is assumed identical for all loaded balls or rollers. Thus the values of the integrals are approximate; however, they are sufficiently accurate for most calculations. Using these integrals,

LE 7.4. J,(E)and Ja(E ) for Single-Row Bearings

0 0.2 0.3 0.4

1 0.9318 0.8964 0.8601

1/ z 0.1590 0.1892 0.2117

1/ z 0.1707 0.2110 0.2462

1 0.9215 0.8805 0.8380

1/ z 0.1737 0.2055 0.2286

1/x 0.1885 0.2334 0.2728

0.5

0.8225

0.2288

0.2782

0.7939

0.2453

0.3090

0.6 0.7 0.8 0.9

0.7835 0.7427 0.6995 0.6529

0.2416 0.2505 0.2559 0.2576

0.3084 0.3374 0.3658 0.3945

0.7480 0.6999 0.6486 0.5920

0.2568 0.2636 0.2658 0.2628

0.3433 0.3766 0.4098 0.4439

0.2523

0.4817

.0.2078 0.1589 0.1075 0.0544 0

0.5775 0.6790 0.7837 0.8909 1

0.5238 1 0.4244 0.2546 0.6000 1.25 1.67 2.5 5 co

0.4338 0.3088 0.1850 0.0831 0

0.2289 0.1871 0.1339 0.0711 0

0.5044 0.6060 0.7240 0.8558 1

0.3598 0.2340 0.1372 0.0611 0

Fr anax

=

(7.70)

J,(E)Z cos a

or (7.71) Figures 7.14 and 7.15 also givevalues of J,, J,, and E vs F, tan aIF, for point and line contact, respectively. ~ 7.7. The ~ 218 angular-contact ~ ~ ball Z bearing e of Example 7.5 is subjected to a radial load of 17,800 N (4000lb) and a thrust load of 17,800 Pa (4000 lb). Estimate the normal load on each ball if the contact angle can be presumed to remain constant at 40".

~

1.o

0.9 0.8

0.7

0.6

0.4

0.3

0.2

0.1

0

~1~~~

0.2

0.4

0.6 0.8 F,. tan alF,,

1.0

7.14. J,(E), JJE), E vs Fr tan a / F a for point-contact bearings.

0.9

0.8 0.7

0.6 W

i \-.

.s" 0.5 i

%

0.4

0.3

0.2

0.1

0

FIG

0.2

0.4

0.6 0.8 1.2 1.0 E; tan aJF,

7.16. J,(E),J,(E),E vs Fr a / F , for line-contact bearings.

Fr tan Fa

From Fig. 7.14, J ,

=

cy

- 17800 tan (40") = o,8391 17800

0.221, J a = 0.263,

E =

0.455,

53 = 16

Ex. 7.5 (7.70)

-

17800 0.221 X 16 X cos (40")

=

6571 N (1477 lb)

The same result could have been obtained by using equation (7.71).

262

D I $ T R I ~ ~ I OOF N INFERNAL LOADING IN $TATICALLS LOADED ~ ~ I N G $

cos-1 (1 - 2€) COS-'

(1 - 2

X

(7.61), (7.65) 0.455) = t- 84'47' (7.62)

[

2

X

6571(1.099 COS 360 -360 2 16

"

1.5

1

6571 1 -

=

0.455 Q - 0.0989)1.5

22.5" (22'30')

cos II,

0 22.5 45 67.5 90 112.5 135 157.5 180

1 0.9239 0.7071 0.3827 0 -0.3827 -0.7071 -0.9239 -1

6571 (1477 lb) 5765 (1296 lb) 3670 (824.2 lb) 1200 (269.6 lb) 0 0 0 0 0

Let the indices 1 and 2 designate the rows of a two-row bearing having zero diametral clearance. Then (7.72) (7.73)

Sal = - 4 3 2

~ubstitutingthese conditions into equations (7.59) and (7.60) yields (7.74) €1

+ €1

=

1

(7.75)

Equation (7.75) pertains only if both rows are loaded. If only one row is loaded, then €1

2 1,

€2

=

0

(7.76)

B E ~ ~ UNDER G S C

O RADIAL~ANI3 THRUST ~ LOAD~

It is further clear from equation (7.4) that =

(>:”

(7.77)

Qmm,

The laws of static equilibrium dictate that (7.78) (7.79)

As before,

F,

=

ZQm,,,J,

(7.80)

COS cy

Fa = ZQmm,Jasin

(7.81)

cy

in which (7.82) (7.83) Table 7.5 gives values of J , and J a as functions of F, tan a/Fa. Figures 7.16 and 7.17 give the same data ingraphical format for point and line contact, respectively.

LE ‘7.5. J a and J , for Double-Row Bearings

0.5 0.5

00

0.6 0.4 2.046 0.7 0.3 1.092 0.8 0.2 0.8005 0.9 0.1 0.6713 1.0 0

0.6000 0.2546 0.4244

0.4577 0

1

1

0,35680.17440.5440.477 0.3036 0.2782 0.281 0.212 0.2758 0.3445 0.125 0.078 0.2618 0,3900 0,037 0.017 0

00

0.4906 0

1

1

0

0

2.389 0.4031 0.1687 0.640 0.570 1.210 0.3445 0.2847 0.394 0.306 0.8232 0.3036 0.3688 0.218 0.142 0.6343 0.2741 0.4321 0.089 0.043 0

0.5238 0.2523 0.4817

264

D

I

S

~

~ OF ~ INTERNAL I O ~ LOADING IN S T ~ T I C A L LLOADED ~ ~ E ~ I N G S

.4

7.16. J,, Jayel, eZyQmax2/Qmaxl, Frz/Frlvs Fr tan a/Fa for double-rowpointcontact bearings.

Ze 7.8. The 22317two-row spherical roller bearings of Example 2.7 supports a 89,000 N (20,000 lb) radial load and a 22,250 N (5000 lb) thrust load simultaneously. Estimate the roller load distribution.

Ex. 2.8

a = 12"

Z = 14

2.7

Ex.

Fr tan a - 89000 X tan (12") = 0.8502 22250 Fa From Fig. 7.17, J , = 0.303, Ja = 0.370, el = 0.220, Fr2/Fr1 = 0.143

=

0.8, ez = 0.2,

F I ~ 7.~ 17. EJ,, Ja7 eZ7Qmax2/Qmaxlt Fr2/Fr1 vs F,tan a / F afor double-row line-contact bearings.

89000 = 14 X Qmml

Qm,

X

0.303 x cos (12")

=

21450 N (4819 lb)

=

0.220

=

~ 1 4 5 0 ( 0 . 6 ~COS 5 t,b

X

21450 = 4719 N (1060 lb)

+ 0.375)1.11

DISTR~UTIO OF ~

~~~

=

LOADING IN S T ~ T I LOADED ~ ~ S El

~~~~

cos-1 (1 - 2€,)

(1- 2

X

0.8) = +- 126'52'

(7.62)

[

4719 I

+12

-

1 (1 - cos 2 x 0.2

~

=

4719(2.5 COS

=

cos-l (1-

==:

$1

+ - 1.5)1*11

2E2)

COS-^ (1 - 2

X

0.2)

I=:

+- 53'8'

A + = - 360' =--

360 - 25.71" 14

0 25.71 51.42 77.13 102.84 128.55 154.26 180

1 (4819 2140 0.9010 19980 (4488 0.6237 15930 (3578 10250 0.2227 -0.2227 (964 4321 -0.6237 -0.9010 -1

Z

(25'43')

lb) lb) lb) (2299 lb) lb) 0 0 0

4719 (1060 lb) 3442 (773 lb) 204 (46 lb) 0 0 0 0 0

If a ball is compressed by a load Q, then since the centers of curvature of the raceway grooves are fixed with respect to the corresponding raceway, the distance between the centers is increased by the amount of the normal approach between the raceways. From Fig. 7.18 it can be seen that

s =A

+ si + ti,

s* = Si" so= s " A

(7.84) (7.85)

If a ball bearing that has a number of rolling elements situated symmetrically about a pitch circle is subjected to a combination of radial, thrust, and moment loads, the following relative displace~entsof inner and outer ring raceways may be defined:

(a)

(b)

7.18. (a)Ball-raceway contact before loading; ( b ) ball-raceway contact under load.

Sa relative axial displacement Sr relative radial displacement e relative angular misalignment

These relative displacements are shown in Fig. 7.19. Consider a rolling bearing prior to the application of load. Figure 7.20 shows the positions of the loci of the centers of the inner andouter race-

7.19.Displacements of an inner ring (outer ring fixed) due to combined radial, asial, and moment loading.

268

D I S T R I B ~ I O NOF INTERNALLOADING IN STATICUY LOADED B E ~ I N ~ S

Bearing axis x

7.20.

Lociof raceway groove curvature radii centers before loading (reprinted

from [7.21).

way groove curvature radii. It can be determined from Fig. 2.2 that the locus of the centers of the inner ring raceway groove curvature radii is expressed by

in which ao is the free contact angle determined by bearing diametral clearance. From Fig. 7.20, then

S&= tni

-

tni

- A COS

% =A

COS

a"

(7.87)

a"

(7.88)

+

In Fig. 7.20, is the angle between the most heavily loaded rolling element and any other rolling element. Because of symmetry 0 5 5 71: If the outer ring of the bearing is considered fixed in space as load is applied to the bearing, then the inner ring will be displacedand thelocus of inner ringraceway grooveradii centers will also be displaced as shown in Fig. 7.21. From Fig. 7.21 it can be determined that s, the distance between the centers of curvature of the inner and outer ring raceway grooves at any rolling element position is given by

+

+,

s

=

[(A sin

ao

+ Sa + $Ii8 cos

or

Y printed from [7.2]).

+)2

+ (A cos

cyo

+ 6, cos +)2]1'2

(7.89)

270

D I S T R I B ~ I O NOF ~E~~

s = A[(sin ao

+ 8, +

LOADING IN S T A T I C ~ L SLOADED B

cos +))" +

ao +

(COS

8, COS

(7.90)

in which 6a aa = A

-

(7.91) (7.92) (7.93)

~ubstitutingequation (7.90) into (7.85) yields

+ (cos ao + 8, cos @)2]1/2

- I} (7.94)

~ubstitutionof equation (7.94) into (7.4) gives

Q = KnAn{[(sin ao + Sa +

giecos

+)2

+ (cos ao + 3,

cos

@)2,1/2

-

1p (7.95)

+,

At any rolling element position the operating contact angle is a. This contact angle can be determined from sin a

=

sin ao + Sa + 9xi3 cos sl/ [(sin ao + Sa + 9xi3 cos $1' + (cos ao + S, cos

(7.96)

+)'I'/'

or __

cos a

=

[(sin ao + Sa +

+

cos ao + 6, cos si3cos +12 + (COS ao + 3,

COS

+)211/2

(7.97)

Equation (7.95) describesthe normal load acting through contact angle a. This normal load maybe resolved into axial and radial components as follows:

Q, = Q sin a

&,

=

& cos + cos a

(7.98) (7.99)

If the applied radial and thrust loads on the bearing are F, and Fa,respectively, then for static equilibrium to exist:

*

$=0 *=+n

~dditional~y, each o f the thrust components produces a momeat about the Y axis (moments about the 2 axis are self-equ~librati~~), w]nich is given by

272

I ) I S T R ~ ~ OF I OI~~ E R N A L LOADING IN STATICALLY LOADED B M I N G S

The foregoing equations were developed by Jones E7.21. Equations (7.104)-( 7.106)are simultaneous nonlinear equations with unknowns Sa, Sr, and 6. They may be solved by numerical methods; for example, the Newton-Raphson method. Having obtained Sa, &, and 6, the maximum rollingelement load may be obtained from equation (7.95) for !,t = 0. Lllax

=

KnA"{[(sina'

+ sa+

+ (cos a' + 8r)2]1'2- 1)"

(7.107)

Solution of the indicated equations generally necessitates the use of a digital computer. In certain cases, however-for example, applications with simple radial, simple thrust or radial andthrust loading with nominal clearance-the simplified methodspresented in the beginning of this chapter will probably provide sufficientlyaccurate calculational results.

Although it is usually undesirable, radial cylindrical rollerbearings and tapered roller bearings can support to a small extent the moment loading due to misalignment, The various types of misalignment are illustrated in Fig. 7.22. Clearly, spherical roller bearings are designed to exclude all moment loads on the bearings and therefore are not included in this discussion. Figure 7.23 illustrates themisalignment of a cylindrical rollerbearing inner ring relative to the outer ring. To commence the analysis, it is assumed that any roller-raceway contact can be subdivided into a number of laminae situated in planes parallel to the radial plane of the bearing. It is also assumed that shear effects between these laminae can be neglected owing to the small magnitudes of the contact deformations that develop. (Only contact deformations are considered.)

In a misaligned cylindrical roller bearing subjected to radial load, at each laminum in a crowned roller-raceway contact, the deformation may be considered to be composed of three components: (1)Amj due to the radial load at roller azimuth location j, (2) e, due to the crown dropat laminum A, and (3) deformation due to the bearing misalignment and roller tilt at roller azimuth location j. These componentsare illustratedschematically in Fig. 7.24. The component due to radial load was the only component considered in the simplified analysis previously discussed; it needs no further ex-

"".

??z%%%# € is alignment (out-of-line) (a)

i%t&%

0

0

-

-

0

fl

0

Off-square or tilted outer ring (b)

?3%zm

?mm

%?z7%%m Cocked or tilted inner ring (c)

Shaft deflection (d)

7.22. Types of misalignment.

planation. The componentdue to crowning can be defined by consider in^ the roller crowning shown by Fig. 7.25. From Fig. 7.25, it can be seen that crown drop cAat a selected laminum is considered as a negative deformation; Le., no roller-raceway loadingcan occur at a laminum until cAis overcome bythe radial and/or the misalignment deformation. Equation (7.108) defines cAin terms of the roller and crowning ~ a r a m e t ~ r s .

FIGURE 7-23. Misalignment of cylindrical roller bearing rings.

0

"C

( h - $)w 5

I

-

I,

~

2

(7.108)

c, = 0

1 - 1, 2

"C

(A - $)w

-2

[

~

I

+ I, 2

1 + I, 2

( ~] ) [Rz 2

c, =

5

112

-

25

(A - $)w

5

-

4)

((A - $>w - 2

2

]

112

1

In (7.10~)? 15 A 5 k . For the bearing misalignment 0 shown in Fig. 7.24, the effective misalignment at roller location azimuth tc(j is rt $0 cos t,$, The plus sign per5 rt d 2 ; the minus sign pertains to +- d 2 5 "C rt w tains to 0 5 (assuming symmetry of loading about the 0 - w diameter). Therefore, the total roller-raceway deformation at roller location j and laminum A is given by equation (7.109).

*

*

s,

=

Aj 3- $ @A - #w cos

*

-

c,

h = 1, k

(7.109)

As discussed in Chapter 6, equation (6.53) is theoretical and relatescontact load to deformation for ideal line contact, while equation (6.54) i s

+

7.24. Components of roller-raceway deformation due to radial load, misalignment, and crowning.

7.25. Typical geometry of a crowned cylindrical roller showing crown radius, roller effective length, and roller straight length.

empirical and pertains to a crowned roller-raceway contact. While the load-deformation characteristic of an individual contact laminum may be described using either equation, the former is applied here since solution of a transcendental equation leads to force and moment equilibrium equations of greater complexity. Considering that the contact is divided into k laminae, each of width w ,therefore contact length is kw . Letting q = QlZ,equation (6.54) becomes (7.110) Rearranging (7.110) to define q yields

(7.111) Equation (7.111) does not consider edge stresses; however, becausethese obtain only over very small areas, they can be neglected with little loss of accuracywhenconsidering equilibrium of loading. ~ubstitutionof equation (7.109) into (7.111) gives

Depending on the degree of loading and misalignment, all laminae in every contact may notbe loaded; in (7.IE),hj is the number of laminae under load at roller location j. Total roller loadingis given by

u ~ t i o of ~ sStatic E q u i l i ~ ~ i u m To determine the individual roller loading, it is necessary to satisfy the requirements of static equilibrium. Hence, for an applied radial load,

~ubstitutingequation (7.113) into (7.114) yields

=o

(7.115)

For an applied coplanar misaligning moment load, the equilibrium condition to be satisfied is 3Qjej cos

$

=

0 Tj

=

0.5;

$

=

I;

$ 9 0, TI- (7.116)

=

0, TI-

where ej is the eccentricity of loading at each roller location.ej, which is illustrated in Fig. 7.26, is given by

77 I

.L

I

F I ~ 7.26. ~ ELoad distribution for a misaligned crowned roller showing eccentricity of loading.

Hence,

The remaining equations to be established are the radial deflection relationships. It is necessary here to determine the relative radial movement of the rings caused by the misalignment as well as that owing to radial loading. To assist in thefirst determination, Fig. 7.27 shows schematically an inner ring-roller assembly misaligned with respect to the

interferencg with

. From thia sketch, it ia evident that one-half aE the roller ineluded angleis descl.ibed by

where

IAL ROLLER B

R

=

~

~

~

S

[(dm- D)2 + z 2 y 2

In developing equations (7.121) and (7.122), the effect of crown drop was investigated and found to be negligible. Expanding equation (7.121) in termsof the trigonometric identity further yields 6,

=

R(cos /3 cos Oj

-

+ sin /3 sin Oj - cos p)

Since ej is small, cos Oj 1, and sin Oj and sin /3 = 112 R, therefore

-

(7.123)

Oj, Moreover Oj = rt: 0 cos lJzj

6, = k4Z0 cos lJzj

(7.124)

The shift of the inner ring center relative to the outer ring center owing to radial loading and clearance, and the subsequent relative radial movement at any roller location are shown in Fig. 7.28. The sum of the relative radial movement of the rings at each roller angular location minus the clearance is equal to the sum of the inner and outer raceway maximum contact deformations at the same angular location. Stating this relationship in equation format: d [ar ct +Z0] COS lJzj - P- 2[Aj rt: 40(A - 4 )COS ~ lJzj - c 2

~ = 0]

(7.125) ~ ~

Equations (7.115), (7.118), and (7.125) constitute a set of 212 + 3 simultaneous nonlinear equations that can be solved forar, 0, and Aj using numerical analysis techniques. Thereafter, the variation of roller loadper

7.88. Displacement of ring centers caused by radial loading showing relative radial movement.

2

D

I

S

~

~ OF ~II N O

~T LOADING ~ ~ IN~S T A T I C ~ Y L

unit length, and subsequently the roller load, may be determined for each roller location using equations (7.112) and (7.113), respectively. Using the foregoing method and digital computation, Harris E7.51 analyzed a 309 cylindrical roller bearing having the following dimensions and loading: Number of rollers Roller effective length Roller straight lengths Roller crown radius Roller diameter Bearing pitch diameter Applied radial load

12 12.6 mm (0.496 in.) 4.78, 7.770, 12.6 mm 1245 mm (49 in.) 14 mm (0.551 in.) 72.39 mm (2.85 in.) 31,600 N (7100 lb)

For the above conditions, Fig.7.29 shows the loading on various rollers for the bearing with ideally crowned rollers [I = 12.6 mm (0.496 in.) and with fully crowned rollers ( I = 0). Fig. 7.30 shows the effect or roller crowning on bearing radial deflection as a function of misalignment.

F

c

When radial cylindrical roller bearings have fixed flanges on bothinner and outer rings, they can carry some thrust load in addition to radial load. Thegreater the amount of radial load applied,the more thrust load that can be carried. As shown by Harris [7.6] and seen in Fig. 7.31, the thrust load causes each roller to tilt an amount 6. Again, it is assumed that a roller-raceway contact can be subdivided into laminae in planes parallel to the radial plane of the bearing. When a radial cylindrical rollerbearing is subjected to applied thrust load, the inner ring shifts axially relative to the outer ring. Assuming deflections owing to roller-end-flange contacts are negligible, then the interference at any axial location (laminurn) is 6, = Aj 3- &(l h - +)Lu - CA

h = 1, kj

(7.126)

where c, is given by equations (7.108). Figure 7.32 illustrates the component deflections in equation (7.126). Substituting equation (7.126) into (7.111) yields

P

P

281

~ I S T R I ~ ~ T IOF O NIIWEXWAL L O ~ I N G IN S T A T I C ~ L S

24

O ~ I3E

~

0.08

22

20

0.05

c-

E E

18 X

.-r'

v

._.16

0.04

4-J

8 ;F:

a"

fii .0

I, = 7.70 mm ( 0.303in. 1 14

-

12

\

\

0.03

,

10

8 5*

0

15

10

20

25

~ i ~ l i g n m e(min.) n~

7.30. Radial deflection vs misalignment and crowning-309 bearing at 31,600 N (7100 lb)radial load.

cylindrical roller

(7.127) "

and at any azimuth -

*, the total roller loading is w0.89

1.24

X

x

A=kj

10-5143~1

v

[Aj

+ &(A

- $)LO

- CA]'"'

(7.128)

Housina Y

Qa)

Shaft

Shaft ” -

7.31.

“4

Shaft

-Q

” ” ” ”

Thrust couple, roller tilting, and inteflerence owing to appliedthrust load.

I ~ 7.32.~ Components E of roller-raceway deflection at opposing raceways due to radial load, thrust load, and crowning.

84

~ I S T R I B ~ I OOF N ~E~~LOADING

IN STATICALLY LOADED B

~

~

~ubstitutingequation (7.128) into (7.129) yields

(7.130) For an applied centric thrust load, the equilibrium condition to be satisfied is

each roller location, the thrust couple is balanced by a radial load coupled caused by the skewed axial load distribution. Thus, hQaj = 2Qjej and F a "-

2

2 j=Z/2+1

h

x

j=1

=

7Qjej = 0

7j =

0.5; 1;

3/j

=

0,

T

3/jfO?T

(7.132)

where ej is the eccentricity of loading indicated by Fig. 7.26 and defined bY

(7.133) h=l

-

~ubstitutionof equations (7.128) and (7.133) into (7.132) yields

(7.134)

G

S

THRUST L

O

~ OF ~ RADIAL G C

~

~

R

I

C

~~

E

~ R

O ~

~G R S

Radial deflection relationships remain to be established. It is necessary to determine the relative radial movement of the bearing rings caused by the thrust loading as well as that due to radial loading. To assist in this derivation, Fig. 7.31 shows schematically a thrust-loaded roller-ring assembly. From this sketch, a roller angle is described by tan q

=

D -

I

(7.135)

The maximum radial interference between a roller and both rings is given by (7.136)

In developing equation (7.136) the effect of crown drop was found to be negligible. Expanding equation (7.136) in terms of the trigonometric identity and recognizing that sj is small and I = D ctn q, yields

stj= 16

(7.137)

Whereas Stj is the radialdeflection due to roller tilting, it can be similarly shown that axial deflection owing to roller tilting is

saj= Dsj

(7.138)

Therefore, the radial interference caused by axial deflection is (7.139) The sum of the relative radial movements of the inner and outer rings at each roller azimuth minus the radial clearance is equal to the sum of the inner andouter raceway maximum contact deformationsat the same azimuth, or

~quations(7.130), (7.134), and (7,140) are a set of simultaneous equations that can be solved for 6, Aj, a,, and iSa. Thereafter, the variation of roller load per unit length and roller load may be determined for each roller a ~ i m ~using t h eq~ations(7.127) and (7.1281, respectively. The axial load on each roller may be determined from

r-

Y

When rollers are subjected to axial loading as shown in Fig. 7.31, due to sliding motions between the roller ends and ring flanges friction forces occur; for exampleFaj= pQaj, in which p is the coefficient of friction. In a m i s a l i ~ e dbearing, each roller which carries load is end and forced against the opposing flangewith a load tion force Fajat the roller end. Due to F4 a moment occurs creating a yawing or skewing motion in addition to the predominant rolling motion about the roller axisand secondary rollertilting. The tilting andskewing motions occurin orthogonal planes which contain the roller axis. Friction forces acting on rolling elements are not introduced until Chapter l.4; however, roller skewingis resisted by the concave curvature of the outer raceway. The resisting forces and accompanying deformations alter the distribution of load along both the outer and inner raceway-roller contacts, Figure 7.33 illustrates the forces which occur ona roller subjected to radial and thrust loading. Frictional stresses T~~~are discussed in detail in Chapter 14. Figure 7.34 shows the roller skewing angle and the roller-outer raceway loading whichresults. The roller-raceway contact deformations which result from skewing, as demonstrated by Harris et al. [7.7], may be described by equation (7.142).

4

(7 142) In (7.142), subscript m refers to the outer and innerraceway contacts; m = 1 and 2 respectively, and deformations due to skewing +mjh are given by

x

7.33. Normal and friction forces acting on a radial and t ~ r u ~ t - l o a roller. ~ed

7.34. Roller-outerracewaycontactshowing roller skewing angle 8 and restoring forces.

(7.143) (7.144)

288

D I S T ~ ~ ~ IOF O ~N E

~ LOADING A L IN STATICALLYLOADED ~

E

~

It can be further seen that equation (7.140) must become

Owing to the unknown variables 4 and A,, the latter replacing Aj, additional equilibrium equations must be established. For equilibrium of roller loading in the radial direction m=2

h=k

m=2

(7.146)

Referring to Fig. 7.34 and considering equilibrium of moments in the plane of roller skewing

lFaj+

xx

m=2 h=k

w2[A - +(k

m=1 h = l

+ l)]rmjh

m=2 A=k

(7.147)

Since the angle pj

-

0, sin pj

-

pj, therefore (7.148)

Using methods described in Chapter 14, it can be determined that the moment loading effect of roller-raceway shear stresses is rather small compared to that of the restoring and roller end forces. Therefore, substituting (7.148) into (7.147) yields

Considering that thecontact deformationsdue to roller radial loading are different for each roller-raceway contact, bearing load equilibrium equations (7.130) and (7.134) must be changed accordingly; hence

~

G

S

"0

(7.150)

and

Equations (7.132), (7.145), (7.146),(7.149), (7.150), and (7.151) constitute a set of simultaneous, nonlinear equations which may be solved for Amj, 4,4, is,,and Sa . Subsequently, the roller-raceway loads Qj and roller end-flange loads Qaj may be determined. The skewingangles determined using the foregoing equations strictly pertain to full complement bearings and bearings having no roller guide flanges. For a bearing with a substantially robust and rigid cage, the skewing angle may be limited by the clearances between the rollers and the cage pockets. For a bearing with guide flanges, the skewing may be limited by the endplay between the roller ends and guide flanges. In general, the latter situationsobtain; however, to the extent that skewing is permitted, the foregoing analysis is applicable.

For radial cylindrical rollerbearings, it is possible to apply general combined loading. The equations for load equilibrium defined above apply; however, the interference at any laminum in the roller-raceway contact is given by

where subscript m = I. refers to the outer raceway and m = 2 refers to the inner raceway. Coefficient vl = -1, and v2 = + l .Contact load per unit length is given by

Similar equations may be developed for tapered roller bearings. In this case, as shown in Chapter 5 , roller end-flange loading occursduring all conditions of applied loading, and the roller and bearing equilibrium equations must be altered accordingly Using Fig. 5.3 to illustrate the geometry and loading of a tapered roller in a bearing, and establishing dimensions r2, r32,and r3%as follows:

r2 r32 r3x

is the radius in a radial plane from the inner ring axis of rotation to the center of the inner raceway contact, is the radius in a radial pla e from the inner ring axis of rotation to the center of the roller end-inner ring flange contact, and is the x-direction distance in an axial plane from the center of the inner raceway contact to the center of the roller end-inner ring flange contact,

the roller load equilibrium equations are m=2

w

21 emcos

m=

A=k

CYm

m=2

m= 1

x

A=k

cm sin am

w

2 qmjh-

(7.154)

A= 1

A= 1

qmjh+ Q~~sin a3 =

o

(7.155)

In (7.154) and (7.155), &3j is the roller end-flange load,and a3is the angle that load &3j makes with a radial plane. Coefficient el = -1, and c2 = +1. The equation for radial plane moment equilibrium of the roller is

where R3 is the radius from the roller axis of rotation to the center of the roller end-flangecontact. Eq~ilibriumof actuating and resisting moments pertaining to roller skewing is given by

The forceand moment equilibrium equations with respect to the bearing inner ringare as follows:

LY SUPPORTED ROLLING ~

E

~

~

G

S

(7.159)

(7.160) In equations (7.158)-(7.160) for J/j 5 = 1.

=

0 or v,coefficient 7 = 0.5; otherwise

Spherical roller bearings are internally self-aligning and therefore cannot carry moment loading. Moreover, for slow or moderate speed applications causing insignificant roller inertial loading and friction, symmetrical (barrel-shaped) rollers in spherical roller bearings will exhibit no tendency to tilt. Therefore, the simpler analytical methods presented earlier in this chapter will yield accurate results. For spherical roller bearings having asymmetrical rollers, however, such as spherical roller thrust bearings (Fig. 1.45), roller tilting and hence skewing is not eliminated. In this case, for the purpose of analysis, the bearing may be considered a special type of tapered roller bearing having fully crowned rollers. Then the methods of analysis discussed in the preceding section may be applied for increased accuracy.

The preceding discussionof distribution of load amongthe bearing rolling elements pertains to bearings having rigidly supported rings. Such bearings are assumed to be supported in infinitely stiff or rigid housings and on solid shafts of rigid material. The deflections considered in the determination of load distribution were contact deformations, that is, Hertzian deflections. This assumption is, in fact, an excellent approximation for most bearing applications. In some radial bearing applications, however, the outer ring of the bearing may be supported at one or two angular positions only, and the

shaft on which the inner ringis positioned may be hollow. The condition of two-point outer ring support, as shown in Figs. 7.35 and 7.36, occurs in theplanet gear bearings of planetary gear power transmission system, and was analyzed by Jones and Harris E7.81. In certain rolling mill applications, the back-up roll bearings may be supported at only one point on the outer ring or possibly at two points as shown in Fig. 7.37. These conditions were analyzed by Harris [7.9]. In certain high speed radial bearings, to prevent skidding it is desirable to preload the rolling elements by using an elliptical raceway, thus achieving essentially twopoint ring loading under conditions of light applied load. The case of a flexible outer ring andan elliptical inner ringwas investigated by Harris and Broschard E7.101. In each of the foregoing applications, the outer ring must be considered flexible to achieve a correct analysis of rolling element loading. In many aircraft applications to conserve weight the power transmission shafting is made hollow. In these cases the inner ring deflections will alter the load distribution from that considering only contact deformation. To determine the load distribution among the rolling elements when one or both of the bearing rings is flexible, it is necessary to determine

FIGURE 7.35. Planet gear bearing.

F L E ~ L ~

~ ~ ROLLING P O R T BEARINGS E ~

7.36. Planet gear bearing showing gear tooth loading.

F1GUR.E 7.37. Cluster mill assembly showing back-up roll bearing loading.

€3~TION OF ~E~~

L O A D ~ GZN ~

T

~

T LOADED I ~ ~€3 L

~

the deflections of a ring loaded at various points around its periphery. This analysis may be achieved by the application of classical energy methods for the bending of thin ratings. As an example of the method of analysis, consider a thin ring subjected to loads of equal magnitude equally spaced at angles A+ (see Fig. 7.38). According to Timoshenko i7.111, the difEerentia1equation describing radial deflection u for bending of a thin bar with a circular center line is (7.161) in which I is the section moment of inertia in bending and E is the modulus of elasticity. It can be shownthat thecomplete solution of equation (7.161) consists of a complementary solution and a particular solution. The complementary solutionis u, = C, sin cf,

+ C,

cos cf,

(7.162)

in which C, and C, are arbitrary constants. Consider that the ring is cut at two positions: at theposition of loading, = $A+, and at theposition = 0, midway between the loads. The loads required to maintain equilibrium over the section are shown in Fig. 7.39.. From Fig. '7.39 it can be seen that since horizontal forces are balanced,

+

+

Qsin = 2F0

+

or

7.88. Thin ring loadedby equally spaced loads of equal magnitude.

(7.163)

1

7.39. Loading of section of thin ring between 0

2

(b

+A+.

Q Fo = 2 sin cjb

(7.164)

The moment at any angle cf, between 0 and &h,b is apparently

M

M0 - Foal- cos cf,)

(7.165)

(4% Mo - (1 - cos cf,)

(7.166)

=

or

M

=

2 sm cf,

Since the section at cf, = 0 is midway between loads,it cannot rotate. theorem f7.111 the angular rotation at any ~ ~ c o r d to i ~Gastigliano's g section is @=:-

au a"

(7.167)

in which U is the strain energy in the beam at the position of loading. ~ ~ o s h ~ E7.113 n k oshows that for a curved beam

= Mb and

since the section is constrained from rotation,

~ u b ~ t i ~ u tequation ing (7.166) into ('7.169)and inte

(7.170) Hence, (7.171) Equation (7.171) may be substituted for M in (7.161) such that the particular solution is (7.172) The complete solution is u = u,

[

Qs3 # sin (b + up = C, sin # + C, cos # + 2E1 2 sin (*A+)

1

A+

]

(7.173) Because the sections at .# = 0 and #

=

*A+ do not rotate, therefore

Hence, the radial deflection at any angle (b between # = 0 and # = i A 3 , is

(7.174) Equation (7.174) maybe expressed in another format as follows: u =

c,

(7.175)

in which C, are influence coefficientsdependent on angular position and ring dimensions.

LY $ ~ ~ O R ROLLING T E ~ ~ ~ I N G $

7

(7.176) Lutz E7.121 using procedures similar to those described above developed influence coefficients for various conditions of point loadingof a thin ring. These coefficients have been expressed in infinite series format for the sake of simplicity of use. For a thin ring loaded by forces of equal magnitude symmetrically located about a diameter as shown in Fig. 7.40, the following equation yields radial deflections:

in which (7.178) The negative sign in (7.178) is used for internal loads and the positive sign is used for external loads. Equation (7.177) definesradial deflection at angle t,bi caused by Qj at position angle t,$. When rollingelement loads Qjare such that a rigid body translation 8, of the ring occurs, in the direction of an applied load, equations (7.177) are not self-sufficient in establishing a solution; however,a directional equilibrium equation may be used in conjunction with (7.177) to determine the translatory movement. Referring to Fig. 7.41 the appropriate equilibrium equation is as follows:

I

magnitude located asymmetrically about

a diameter.

8

UTIQN OF ~E~~

~1~~~

LOADING IN STATICALLY L Q A D E ~B

~

~

N

7.41. Thin ring showing e q u i l i b ~ u of~forces,

Fr cos 3/i

-

Qj cos 3ij

=

0

(7.179)

In the planet gear bearing application demonstrated in Fig. 7.36 the gear tooth loads maybe resolved into tangential forces, radial forces, and moment loads at Ifi = 90" (see Fig. 7.42). The ring radial de~ections at angle Ifii due to tangential forces Ft are given by (7.180)

in which

7.42.

ring.

Resolution of gear tooth loadingon outer

~

S

mr m& 2 m(m2 - I ) 2

m=m sin - cos

g i

x

2s3 =TEI m=2

(7.181)

Equations (7.180) are not self-sufficient and an appropriate equilibrium equation must be used to define a rigid ring translation. The separating forces Fs are self-equilibratin~ and thusdo not cause a rigid ring translation. The radial deflections at angles t,bi are given by sui = ,giFS

(7.182)

in which

Note that equations (7.183) are special cases of (7.178) in which position angle J/j is 90" and loads Qj are external. Similarly, the moment loads applied at sr/ = 90" are self-equilibrating. The radial deflections are given by

in which

To find the ring radial deflections at any regular position due to the combination of applied and resisting loads, the principle of superposition is used. Hence for the planet gear bearing, the radial deflection at any angular position t,bi is the sum o f the radial deflections due to each individual load, that is,

or

ui

=

,CiFS + MCiM+ tCiFt + zQCij

(7,187)

lements do the

A load maynot be transmitted through a rolling element unless the outer ring deflects suf~cientlyto consume the radial clearance at theangular position occupied bythe rolling element. Furthermore, because a contact deformatio~is caused by loading of the rolling element, the ring deflections cannot be determined without considering these contact deformations. Therefore the loading of a rolling element at angular position Jlj depends on the relative radial clearance. The relative radial approach of the rings includes the translatory movement of the center of the outer ring relative to the initial center of that ring, which position is fixed in space. Hence forthe planet gear bearing the relative radial approach at angular position +i is

sicos= 6,

Jli

+ ui

(7.188)

From equation (7.4) the relative radial approach is related t o the rolling element load as follows: Qj =r

Qj

=

IC( Sj - rj)"

sj > rj

0

Sj 5 rj

(7.189)

in which rj is the radial clearance at angular position sum of Pd/2 and the condition of ring ellipticity.

Jlj. Here rj is the

olling Elem~nt Using the example of the planet gear bearing, the complete loading of the outer ring is shown in Fig, 7.43, which alsoillustrates the rigid ring translation 6,. Combination of equations (7.187)-( 7.189) yields i=Z/2+2

si - 6,cos

Jli

-

*CiFS- MciM- ,CiFt -

iK j=2

*Cii(Sj-

rj)" =

0

(7,190) The required equilibrium equation is (7.191) considering symmetry about the diameter parallel to the load. In equation (7.191), 3 = 0.5 if the rolling element is located at $ = 0" or at Jlj = 180"; otherwise 3 = 1.

301

FLEXIBLY S ~ ~ O R T EROLLING D BEARINGS

~

I 7.43. ~Total loading ~ ofEouter ring in planet gear bearing.

Equations (7.190) and (7.191) constitute a set of simultaneous nonlinear equations which may be solved bynumerical analysis. The NewtonRaphson method is recommended. Using this method, the unknowns Sj and hence Qj can be determined at each rolling element location. Figure 7.44 shows a typical distribution Planet gear bearing

Rigidring bearing 7.44.

bearing.

Comparison of load distribution for a rigid ring bearing and planet gear

8,000

35,000

7,000

30,000

6,000 25,000

5,000

20,000 -c) v -0

.8

-tc

N

4,000

0

n:

15,000

3,000

10,000 2,000

5,000 1,000

Roller position (degrees t 1 7.46. Roller load vs number of rollers and position. 222,500 N (50,000 lb) at dimensions constant. Outer ring section thickness increasesas thenumber of rollers is increased and rollerdiameter is subsequently decreased. st 30°, inner

required to solve the displacements and load distribution accurately in a rolling bearing mounted in a flexible support. Figure 7.48 from Zhao E7.141 shows the grids used to analyze a flexibly mounted cylindrical roller bearing assuming both solid and hollow rollers. The load distribution would be similar to that indicated in Fig. 7.47.

D I $ ~ I B ~ OF I OI ~N T E LOADING ~ ~ IN $ T ~ T I C ~ LLOADED Y BE~ING$

7.47. Photoelastic studyof a roller bearing supportingloads aligned at approximately +- 30" to the bearing axis.

(4

7.48. Finite element meshes for analyzing (a)cylindrical roller bearing rings, (b) solid rollers, (e) hollow rollers, and ( d ) contact zone. From [7.14].

REFERENCES

Bourdon et al. E7.151 and E7.161 provide a method to define stiffness matrices for use in standard finite element models to analyze rolling bearing loads and deflections, and the loading and deflections of the mechanisms in which they are employed. For flexible mechanisms and bearing support system