Mechanical Engineers Handbook Episode 6 docx

Cylinder roller bearings provide a greater force than ball bearings of thesame size because of the greater contact area.. Tapered cone roller bearings combined the advantages of ball and

When Castigliano's theorem is applied, the de¯ection of the helicalspring is

3.4.5 SPRING ENDSFor helical springs the ends can be speci®ed as (Fig 3.5) (a) plain ends; (b)plain and ground ends; (c) squared ends; (d) squared and ground ends Aspring with plain ends (Fig 3.5a) has a noninterrupted helicoid, and the endsare the same as if a long spring had been cut into sections A spring withplain and ground ends (Fig 3.5b) or squared ends (Fig 3.5c) is obtained bydeforming the ends to a zero-degree helix angle Springs should always beboth squared and ground (Fig 3.5d) because a better transfer of the load isobtained Table 3.2 presents the type of ends and how that affects thenumber of coils and the spring length In Table 3.2, Na is the number ofactive coils, and d is the wire diameter.

Figure 3.5

Table 3.2 Types of Spring Ends

Source: J E Shigley and C R Mischke, Mechanical Engineering Design McGraw-Hill, New York, 1989 Used with permission.

EXAMPLE An oil-tempered wire is used for a helical compression spring The wire

diameter is d ˆ 0:025 in, and the outside diameter of the spring is D0ˆ0:375 in The ends are plain and the number of total turns is 10.5 Find:

The torsional yield strengthThe static load corresponding to the yield strengthThe rate of the spring

The de¯ection that would be caused by the static load foundThe solid length of the spring

The length of the spring so that no permanent change of the free lengthoccurs when the spring is compressed solid and then releasedThe pitch of the spring for the free length

SolutionFrom Eq (3.3) the torsional yield strength, for hardened and temperedcarbon and low-alloy steel, is

Ssyˆ 0:50Sut:The minimum tensile strength given from Eq (3.1) is

Sut ˆdAm;where, from Table 3.1, the constant A ˆ 146 kpsi and the exponent m ˆ0:193

The minimum tensile strength is

Sut ˆ A

dmˆ 146…0:025†0:193ˆ 297:543 kpsi:

The torsional yield strength is

The spring rate, Eq (3.22) for N ˆ Na is

3.5 Torsion SpringsHelical torsion springs (Fig 3.6) are used in door hinges, in automobilestarters, and for any application where torque is required Torsion springs are

of two general types: helical (Fig 3.7) and spiral (Fig 3.8) The primary stress

in torsion springs is bending The bending moment Fa is applied to each end

of the wire The highest stress acting inside of the wire is

where the factor for inner surface stress concentration Ki is given in Fig 3.9,and I is the moment of inertia The distance from the neutral axis to theextreme ®ber for round solid bar is c ˆ d=2, and c ˆ h=2 for rectangular bar.For a solid round bar section, I ˆ pd4=64, and for a rectangular bar,

I ˆ bh3=12 Substituting the product Fa for bending moment and theequations for section properties of round and rectangular wire, one maywrite for round wire

where M is the bending moment, L is the beam length, E is the modulus ofelasticity, and I is the momentum of inertia.

Equation (3.28) can be used for helical and spiral torsion springs Helicaltorsion springs and spiral springs can be made from thin rectangular wire

Round wire is often used in noncritical applications

Figure 3.6

Figure 3.7

3.6 Torsion Bar SpringsThe torsion bar spring, shown in Fig 3.10, is used in automotive suspension.The stress, angular de¯ection, and spring rate equation are

where T is the torque, r ˆ d=2 is the bar radius, l is the length of the spring,

G is the modulus of rigidity, and J is the second polar moment of area

For a solid round section, J is

as a triangular plate (Fig 3.12a) cut into n strips of width b or stacked in agraduated manner (Fig 3.12b)

Figure 3.10

Figure 3.11

To support transverse shear Ne, extra full-length leaves are added on thegraduated stack, as shown in Fig 3.13 The number Ne is always one lessthan the total number of full length leaves N.

The prestressed leaves have a different radius of curvature than thegraduated leaves This will leave a gap h between the extra full-length leavesand the graduated leaves before assembly (Fig 3.14)

Figure 3.12

Figure 3.13

3.7.1 BENDING STRESS seThe bending stress in the extra full-length leaves installed without initialprestress is

d ˆ 12Fl3

bt2E…3Ne‡ 2Ng†; …3:39†

where E is the modulus of elasticity (psi)

3.7.4 BENDING STRESS, sThe bending stress of multileaf springs without extra leaves or with extra fulllength prestressed leaves that have the same stress after the full load hasbeen applied is

3.7.5 GAPThe gap between preassembled graduated leaves and extra full-length leaves(Fig 3.14) is

3.8 Belleville SpringsBelleville springs are made from tapered washers (Fig 3.15a) stacked inseries, parallel, or a combination of parallel±series, as shown in Fig 3.15b.The load±de¯ection and stress±de¯ection are

where F is the axial load (lb), d is the de¯ection (in), t is the thickness of thewasher (in), h, is the free height minus thickness (in), E is the modulus ofelasticity (psi), s is the stress at inside circumference (psi), do is the outsidediameter of the washer (in), diis the inside diameter of the washer (in), and m

is the Poisson's ratio The constant M, C1, and C2 are given by the equations

C2 ˆp log 6

e…do=di†

do=diÿ 12

Under certain combinations of force, speed, ¯uid viscosity, and bearinggeometry, a ¯uid ®lm forms and separates the contacting surfaces in a slidingbearing, and this is known as a hydrodynamic ®lm An oil ®lm can also bedeveloped with a separate pumping unit that supplies pressurized oil to thebearing, and this is known as a hydrostatic ®lm

The surfaces in a bearing may also be separated by balls, rollers, orneedles; these are known as rolling bearings Because shaft speed isrequired for the development of a hydrodynamic ®lm, the starting friction

in hydrodynamic bearings is higher than in rolling bearings To minimizefriction when metal-to-metal contact occurs, low-friction bearing materialshave been developed, such as bronze alloys and babbitt metal

The principal advantage of these bearings is the ability to operate atfriction levels considerably lower at startup, the friction coef®cient having thevalues m ˆ 0:001±0:003 Also, they have the following advantages overbearings with sliding contact: they maintain accurate shaft alignment forlong periods of time; they can carry heavy momentary overloads withoutfailure; their lubrication is very easy and requires little attention; and they areeasily replaced in case of failure

Rolling bearings have the following disadvantages: The shaft and housedesign and processing are more complicated; there is more noise for thehigher speeds; the resistance to shock forces is lower; the cost is higher.4.2 Classi®cation

The important parts of rolling bearings are illustrated in Fig 4.1: outer ring,inner ring, rolling element and separator (retainer) The role of the separator

is to maintain an equal distance between the rolling elements The races arethe outer ring or the inner ring of a bearing The raceway is the path of therolling element on either ring of the bearing

Rolling bearings may be classi®ed using the following criteria (Fig 4.2):

j The rolling element share: ball bearings (Fig 4.2a±f), roller bearings(cylinder, Figs 4.2g,h; cone, Fig 4.2i; barrel, Fig 4.2j), and needlebearings (Fig 4.2k)

j The direction of the principal force: radial bearings (Figs 4.2a,b,g,h),thrust bearings (Figs 4.2d,e), radial-thrust bearings (Figs 4.2c,i), orthrust-radial bearings (Fig 4.2f)

j The number of rolling bearing rows: rolling bearings with one row(Figs 4.2a,d,g,k), with two rows (Figs 4.2b,e,h), etc

The radial bearing is primarily designed to support a force perpendicular

to the shaft axis The thrust bearing is primarily designed to support a forceparallel to the shaft axis

Single row rolling bearings are manufactured to take radial forces andsome thrust forces The angular contact bearings provide a greater thrustcapacity Double row bearings are made to carry heavier radial and thrustforces The single row bearings will withstand a small misalignment orde¯ection of the shaft The self-aligning bearings (Fig 4.2f), are used forsevere misalignments and de¯ections of the shaft

Cylinder roller bearings provide a greater force than ball bearings of thesame size because of the greater contact area This type of bearing will nottake thrust forces Tapered (cone) roller bearings combined the advantages

of ball and cylinder roller bearings, because they can take either radial orthrust forces, and they have high force capacity

Needle bearings are used where the radial space is limited, and when theseparators are used they have high force capacity In many practical casesthey are used without the separators

The pitch diameter can be calculated exactly as

In general the ball bearings are manufactured with a clearance betweenthe balls and the raceways The clearance measured in the radial plane is thediametral clearance, sd, and is computed with the relation (Fig 4.3)

where D is the ball diameter

Because a radial ball bearing has a diametral clearance in the no-loadstate, the bearing also has an axial clearance Removal of this axial clearanceFigure 4.2

causes the ball±raceway contact to assume an oblique angle with the radialplane Angular-contact ball bearings are designed to operate under thrustforce, and the clearance built into the unloaded bearing along with theraceway groove curvatures determines the bearing free contact angle.

Because of the diametral clearance for radial ball bearing there is a freeendplay, sa, as shown in Fig 4.4 In Fig 4.4 the center of the outer ringraceway circle is Oe, the center of the inner ring raceway circle is Oi, and thecenter of the ball is O

The distance between the centers Oe and Oi is

If the raceway groove curvature radius is r ˆ f D, where f is a sionless coef®cient, then

dimen-A ˆ … feˆ fiÿ 1†D ˆ BD; …4:5†where B ˆ fe‡ fiÿ 1 is de®ned, as the total curvature of the bearing In thepreceding formula, reˆ feD and ri ˆ fiD, where fe and fi are adimensionalcoef®cients

The free contact angle a0is the angle made by the line passing throughthe points of contact of the ball and both raceways and a plane perpendicular

to the bearing axis of rotation (Fig 4.4) The magnitude of the free contactangle can be written

The diametral clearance can allow the ball bearing to misalign slightlyunder no load The free angle of misalignment y is de®ned as the maximumangle through which the axis of the inner ring can be rotated with respect tothe axis of the outer ring before stressing bearing components,

and ye (Fig 4.5b) is the misalignment angle for the outer ring,

cos yeˆ 1 ÿ sd‰…2feÿ 1†D ÿ sd=4Š

2dm‰dmÿ …2feÿ 1†D ‡ sd=2Š: …4:9†

With the trigonometric identity

cos yi‡ cos yeˆ 2 cos‰…yi‡ ye†=2Š cos…yiÿ ye†=2Š …4:10†

and with the approximation yiÿ ye  0, the free angle of misalignmentbecomes

4.4 Static Loading

In Fig 4.6a is shown a single row radial thrust (angular contact) ball bearing

The contact angle a is the angle of the axis of contact between balls and

Figure 4.6

races For a single row radial ball bearing, the angle a is zero If Fr is theradial force applied to the ball, then the normal force to be supported by theball is

The normal and axial forces for the inner ring are

Fi ˆcos aFri

i and Fai ˆ Dritan ai; …4:15†where Fri is the radial force acting on the inner ring The normal and axialforces for the outer ring are

Feˆcos aFre

e and Fae ˆ Fretan ae; …4:16†where Fre is the radial force acting on the outer ring The normal and axialforces for the frontal face are

Ff ˆ Frfcos af and Faf ˆ Frf tan af; …4:17†where Frf is the radial force acting on the frontal face

The equilibrium equations for the radial and axial directions are

between 35 and 50 mm range from ‡0:0000 to ÿ0:0005 in for ABEC grade I,and from ‡0:0000 to ÿ0:0001 in for ABEC grade 9 Tolerances on otherdimensions are comparable Similarly, the AFBMA Roller Bearing EngineersCommittee has established RBEC standards I and 5 for cylindrical rollerbearings.

The bearing manufacturers have established standard dimensions (Fig

4.7 and Table 4.1) for ball and straight roller bearings, in metric sizes, whichde®ne the bearing bore d, the outside diameter d0, the width w, the ®llet sizes

on the shaft and housing shoulders r, the shaft diameter dS, and the housingdiameter dH

For a given bore, there is an assortment of widths and outside diameters

Furthermore, the outside diameters selected are such that, for a particularoutside diameter, one can usually ®nd a variety of bearings having differentbores That is why the bearings are made in various proportions for differentseries (Fig 4.8): extra-extra-light series (LL00), extra-light series (L00), lightseries (200), and medium series (300)

Figure 4.7

Table 4.1 Bearing Dimensions

4.6 Bearing SelectionBearing manufacturers' catalogs identify bearings by number, give completedimensional information, list rated load capacities, and furnish detailsconcerning mounting, lubrication, and operation.

Often, special circumstances must be taken into account Lubrication isespecially important in high-speed bearing applications, the best being a ®neoil mist or spray This provides the necessary lubricant ®lm and carries awayfriction heat with a minimum ``churning loss'' within the lubricant itself Forball bearings, nonmetallic separators permit highest speeds

The size of bearing selected for an application is usually in¯uenced bythe size of shaft required (for strength and rigidity considerations) and by theavailable space In addition, the bearing must have a high enough load rating

to provide an acceptable combination of life and reliability

4.6.1 LIFE REQUIREMENTThe life of an individual ball or roller bearing is the number of revolutions (orhours at some constant speed) for which the bearings run before the ®rstevidence of fatigue develops in the material of either the rings or of any ofthe rolling elements Bearing applications usually require lives different fromthat used for the catalog rating Palmegran determined that ball bearing lifevaries inversely with approximately the third power of the force Later studieshave indicated that this exponent ranges between 3 and 4 for various rolling-element bearings Many manufacturers retain Palmegren's exponent of 3 forball bearings and use 10=3 for roller bearings Following the recommenda-tion of other manufacturers, the exponent 10=3 will be used for both bearingtypes Thus the life required by the application is

L ˆ LR…C =Fr†3:33; …4:22†

where C is the rated capacity (from Table 4.2), LRis the life corresponding tothe rated capacity (i.e., Lr ˆ 9  107 revolutions), and Fr is the radial forceinvolved in the application The required value of the rated capacity for theapplication is

Creq ˆ Fr…L=LR†0:3: …4:23†

For a group of apparently identical bearings the rating life, LR, is the life

in revolutions (at a given constant speed and force) that 90% of the grouptested bearings will exceed before the ®rst evidence of fatigue develops

Different manufacturers' catalogs use different values of LR (some use

LR ˆ 106 revolutions)

Figure 4.8

4.6.2 RELIABILITY REQUIREMENTTests show that the median life of rolling-element bearings is about ®ve timesthe standard 10% failure fatigue life The standard life is commonly desig-nated as the L10 life (sometimes as the B10 life), and this life corresponds to10% failures It means that this is the life for which 90% have not failed, and

Table 4.2 Bearing Rated Capacities,C, for 90  106 Revolution Life (LR) with 90% Reliability

it corresponds to 90% reliability …r ˆ 90%† Thus, the life for 50% reliability

is about ®ve times the life for 90% reliability Using the general Weibullequation together with extensive experimental data, a life adjustmentreliability factor Kr, is recommended The life adjustment reliability factor,

Kr, is plotted in Fig 4.9 This factor is applicable to both ball and rollerbearings The rated bearing life for any given reliability (greater than 90%) isthus the product KrLR Incorporating this factor into Eqs (4.22) and (4.23)gives

Standard values of load angle a for angular ball bearings are 15, 25, and

35 Only the 25 angular ball bearings will be treated next The radialequivalent force, Fe, for angular ball bearings with a ˆ 25 is:

j For 0:00 < Fa=Fr< 0:68 ) Feˆ Fr

j For 0:68 < Fa=Fr< 10:0 ) Feˆ Fr‰1 ‡ 0:87…Fa=Frÿ 0:68†Š

j For Fa=Fr> 10:0 ) Feˆ 0:911Fa

4.6.4 SHOCK FORCEThe standard bearing rated capacity is for the condition of uniform forcewithout shock, which is a desirable condition In many applications there arevarious degrees of shock loading This has the effect of increasing thenominal force by an application factor, Ka In Table 4.3 some representativesample values of Ka are given The force application factor in Table 4.3serves the same purpose as factors of safety

If we substitute Fe for Fr and adding Ka, Eq (4.24) gives

When more speci®c information is not available, Table 4.4 may be used

as a guide for the life of a bearing in industrial applications Table 4.4contains recommendations on bearing life for some classes of machinery.The information has been accumulated by experience

EXAMPLE 1

(FROM REF 8) Select a ball bearing for a machine for continuous 24-hour service The

machine rotates at the angular speed of 900 rpm The radial force is

Fr ˆ 1 kN, and the thrust force is Faˆ 1:25 kN, with light impact

Table 4.3 Application FactorsKa

SolutionBoth radial, a ˆ 0, and angular, a ˆ 25, ball bearings will be chosen Theequivalent radial force for radial and angular ball bearings is for Fa=Fr ˆ 1:25,

j For 0:35 < Fa=Fr< 10:0 ) Feˆ Fr‰1 ‡ 1:115…Fa=Frÿ 0:35†Š ˆ 2:4 kN(radial bearing)

j For 0:68 < Fa=Fr< 10:0 ) Feˆ Fr‰1 ‡ 0:87…Fa=Frÿ 0:68†Š ˆ 1:8 kN(angular bearing)

From Table 4.3 choose (conservatively) Kaˆ 1:5 for light impact FromTable 4.4 choose (conservatively) 60,000 hour life The life in revolutions is

j From Table 4.2 with 10.55 kN for L00 series ) C ˆ 11:6 kN and

d ˆ 70 mm bore From Table 4.1 with 70 mm bore and L00 series,the bearing number is L14

Table 4.4 Representative Bearing Design Lives

Design life

Machines used intermittently, where service

Machines intermittently used, where reliability is

Machines for continuous 24-hour service where

Source : R C Juvinall and K M Marshek, Fundamentals of Machine Component Design John Wiley & Sons, New York, 1991 Used with permission.

j From Table 4.2 with 10.55 kN for 200 series ) C ˆ 12:0 kN and

d ˆ 55 mm bore From Table 4.1 with 55 mm bore and 200 series,the bearing number is 211

j From Table 4.2 with 10.55 kN for 300 series ) C ˆ 10:6 kN and

d ˆ 35 mm bore From Table 4.1 with 35 mm bore and 300 series,the bearing number is 307

For angular contact bearings the appropriate choices would be L11, 207,and 306

The ®nal selection would be made on the basis of cost of the totalinstallation, including shaft and housing m

EXAMPLE 2 Figure 4.10 shows a two-stage gear reducer with identical pairs of gears An

electric motor with power H ˆ 2 kW and n1ˆ 900 rpm is coupled to theshaft a On this shaft there is rigidly connected the input driver gear 1 withthe number of teeth N1ˆ Np ˆ 17 The speed reducer uses a countershaft bwith two rigidly connected gears 2 and 20, having N2 ˆ Ng ˆ 51 teeth and

N20ˆ Npˆ 17 teeth The output gear 3 has N3ˆ Ng ˆ 51 teeth and is rigidly

®xed to the shaft c coupled to the driven machine The input shaft a andoutput shaft c are collinear The countershaft b turns freely in bearings A and

B The gears mesh along the pitch diameter and the shaft are parallel Thediametral pitch for each stage is Pd ˆ 5, and the pressure angle is f ˆ 20.The distance between the bearings is s ˆ 100 mm, and the distance

l ˆ 25 mm (Fig 4.10) The gear reducer is a part of an industrial machineintended for continuous one-shift (8 hours per day) Select identical extra-light series (L00) ball bearings for A and b

Figure 4.10

SolutionThe pitch diameters of pinions 1 and 20are d1ˆ d20 ˆ dp ˆ Np=Pd ˆ 17=5 ˆ3:4 in The pitch diameters of gears 2 and 3 are d2ˆ d3 ˆ dgˆ Ng=Pdˆ51=5 ˆ 10:2 in The circular pitch is p ˆ p=Pd ˆ 3:14=5 ˆ 0:63 in.

Torque Carried by Each Shaft

The relation between the power Haof the motor and the torque Main shaft ais

The force on the countershaft gear 2 at P is

F12ˆ ÿF21 ˆ Fr 12 ‡ Ft12k ˆ ÿ179:2 ‡ 492:34k N: …4:30†The forces on the countershaft pinion 20at R are three times as large, that is,

Ft0 ˆMb

rp ˆ

63:660:0431ˆ 1477 N

Fr0 ˆ Ft0tan f ˆ 1477 tan 20 ˆ 537:6 Nand

F320ˆ Fr 320 ‡ Ft320k ˆ ÿ537:6 ÿ 1477k N: …4:31†The unknown forces applied to bearings A and B can be written as

Figure 4.11