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PowerPoint Images Chapter 11
Mechanical Engineering Design Seventh Edition
Shigley • Mischke • Budynas Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Mechanical Design Lab. with Advanced
Materials
Chapter 11 Rolling –Contact Bearings Rolling-contact Rolling contact (or rolling) bearings, bearings antifriction bearings transfer load through elements in rolling contact rather than sliding contact. contact The starting friction is about twice the running friction, but still till is i negligible li ibl in i comparison i with ith the th starting t ti friction f i ti off a sleeve bearing. Load, speed, and the operating viscosity of the lubricant do affect the frictional characteristics of a rolling bearing. Bearings the designers specify have already been designed. Bearing specialists must consider fatigue loading, friction, heat, corrosion, kinematic problems, material properties, lubrication, machining tolerances, assembly, use, and cost. Mechanical Design Lab. with Advanced Materials
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Chapter 11 Rolling –Contact Bearings With i h the h rollers ll usedd by b the h Assyrians to move massive stones in i 1100 BC. And A d llater, with i h crude d cart wheels, man strived to overcome friction’s f i i ’ drag. d The simple Th i l ball b ll bearings b i for f 19th century bicycles marked man’s ’ fi first important i victory. i Fi Figure. Th evolution The l ti off rolling lli bearings. b i Mechanical Design Lab. with Advanced Materials
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Chapter 11 Rolling –Contact Bearings
Figure. A large spherical roller bearing assembly for a steel rolling mill application.
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Chapter 11 Rolling –Contact Bearings
Figure A slewing bearing for an English Channel tunneling machine. Figure. machine Mechanical Design Lab. with Advanced Materials
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Chapter 11 Rolling –Contact Bearings Figure. Schematic Fi S h ti diagram of the internal d i off the design th slewing l i bearing of previous figure. fi (a) inner ring, ring (b) outer ring, (c) rollers, rollers (d) spacers, (e) threaded holes, holes (f) integral seals (SKF photograph) photograph). Mechanical Design Lab. with Advanced Materials
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Chapter 11 Rolling –Contact Bearings
Figure. A single-row i l d deep-groove ( (Conrad) d) assembly, bl radial di l ball bearing. Mechanical Design Lab. with Advanced Materials
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Chapter 11 Rolling –Contact Bearings
Figure. Diagram illustrating the method of assembly of a deepgroove (Conrad-type) ball bearing. Figure. g Photograph g p showingg the deep-groove pg (Conrad-type) ball bearing components just prior to snapping the inner ring to the position concentric with the outer ring. Mechanical Design Lab. with Advanced Materials
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Chapter 11 Rolling –Contact Bearings
Figure. A single-row i l d deepgroove ball bearing assembly h i a spherical having h i l outer surface to make it externally aligning. li i
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Chapter 11 Rolling –Contact Bearings
Figure. A single-row Fi i l d deep-groove b ll bearing ball b i having h i two t seals to retain lubricant (grease) and prevent ingress of dirt i t the into th bearing b i Mechanical Design Lab. with Advanced Materials
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Chapter 11 Rolling –Contact Bearings
Figure. A single-row deep-groove ball bearing having two shields to exclude dirt from the bearing. bearing Mechanical Design Lab. with Advanced Materials
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Chapter 11 Rolling –Contact Bearings
Figure. A single-row single row deep-groove deep groove ball bearing assembly having shields and seals. This bearing might be used in an agricultural application. Mechanical Design Lab. with Advanced Materials
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Chapter 11 Rolling –Contact Bearings
Figure. An angular angular-contact contact ball bearing. Mechanical Design Lab. with Advanced Materials
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Chapter 11 Rolling –Contact Bearings A radial di l ball b ll bearing b i is i generally designed to have a di diametral l clearance l in i the h noload state with an axial play. Removal of this axial freedom causes the t e ball-raceway ba aceway co contact tact to assume an oblique angle with thee radial d plane; p e; hence, e ce, a contact co c angle different from zero degrees will occur.
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Chapter 11 Rolling –Contact Bearings
Figure. Angularcontact ball b ll bearings. b i
(a) Small angle
(b) Large angle
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11-1 Bearing Types (1) Fig. 11.1 Nomenclature of a ball bearing Bearings are manufactured to take pure radial loads, loads pure thrust loads, or a combination of the two kinds of loads. Mechanical Design Lab. with Advanced Materials
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11-1 Bearing Types (2)
Fig. 11.2
Wrong drawing
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11-1 Bearing Types (3)
Shield: It is not a complete closure, closure but does offer a measure of protection against dirt. Seals: When the seals are on both sides, the bearings are lubricated at the factory, and supposed to be lubricated for life. Instrument bearings: They are high-precision and available in stainless steel and high-temperature materials. materials Non-precision bearings: They are used with no separator and sometimes having split or stamped sheet-metal races. Ball bushings: They are used with either rotation or sliding motion or both. Bearings are often used with flexible rollers. Mechanical Design Lab. with Advanced Materials
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11-1 Bearing Types (4) Roller bearings require almost perfect geometry of the raceways not to make rollers skew and get out of line, which requires heavy retainers.
Fig. 11.3
Helical rollers are made by winding rectangular material into rollers after which they are rollers, hardened and ground. Because of the inherent flexibility flexibility, they will take considerable misalignment misalignment. Mechanical Design Lab. with Advanced Materials
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11-2 Bearing Life (1)
If a bearing is clean and properly lubricated lubricated, is mounted and sealed against the entrance of dust and dirt, is maintained in this condition, and is operated at reasonable temperatures, then metal fatigue will be the only cause of failure. Under ideal conditions, the fatigue failure consists of spalling of the load-carrying surfaces. The ABMA (American Bearing Manufactures Association) standard states that the failure criterion is the first evidence of fatigue. g The fatigue criterion used by the Timken Company laboratories is the spalling or pitting of an area of 0.01 in2. Timken company also observes that the useful life of the bearing may extend considerably beyond this point. Mechanical Design Lab. with Advanced Materials
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11-2 Bearing Life (2) Rating life, minimum life, L10 life, or B10 life: number of revolutions (or hours at a constant speed) that 90 % of a group off bearings b i will ill achieve hi or exceedd before b f failure. f il Median M di lif life: 50th percentile til life lif off a group off bearings. b i (4 ~5 times the L10 life) Common life measures (bearing life-stochastic variable) (1) Number N b off revolutions l ti off th the inner i ring i (outer ( t ring i stationary) until the first tangible evidence of fatigue.
(2) Number of hours of use at a standard angular speed until the first tangible g evidence of fatigue. g Mechanical Design Lab. with Advanced Materials
11-3 Bearing Load Life at Rated Reliability (1) The load - life function at 0.90 reliability (or at constant reliability) FL1 /a = constant
(11 - 1)
a = 3 for ball bearings g a = 10/3 for roller bearings ( li di icall andd ttapered roller) (cylindiri ll ) F 1L11/a = F 2L12/a
(11 - 2)
Fig. 11.4 Mechanical Design Lab. with Advanced Materials
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11-3 Bearing Load Life at Rated Reliability (2) A bearing manufacturer may choose a rated cycle value of 106 revolutions ((or in the case of the Timken Company, p y 90·106 revolutions) Otherwise declared in the manufacture Otherwise, manufacture’ss catalog as their ratingg life- catalogg load ratingg C10.
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11-3 Bearing Load Life at Rated Reliability (3) C10 (LR nR 60 )
1/ a
= FD (LD nD 60 )
1/ a
1/ a
⎛ LD nD 60 ⎞ ⎟⎟ C10 = FD ⎜⎜ ⎝ LR nR 60 ⎠ C10 : catalog t l rating ti (lbf or kN)
(11 - 3)
FD : desired radial load (lbf or kN) LR : rating life (hrs) LD : desired life (hrs) nR : rating speed (rev/min) nD : desired d i d speedd (rev/min) ( / i ) Mechanical Design Lab. with Advanced Materials
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Example 11-1 Consider SKF, SKF which rates its bearings for 11.0 0·10 106 revolutions, so that L10 life is 60LRnR=106 revolutions. However, Timken uses 90· 106 revolutions.
If you desire a life of 5000 h at 1725 rpm with a load of 400 lbf with a reliability of 90 %, for which catalog rating would you search in an SKF catalog? From Eq. (11 - 3), ⎛ LD nD 60 ⎞ ⎡ 5000(1725)60 ⎤ ⎟⎟ = 400 ⎢ = 3211 lbf = 14.3 kN C10 = FD ⎜⎜ 6 ⎥ 10 ⎦ ⎣ ⎝ LR nR 60 ⎠ You are to select a bearing whose C10 rate is larger 1 /a
1/ 3
than 3211 lbf (14.3 kN) from the SKF catalog. Mechanical Design Lab. with Advanced Materials
11-4 Bearing Survival: Reliability versus Life
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At constant t t load, l d the th Weibull W ib ll is i by b far f the th mostt popular, l largely l l because b of its ability to adjust to varying amounts of skewness.
⎡ ⎛ x − x ⎞b ⎤ 0 ⎟⎟ ⎥ R = exp⎢− ⎜⎜ ⎢⎣ ⎝ θ − x0 ⎠ ⎥⎦ where
(11 - 4)
R = reliability. y x = life measure dimensionless variate, L/L10 . x0 = guaranteed d, or minimum i i value l off x.
θ = characteristic parameter corresponding to the 63.2 % of x. ⎡ ⎛ x − x ⎞b ⎤ 0 ⎟⎟ ⎥ F = 1 − R = 1 − exp⎢− ⎜⎜ ⎢⎣ ⎝ θ − x0 ⎠ ⎥⎦
(11 - 5) Mechanical Design Lab. with Advanced Materials
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Example 11-2 (1) Consider C id the h distribution di ib i properties i off a 02-30 02 30 mm ddeepgroove ball bearing if the Weibull parameters are x0=0.02, (θ − x0) = 44.439, 439 andd b = 1.483. 1 483 Fi Findd the h mean, median, di 10th percent life, standard deviation, and coefficient of variation. Solution : From Eq. (2 - 28) and Table A - 34, 1 ⎞ ⎛ 1⎞ ⎛ μx = x0 + (θ − x0 )Γ⎜1 + ⎟ = 0.02 + 4.439Γ⎜1 + ⎟ = 4.033 ⎝ b⎠ ⎝ 1.483 ⎠ From Eq. (2 - 26) 1/ b
x0.50
⎛ 1⎞ = x0 + (θ − x0 )⎜ ln ⎟ ⎝ R⎠
1 / 1.483
⎛ 1 ⎞ = 0.02 + 4.439⎜ ln ⎟ ⎝ 0.5 ⎠
= 3.487
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Example 11-2 (2) The 10th percentile life x is 1 / 1.483
x0.10
L10 ⎛ 1 ⎞ = = x0 + (θ − x0 )⎜ ln ⎟ L10 ⎝ 0.9 ⎠
1 / 1.483
⎛ 1 ⎞ = 0.02 + 4.439⎜ ln ⎟ ⎝ 0.9 ⎠
≅ 1 ((as it should be)) From Eq. (2 - 29) 1/ 2
1 ⎞⎤ ⎡ ⎛ 2⎞ 2⎛ σˆ x = (θ − x0 )⎢Γ⎜1 + ⎟ − Γ ⎜1 + ⎟⎥ ⎝ b ⎠⎦ ⎣ ⎝ b⎠
1/ 2
2 ⎞ 1 ⎞⎤ ⎡ ⎛ 2⎛ = 4.439 ⎢Γ⎜1 + ⎟ − Γ ⎜1 + ⎟⎥ ⎝ 1.483 ⎠⎦ ⎣ ⎝ 1.483 ⎠ σˆ x 2.753 = = 0.683 Cx = μx 4.033
= 2.753
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11-5 Relating Load, Life, and Reliability (2) Fig. 11.5 The catalogue Th l information is plotted as point i A, A whose h coordinates are C10 and d x10=L L10/L10=1, 1 a point on the 0.90 reliability li bili contour.
⎡ ⎛ x − x ⎞b ⎤ 0 ⎟⎟ ⎥ R = exp ⎢− ⎜⎜ θ − x ⎢⎣ ⎝ 0 ⎠ ⎥ ⎦
(11 - 4)
↓
↑
FL1/a = constant
((11 - 1))
=1
The design point is at D, D with the coordinates FD and xD, a point that is on the R=RD reliability contour. The designer must move from point D to a point A via point B as follows. follows Mechanical Design Lab. with Advanced Materials
11-5 Relating Load, Life, and Reliability Along DB (constant reliability contour), 1/ a
⎛ xD ⎞ FB = FD ⎜⎜ ⎟⎟ ( ) (a) ⎝ xB ⎠ Along BA (constant load) ⎡ ⎛ x − x ⎞b ⎤ 0 ⎟⎟ ⎥ RD = exp⎢− ⎜⎜ B ⎢⎣ ⎝ θ − x0 ⎠ ⎥⎦ or 1/ b
⎛ 1 ⎞ ⎟⎟ xB = x0 + (θ − x0 )⎜⎜ ln ⎝ RD ⎠ S b tit ting this Substituti thi into i t Eq. E (a) ( ) yields i ld 1/ a
⎛ ⎞ xD ⎟ FB = F A= C10 = FD ⎜⎜ 1/ b ⎟ ⎝ x0 + (θ − x0 )(ln 1 / RD ) ⎠
((11 - 6))
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11-5 Relating Load, Life, and Reliability
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When F = 1 − RD ≤ 0.1 0 1 (or RD ≥ 0.9) 0 9) ln
1 1 = ln = ln(1 + F + ⋅ ⋅ ⋅) ≅ F = 1 − RD RD 1− F 1/ a
⎛ ⎞ xD ⎟ FB = C10 ≅ FD ⎜⎜ 1/ b ⎟ ⎝ x0 + (θ − x0 )(1 − RD ) ⎠
(11 - 7)
Loads are often nonsteady, so that the desired load is multiplied by an application factor af. The steady load afFD does the same damage as the variable load FD does to the h rolling lli surfaces. f Shafts generally have two bearings. Then R is related to R=RA· RB 0 5. In sizing bearings, bearings one can begin by making RA=RB=(R)0.5 Mechanical Design Lab. with Advanced Materials
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Example 11-3
The design Th d i load l d on a ball b ll bearing b i is i 413 lbf andd an application li i factor of 1.2 is appropriate. The speed of the shaft is to be 300 rpm, the h life lif to bbe 30 kh with i h a reliability li bili off 0.99. 0 99 What Wh is i the h C10 catalog entry to be sought (or exceeded) when searching f a deep-groove for d b i in bearing i a manufacturer’s f ’ catalog l on the h basis of 106 revolutions for rating life ? The Weibull parameters are x0=0.02, 0 02 (θ - x0)=4.439, ) 4 439 andd b=1.483. b 1 483 L 60 LD nD 60(30000 )300 xD = = = = 540 6 L10 60 LR nR 10
For a ball bearing, g, a = 3. From Eq. q ((11 - 7)) 1/ 3
⎛ ⎞ 540 ⎟ FB = C10 ≅ (1.2)(413)⎜⎜ 1 / 1.483 ⎟ ⎝ 0.02 + 4.439(1 − 0.99 ) ⎠
= 6696 lbf
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11-6 Combined Radial and Thrust Loading (1)
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Fr: radial load Fa: axial load Fe: equivalent radial load that does the same damage as the combined radial and thrust loads together. V: rotation factor (V (V=11.0 0 when the inner ring rotates and V V=11.2 2 when the outer ring rotates). Self-aligning Self aligning bearings are an exception (V=1 for rotation of either ring.) Mechanical Design Lab. with Advanced Materials
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11-6 Combined Radial and Thrust Loading (2) Fe =1 VFr
Fa when ≤ e (11 - 8a) VFr
Fe F = X +Y a VFr VFr
when
Fa > e (11 - 8b) VFr
X : the ordinate intercept p Y : the slope of the line for
Fa > e, VFr
Fe = X iVFr + Yi Fa where i = 1 i=2
or (11 - 9)
when when
Fa ≤e VFr Fa > e. VFr
Table 11 − 1 lists values of X 1, Y1, X 2 and Y2 as a function of e,
Fig. 11.6
which in turn is a function of Fa /C0 . Mechanical Design Lab. with Advanced Materials
Table11-1 Equivalent radial load factors for ball bearings
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11-6 Combined Radial and Thrust Loading (3) Fig. 11.7
The ABMA has established standard boundary conditions for bearings, which define the bearing bore, the OD, the width, and the fillet sizes on the shaft and housing shoulders. Mechanical Design Lab. with Advanced Materials
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11-6 Combined Radial and Thrust Loading (4)
Fig. Size comparison of deep-groove ball bearing dimension series.
The bearings are identified by a two-digit two digit number called the dimension-series code: The first number is the width series, series 0,1,2,3,4,5, 0 1 2 3 4 5 and 6. 6 The second number is the diameter series (outside), 8 9 0 1 2 3 and 44. 8,9,0,1,2,3, Mechanical Design Lab. with Advanced Materials
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Table 11-2 Dimensions and load ratings for single-row 02-series deep-groove and angular-contact ball bearings
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Table 11-3 Dimensions and basic load ratings for cylindrical roller bearings
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11-6 Combined Radial and Thrust Loading (5) The housing Th h i andd shaft h f shoulder diameters listed in the h tables bl should h ld be b usedd whenever possible to secure adequate d support for f the h bearing and to resist the maximum i thrust h loads. l d
Fig. 11.8 Fig 11 8 Shaft and housing shoulder diameters ds and dH to ensure good bearing support.
Table b e 11-4 co containss recommendations on bearing life for some classes of machinery and Table 11-5 for the load-application pp factors. Mechanical Design Lab. with Advanced Materials
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Table 11-4 Bearing –life recommendations for various classes of machinery
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Table 11-5 Load-application factors
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11-6 Combined Radial and Thrust Loading (6)
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The static load rating C0 of a rolling element bearing is defined as that load applied to a nonrotating bearing that will result in permanent deformation of 0.01 % at the weaker of the inner or outer raceway contacts occurring at the position of the maximum loaded rolling element. For a ball bearing of 10 mm diameter this corresponds to a plastic deformation of the two bodies of 1 μm. Experience has shown that such minor deformations neither impair the quiet running of a bearing at normal loading, nor shorten its fatigue life. Mechanical Design Lab. with Advanced Materials
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11-6 Combined Radial and Thrust Loading (6) The static load rating is given in bearing catalog tables. C0 = M ⋅ nb ⋅ d b2
(ball be (b bearings) gs)
C0 = M ⋅ n r ⋅ lc ⋅ d
(roller bearings)
where C0 = bearing static load rating, lbf (kN) nb = number of balls. n r = number of rollers
Values of M Type
in, lbf
mm, kN
d b = diameter of balls, in (mm)
Radial ball
1.78·103
5.11·103
d = diameter of rollers, rollers in (mm)
Ball thrust
7.10·103
20.4·103
Radial roller
3.13·103
8.99·103
Roller thrust
14 2·10 14.2 103
40 7·10 40.7 103
lc = length of contact line, in (mm)
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Example 11-4 (1)
An SKF 6210 angular-contact ball bearing has an axial load Fa of 400 lbf and a radial load Fr of 500 lbf applied with the outer ring stationary. The basic static load rating C0 is 4450 lbf and the basic load rating C10 is 7900 lbf. Estimate the L10 life at a speed of 720 rpm. Solution : V = 1 and Fa /C0 = 400/4450 = 0.090. Interpolate for e in Table11 - 1 : Fa /C0
e
0.084 0 090 0.090
0.28 e
0.110
0.30
from which e = 0.285 0 285
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Example 11-4 (2) Fa / (VFr ) = 400/(1⋅ 500) = 0.8 > e = 0.285. Thus X 2 = 0.56 from Table 11 - 1, and Fa /C0
Y2
0.084
1.55
0.090
Y2
from which Y2 = 1.527
0.110 1.45 From Eq. q ((11 - 9), ) Fe = X 2VFr + Y2 Fa = 0.56(1)500 + 1.527(400) = 890.8 lbf With LD = L10 and FD = Fe , from Eq. q ((11 - 3)) 60 LR nR L10 = 60nD
a
3
⎛ C10 ⎞ 106 ⎛ 7900 ⎞ ⎜⎜ ⎟⎟ = ⎜ ⎟ = 16150 h 60(720) ⎝ 890.8 ⎠ ⎝ Fe ⎠ Mechanical Design Lab. with Advanced Materials
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11-7 Variable Loading (1) Bearing B i loads l d are frequently f tl variable and occur in some id tifi bl patterns: identifiable tt 1. Piecewise constant l di in loading i a cyclic li pattern tt 2. Continuously variable loading in a repeatable cyclic c clic pattern 3 Random variation. 3. ariation Equation (11-1) (11 1) can be written: FaL=constant=K
Fig. 11.9: The linear damage hypothesis says that in the case of load F1, the area under the curve from L=0 to L=LA is a measure of a D = F the damage 1 LA . Mechanical Design Lab. with Advanced Materials
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11-7 Variable Loading (2) The damage Th d done d by b loads l d Fe1, Fe2, and Fe3 is D = Fea1 l1 + Fea2 l 2 + Fea3l 3
(b)
where l i is the number of revolutions at life Li . The equivalent steady load Feq when run for l1 +l 2 +l 3 revolutions does the same damage D. D = Feqa (l1 + l 2 + l 3 )
(c)
From (b) and (c) ⎛F l +F l +F l ⎞ ⎟⎟ Feq = ⎜⎜ l1 + l 2 + l 3 ⎝ ⎠ a e1 1
a e2 2
= (∑ f i Feia )
1 /a
a e3 3
1 /a
Fig. 11.10
(11 - 10)
where f i is the fraction of revolution run upp under load Fei . Mechanical Design Lab. with Advanced Materials
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11-7 Variable Loading (3) Si Since l i = ni ti , where h ni is i the th rotationa t ti l speed d att load l d Fei and ti is the duration of that speed, ⎛ ∑ ni ti F ⎞ ⎟ ((11 - 11)) Feq = ⎜⎜ ⎟ ⎝ ∑ ni ti ⎠ When the character of the individual loads change, g , Eq. q ((11 - 10)) a ei
1 /a
can be written with an application factor a fi prefixed to each Fei Feq =
[∑ f (a
K Leq = a Feq
i
fi
Fei
)]
a 1 /a
(11 - 12)
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11-7 Variable Loading (4)
F Leq = F l + F l + F l a eq
a e1 1
a e2 2
a e3 3
Since K = Fea1 L1 = Fea2 L2 = Fea3 L3 , K =F l +F l +F l a e1 1
a e2 2
a e3 3
li K K K = l 1+ l 2+ l3 = K∑ L1 L2 L3 Li or
li ∑ L =1 i
((11 - 13))
Palmgren − Miner Rule. Mechanical Design Lab. with Advanced Materials
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Example 11-5
A ball bearing is run at four piecewise continuous steady loads as shown in the following table. Columns (1), (2), and (5) to (8) are given. given Find the equivalent load Fe. (1) (2) (3) (4) (5) (6) (7) (8) (9) Time fraction
rpm
01 0.1
2000
200
0.1
3000
03 0.3 0.5
(1)ⅹ(2) (3)/Σ(3)
Fri
Fai
Fei
afi
afiFei
0 077 0.077
600
300
794
1 10 1.10
873
300
0.115
300
300
626
1.25
795
3000
900
0 346 0.346
750
300
878
1 10 1.10
966
2400
1200 2600
0.462 1 000 1.000
375
300
668
1.25
835
From Eq. (11 - 10), with a = 3, the equivalent load Feq is
[
Feq = 0.077(873) + 0.115(795) + 0.346(966) + 0.462(835) 3
3
3
]
3 1/ 3
= 884 lbf
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11-7 Variable Loading (5) Th differenti The diff tiall damage d d done b Fa by during rotation through the angle dθ (i.e., a cam) is φ
D = ∫ dD = ∫ F a dθ = Feqaφ 0
1/ a
⎡1 φ a ⎤ Feq = ⎢ ∫ F dθ ⎥ ⎣φ 0 ⎦ K Leq = a (11 - 14) Feq
Fig. 11.11 Mechanical Design Lab. with Advanced Materials
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Example 11-6 (1)
The operation of a particular rotary pump involves a power demand of P = P + A′ sin θ where P is the average power. The bearings feel the same variation as F = F + A sin i θ. Develop an application factor a f for this applicatio pp n of ball bearings. g Mechanical Design Lab. with Advanced Materials
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Example 11-6 (1) Solution : From Eq. (11 - 14), with a = 3, ⎛ 1 Feq = ⎜ ⎝ 2π ⎡ 1 =⎢ ⎣ 2π
∫
2π
0
∫
2π
0
1/ a
⎞ F dθ ⎟ ⎠ a
⎡ 1 =⎢ ⎣ 2π
⎤ ( ) ∫ F + A sin θ dθ ⎥⎦ 2π
3
1/ 3
0
1/ 3
⎤ F 3dθ + 3F 2 A∫ sin θdθ + 3F A2 ∫ sin 2 θdθ + A3 ∫ sin 3 θdθ ⎥ 0 0 0 ⎦ 2π
⎡ 1 ⎤ = ⎢ (2πF 3 + 0 + 3πF A2 + 0 )⎥ ⎣ 2π ⎦
2π
1/ 3
⎡ 3⎛ A⎞ = F ⎢1 + ⎜ ⎟ ⎢⎣ 2 ⎝ F ⎠
2π
2 1/ 3
⎤ ⎥ ⎥⎦
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Example 11-6 (2) In terms off F , the h applicatio li i n factor f i is 2 1/ 3
⎡ 3⎛ A⎞ ⎤ a f = ⎢1 + ⎜ ⎟ ⎥ ⎢⎣ 2 ⎝ F ⎠ ⎦⎥ W can present the We th result lt in i tabular t b l form f : A/F
af
0 0.2
1 1.02
0.4
1.07
06 0.6
1 15 1.15
0.8
1.25
1.0
1.36 Mechanical Design Lab. with Advanced Materials
11-8 Selection of Ball and Cylindrical Roller Bearings (1) Example E l 11-7 11 7 The second shaft on a parallel-shaft ll l h f 25 25-hp h foundry crane speed reducer d contains i a Roller bearing helical gear with a pitch ↓ di diameter off 8.08 8 08 in. i Helical e c gears ge s transmit s components of force in the tangential, g , radial,, and axial directions.
↑ Ball bearing
af =1.2 Fig. 11.12
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11-8 Example 11-7 (2) The components of the gear force transmitted to the second shaft are shown in Fig. 11-12, at point A. The bearing reactions at C and D, D assuming simple-supports, simple supports are also shown. A ball bearing is to be selected at C to accept the thrust, thrust and a cylindrical roller bearing is to be utilized at D. The life goal of the speed reducer is 10 kh, kh with R ≥ 0.96 0 96 for the ensemble of all four bearings (both shafts) for the Weibull parameters of Ex. Ex 11-3. 11 3 The application factor is to be 1.2. (1) Select the roller bearing for location D. D (2) Select the ball bearing (angular contact) for location C, assuming the inner ring rotates rotates. Mechanical Design Lab. with Advanced Materials
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11-8 Example 11-7 (3)-Solution The torque transmitted is T = 595(4.04) 595(4 04) = 2404 lbf ⋅ in. in The speed at the rated horse - power, given by Eq. (4 - 40) is 63025H 63025( 25) nD = = = 655.4 rpm 2404 T ( H = 550 lbf ⋅ ft/s = 550 ⋅ 12 ⋅ 60 lbf ⋅ in/min = 550 ⋅ 12 ⋅ 60/ (2π ) lbf ⋅ in ⋅ rpm = 63025 lbf ⋅ in ⋅ rpm ) The radial load at D is 106.6 2 + 297.52 = 316.0 lbf The radial load at C is 355.6 2 + 297.52 = 464.4 lbf
The individual bearing reliabilities, if equal, must be at least 4 0.96 = 0.99. The dimensionless design life for both bearings is L 60 LD nD 60(10000)655.4 = = = 393.2 xD = 6 L10 60 LR nR 10 Mechanical Design Lab. with Advanced Materials
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11-8 Example 11-7 (4) ((a)) From F E Eq. (11 - 7), 7) the th Weibull W ib ll parameter t s off Ex. E 11 - 3, 3 an application factor of 1.2, and a = 10/3 for the roller bearing at D, the catalog rating should be equal to or greater than ⎡ ⎤ xD C10 = a f FD ⎢ 1 /b ⎥ ⎣ x0 + (θ − x0 )(1 − RD ) ⎦
1 /a
3 / 10
⎡ ⎤ 393.2 = 16.0 kN = 1.2(316.0) ⎢ 1 / 1.483 ⎥ ⎦ ⎣ 0.02 + 4.439(1 − 0.99 ) The absence of a thrust component makes the selection procedure simple. Choose a 02 - 25 mm series, or a 03 - 25 mm series cylindrical roller bearings g from Table 11 - 3. Mechanical Design Lab. with Advanced Materials
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11-8 Example 11-7 (6) component. (b) The Th ball b ll bearing b i at C involves i l a thrust h This selection procedure requires an iterative procedure. Assuming Fa /(VFr ) > e, 1 Choose Y2 from Table 11 - 1. 1. 1 2. Find C10 .
3 Tentativel 3. T i ly identify id if a suitable i bl bearing b i from f T bl 11 - 2, Table 2 note C0 . 4. Using Fa /C0 enter Table 11 - 1 to obtain a new value of Y2 . 5. Find C10 . 6. If the same bearing is obtained, 6 obtained stop. stop 7. If not, take next bearing and go to step 4. Mechanical Design Lab. with Advanced Materials
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11-8 Example 11-7 (7) As a first approximation, take the middle entry from Table 11 - 1 : X 2 = 0.56
Y2 = 1.63
From Eq. (11 - 8b), with V = 1, Fe 344 Y Fa =X+ = 0.56 + 1.63 = 1.77 V Fr VFr (1)464.4 Fe = 1.77VFr = 1.77(1)464.4 = 822 lbf
or 3.66 kN
From Eq. (11 - 7), with a = 3, 1/ 3
⎡ ⎤ 393 2 393.2 C10 = 1.2(3.66) ⎢ = 53.4 kN 1 / 1.483 ⎥ ⎦ ⎣ 0.02 + 4.439(1 − 0.99 ) From Table 11 - 2, angular - contact bearing 02 - 60 mm has C10 = 55.9 kN. C0 is 35.5 kN. Mechanical Design Lab. with Advanced Materials
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11-8 Example 11-7 (8) St 4 becomes, Step b with ith Fa in i kN, kN Fa 344( 4.45)10− 3 = = 0.0431 C0 35.5
which makes e from Table 11 - 1 approximately 0.24. 0 24 Now Fa /(VFr ) = 344/(1 ⋅ 464.4) = 0.74, which is greater than 0.24, so we find Y2 by interpolation : Fa /C0
Y2
0.042 0 0431 0.0431
1.85 Y2
0.056
1.71
f from which hi h Y2 = 1.84 1 84
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11-8 Example 11-7 (9) From Eq. Eq (11 - 8b) Fe 344 = 0.56 + 1.84 = 1.92 VFr 464.4 Fe = 1.92VFr = 1.92(1)464.4 = 892 lbf
or 3.97 kN
The prior calculation for C10 changes only in Fe , so 3.97 C10 = 53.4 = 57.9 kN 3 66 3.66 From Table 11 - 2, an angular contact bearing 02 - 65 mm has C10 = 63.7 kN and C0 = 41.5 kN. Again Fa 344(4.45)10 −3 = = 0.0369 C0 41.5 which makes e from Table 11 - 1 approximately 0.23. 0 23 Mechanical Design Lab. with Advanced Materials
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11-8 Example 11-7 (10) Now Fa /(VFr ) = 344/(1 ⋅ 464.4) 464 4) = 0.74, 0 74 which is greater than 0.23, 0 23 so we find Y2 by interpolation : Fa /C0
Y2
0.028 0.0369
1.99 Y2
0.042
1.85
from which Y2 = 1.90
From Eq. (11 - 8b) 344 Fe = 0.56 + 1.90 = 1.967 VFr 464.4 Fe = 1.967VFr = 1.967(1)464.4 = 913.5 lbf
or 4.065 kN
Since C10 changes only in Fe , so 44.065 065 53.4 = 59.4 kN 3.66 From Table 11 - 2, an angular contact bearing 02 - 65 mm is still selected,
C10 =
so the iteration is complete. Mechanical Design Lab. with Advanced Materials
11-9 Selection of Tapered Roller Bearings (1) Four Components: Cone (inner ring) Cupp (outer ( ring)g) sliding surface Tapered p rollers Cage (spacer, retainer) Fig. g 11.13 Mechanical Design Lab. with Advanced Materials
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11-9 Selection of Tapered Roller Bearings (2) A tapered roller bearing can carry both radial and thrust loads. loads The radial load will induce a thrust reaction within the bearing because of the taper. To avoid the separation of the races and the rollers, rollers this thrust must be resisted by an equal and opposite force(At least two tapered roller bearings on a shaft should be used).
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11-9 Selection of Tapered Roller Bearings (3)
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Fig. 11.14
The radial Th di l loads l d act perpendicular to the shaft axis i through h h points i A0 andd B0. The geometric spread ag for the direct mounting (cone backss facing bac ac g each eac other) ot e ) iss greater than for indirect mounting ou g (co (conee fronts o s facing c g each other), however, the system y stability y is the same (ae is the same in both cases).
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11-9 Selection of Tapered Roller Bearings (4) A radial load will induce a thrust reaction reaction. The load zone includes about half the rollers and subtends an angle of approximately 180o. Using g the symbol y Fa(180) for the induced thrust load from a radial load with a 180o load zone, Timken provides
0.47 Fr Fa (180 ) = (11 - 15) K where K is a geometric factor K = 0.389co 0.389cotα0 where α0 is half the included cup angle. Mechanical Design Lab. with Advanced Materials
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11-9 Selection of Tapered Roller Bearings (5)
69
From Eq. Eq (11 - 7) for a three - parameter Weibull model, model 1/ a
⎛ ⎞ xD ⎜ ⎟ C10 ≅ FD ⎜ 1/ b ⎟ ⎝ x0 + (θ − x0 )(1 − RD ) ⎠
or a
1 /b ⎛ C10 ⎞ ⎟⎟ xD = x0 + (θ − x0 ) (1 − RD ) ⎜⎜ ⎝ FD ⎠ Timken uses a two - pparameter Weibull model with x0 = 0, θ = 4.48, and b = 3/2.
For a Timken tapered roller bearing with a = 10/3, and L10 = 90 ⋅ 106 revolutions 10 / 3
2 / 3 ⎛ C10 ⎞ 6 ⎟⎟ xD = LD / 90 ⋅ 10 = 4.48 (1 − RD ) ⎜⎜ ⎝ FD ⎠
LD = 4.48 (1 − RD )
2/3
10 / 3
⎛ C10 ⎞ ⎜⎜ ⎟⎟ ⎝ FD ⎠
90 ⋅ 106
Mechanical Design Lab. with Advanced Materials
11-9 Selection of Tapered Roller Bearings (6) Ti k wriites this Timken hi equation i as ⎛ C10 ⎞ ⎟⎟ LD = a1a2 a3a4 ⎜⎜ ⎝ FD ⎠
10 / 3
90 ⋅ 106
(11 - 16)
a1 = 4.48(1 − RD )
2/3
a 2 = bearing material a3 = a3k a3l a3m a3k = load zone a3l = lubricant ( = fT f v ), ) Temperature factor fT in Fig. 11 − 16, Viscosity factor f v in Fig. 11 − 17. a3m = alignment a4 = 1 (spall size is 0.01in 0 01in 2 ) Mechanical Design Lab. with Advanced Materials
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11-9 Selection of Tapered Roller Bearings (7)
71
For the usual case, case a2 = a3k = a3m = 1, ⎡ ⎤ LD C10 = a f P ⎢ (11 - 17) 6 ⎥ 2/3 ⎣ 4.48 fT f v (1 − RD ) 90 ⋅ 10 ⎦ where FD is replaced by a f P and LD is in revolutions The load P is the dynamic equivalent load of the combination Fr and Fa of Sec. 11 - 6. The particular values of X and Y are given in Table 11 - 6.
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11-9 Selection of Tapered Roller Bearings (8)
Fig 11.15 Fig. 11 15
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11-9 Selection of Tapered Roller Bearings (9)
Fig. 11.15
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11-9 Selection of Tapered Roller Bearings (10) Fig. 11.16
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11-9 Selection of Tapered p Roller Bearings (11) Fig. i 11.17 11 1
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Table 11-6 Dynamic equivalent radial load equations for P76 ((from Timken Company) p y)
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77 Table 11-6 Dynamic equivalent radial load equations for P (from Timken Company)
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78 Table 11-6 Dynamic equivalent radial load equations for P (from Timken Company)
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Example 11-8 (2)
The shaft Th h ft depicted d i t d in i Fig. Fi 11-18a 11 18 carries i a hhelical li l gear with ith a tangential force of 3980 N, a separating force of 1770 N, and a thrust th t force f off 1690 N att th the pitch it h cylinder li d with ith directions di ti shown. Th pitch The it h diameter di t off the th gear is i 200 mm. The shaft runs at a speed of 1050 rpm, and the span (effective spread) bet between een the direct-mount di t t bearings is 150 mm mm. The design life is to be 5000 h and an application factor of 1 is appropriate appropriate. The lubricant will be ISO VG 68 (68 cSt at 40 oC) oil with an estimated operating temperature of 55 oC. C If the reliability of the bearing set is to be 0.99, select suitable single row tapered single-row tapered-roller roller Timken bearings. bearings Mechanical Design Lab. with Advanced Materials
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Example 11-8 (1)
Fig. 11.18
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Example 11-8 (3) Solution : The reactions in the xz plane are 3980(50) RzA = = 1327 N 150 3980(100) RzB = = 2653 N 150 The reactions in the xy plane are 1770(50) 1690(100) + = 1727 N R yA = 150 150 1770(100) 1690(100) R yB = − = 53.3 N 150 150 2 zA
2 1/ 2 yA
FrA = (R + R
= (1327 + 1717
2 zA A
2 1/ 2 yA
2
FrB = (R + R
) )
2
= (2653
) + 53.3 )
2 1/ 2
= 2170 N
2 1/ 2
= 2654 N
Trial 1 : For the left bearing, we will use K A = K B = 1.5 to start. From Table 11 - 6,, notingg that m = +1 for direct mounting g and Fae to the right g is p positive Mechanical Design Lab. with Advanced Materials
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Example 11-8 (4) 0.47 FrA 0.47 FrB − mFae KB KA
0.47(2170) 0.47(2654 ) − (+1)(− 1690) 1 .5 1.5 680 < 2522 Th f Therefore, we use the th upper sett off equations ti i Table in T bl 11 - 6 to t find fi d the th thrust th t loads l d : FaA =
0.47 FrB 0.47(2654) − mFae = − (+1)(−1690) = 2522 N 1 .5 KB
FaB =
0.47 FrB 0.47(2654) = = 832 N KB 1.5
The dynamic equivalent PA and PB are PA = 0.4 FrA + K A FaA = 0.4(2170) FrA + 1.5(2522) = 4651 N PB = FrB = 2654 N From Fig.11 - 16 for 1050 rpm at 55o C, fT = 1.31. From Fig.11 - 17, f v = 1.01. For use in Eq. Eq (11 - 16), 16) a3l = fT f v = 1.31(1.01) 1 31(1 01) = 1.32. 1 32 Mechanical Design Lab. with Advanced Materials
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Example 11-8 (5) E i Estimate RD as 0.99 0 99 = 0.995 for f eachh bbeariing. For bearing A, from Eq. (11 - 17) the catalog entry C10 should equal or exceed ⎡ ⎤ LD C10 = a f P ⎢ 2/ 3 6 ⎥ ( )⎦ ( ) . f f − R ⋅ 4 48 1 90 10 T v D ⎣
3 / 10
( LD in revolutions) 3 / 10
⎡ ⎤ 5000(1050)60 C10 = (1)(4651) ⎢ = 11470 N 2/ 3 6 ⎥ ⎣ 4.48 ⋅ 1.32(1 − 0.995) (90 ⋅ 10 )⎦ From Fig.11 - 15, tentatively select 15100 cone and 15245 cup, which will work : K A = 1.67 , C10 = 12100 N For bearing B, from Eq. (11 - 17) the catalog entry C10 should equal or exceed 3 / 10
⎡ ⎤ 5000(1050)60 C10 = (1)(2654) ⎢ = 6540 N 2/ 3 6 ⎥ ⎣ 4.48 ⋅ 1.32(1 − 0.995) (90 ⋅ 10 )⎦ Tentatively select the bearing identical to bearing A, which will work : K B = 1.67 , C10 = 12100 N Mechanical Design Lab. with Advanced Materials
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Example 11-8 (6) Trial 2 : Use K A = K B = 1.67 from the tentative bearing selection. 0.47 FrA 0.47 FrB − mFae KB KA
0.47(2170) 0.47(2654 ) − (+1)(− 1690) 1.67 1.67 611 < 2437 Th f Therefore, we use the th upper sett off equations ti i Table in T bl 11 - 6 to t find fi d the th thrust th t loads l d : FaA =
0.47 FrB 0.47(2654) − mFae = − (+1)(−1690) = 2437 N 1.67 KB
FaB =
0.47 FrB 0.47(2654) = = 747 N 1.67 KB
The dynamic equivalent PA and PB are PA = 0.4 FrA + K A FaA = 0.4(2170) + 1.67(2437) = 4940 N PB = FrB = 2654 N Mechanical Design Lab. with Advanced Materials
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Example 11-8 (7) For bearing A, from Eq. Eq (11 - 17) 3 / 10
⎡ ⎤ 5000(1050)60 C10 = (1)(4938) ⎢ = 12170 N 2/ 3 6 ⎥ ⎣ 4.48 ⋅ 1.32(1 − 0.995) (90 ⋅ 10 )⎦ Although this catalog entry exceeds slightly the tentative selection for bearing A, we will keep it. For bearing B, PB =F rB= 2654 N. 3 / 10
⎡ ⎤ 5000(1050)60 C10 = (1)(2654) ⎢ = 6543 N 2/ 3 6 ⎥ ⎣ 4.48 ⋅ 1.32(1 − 0.995) (90 ⋅ 10 )⎦ Select cone and cup 15100 and 15245, respectively, for both bearing A and B. From Fig.11 - 15 the effective load center is located at a = −5.8 mm. Thus the shoulder - to - shoulder dimension should be 150 − 2(5.8) = 138. 4 mm.
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11-10 Design Assessment for Selected Rolling-Contact Bearings (1)
86
Design D i A Assessment ffor a R Rolling lli Contact C Bearing B i includes, i l d at a minimum: (1) Bearing reliability for the load imposed and life expected. (2) Shouldering Sh ld i on shaft h f andd housing h i satisfactory. if (3) Jou Journal a finish, s , ddiameter a ete aand d to tolerance e a ce co compatible. pat b e. (4) Housing finish, diameter and tolerance compatible. (5) Lubricant type according to manufacturer’s recommendations;; lubricant paths p and volume supplied pp to keep operating temperature satisfactory. (6) Preloads, P l d if required, i d are supplied. li d Mechanical Design Lab. with Advanced Materials
11-10 Design Assessment for Selected Rolling-Contact Bearings (2)
Bearing Reliability:
From Eq.11 - 10, 1/ a
⎛ ⎞ xD ⎟ FB = C10 = FD ⎜⎜ (11 - 6) 1/ b ⎟ ⎝ x0 + (θ − x0 )(ln 1 / RD ) ⎠ R l i FD with Replacing ith a f FD , the th reliabili li bility t RD = R is i expressedd as b a ⎧ ⎡ ⎫ ⎤ a F ⎛ ⎞ ⎪ ⎢x ⎜ f D ⎟ − x ⎥ ⎪ 0 ⎪ ⎢ D ⎝⎜ C10 ⎟⎠ ⎪ ⎥ R = exp⎨− ⎢ ⎥ ⎬ θ − x 0 ⎪ ⎢ ⎥ ⎪ ⎪ ⎢ ⎥⎦ ⎪ ⎣ ⎩ ⎭
(11 - 18)
Wh R ≥ 0.9, When 0 9 using i Eq. E (11 - 7) 1/ a
⎛ ⎞ xD ⎟ FB = C10 ≅ FD ⎜⎜ 1/ b ⎟ ⎝ x0 + (θ − x0 )(1 − RD ) ⎠
(11 - 7)
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11-10 Design Assessment for Selected Rolling-Contact Bearings (3)
[
⎡ x D (a f FD )/ C10 R ≅ 1− ⎢ θ − x0 ⎢⎣
]
a
− x0 ⎤ ⎥ ⎥⎦
88
b
(11 - 19)
In tapered roller bearings, or other bearings for a two - parameter Weibull distribution, E (11 - 18) becomes, Eq. b f x0 = 0 , θ = 4.48, b = 3 / 2, for b ⎧ ⎡ ⎧ ⎡ ⎤ ⎫ ⎪ ⎢ ⎪ ⎢ ⎥ ⎪ ⎪⎪ ⎢ ⎪⎪ ⎢ ⎥ ⎪⎪ xD xD R = exp⎨− ⎢ = exp⎨− ⎢ a ⎥ ⎬ ⎛ C10 ⎪ ⎢ ⎛ C10 ⎞ ⎥ ⎪ ⎪ ⎢ ⎜ ⎟ ⎪ ⎢θ ⎜ a F ⎟ ⎥ ⎪ ⎪ ⎢ 4.48 f T f v ⎜⎜ a F ⎝ f D ⎩⎪ ⎣ ⎝ f D ⎠ ⎦ ⎭⎪ ⎩⎪ ⎣ and Eq. q ((11 - 19)) becomes b
⎤ ⎥ ⎥ 10 / 3 ⎥ ⎞ ⎥ ⎟ ⎥ ⎟ ⎠ ⎦
⎧⎪ ⎫⎪ ⎧⎪ ⎫⎪ xD xD R ≅ 1− ⎨ =1 − ⎨ a ⎬ 10 / 3 ⎬ ( ) ( ) θ C / a F 4 . 48 f f C / a F ⎪⎩ ⎪⎭ ⎪⎩ ⎪⎭ 10 f D T v 10 f D
[
]
[
]
3/ 2
⎫ ⎪ ⎪⎪ ⎬ ⎪ ⎪ ⎪⎭
(11 - 20)
3/2
(11 - 21)
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Example 11-9
In Ex.11 − 3, the minimum required load rating for 99 % reliability, at x D = L/L10 = 540, is C10 = 6696 lbf = 30.4 kN. From Table 11 - 2, 2 a 02 - 40 mm deep - groove ball bearing would satisfy the requirement. If the bore in the application had to be 70 mm or larger (selecting a 02 - 70 mm deep - groove ball bearing), what is the resulting reliability?
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Example 11-9 Solution : From Table 11 - 2, for a 02 - 70 mm deep - groove ball bearing, C10 = 61.8 61 8 kN = 13890 lbf. lbf Using Eq. (11 - 19), recalling from Ex. 11 - 3 that a f = 1.2, FD = 413 lbf, x0 = 0.02, 0 02 (θ − x0 ) = 4.439, 4 439 and d b = 1.489 1 489
[
⎡ x D (a f FD )/ C10 R ≅ 1− ⎢ θ − x0 ⎢⎣
]
a
− x0 ⎤ ⎥ ⎥⎦
b
(11 - 19) 1.489
⎡ ⎤ ⎡1.2 ⋅ 413 ⎤ − 0.02 ⎥ ⎢ 540 ⎢ ⎥ ⎣ 13890 ⎦ ⎥ = 0.999965 = 1− ⎢ ⎢ ⎥ 4.439 ⎥ ⎢ ⎣ ⎦ Which as expected, Which, expected is much higher than 0.99 0 99 from Ex. Ex 11 - 3. 3 3
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Example 11-10 In Ex. 11 - 8 bearings b i A and d B (cone ( 15100 and d cup 15245)) have h C10 = 12100 N. What is the reliability of the pair of bearings A and B ? Solution :
Th desired The d i d life lif x D = 5000(1050)60/ (90 ⋅ 10 6 ) = 3.5 Using Eq. (11 - 21) for bearing A, where from Ex. 11 - 8, FD = PA = 4938 N, f T f v = 1.32, 1 32 andd a f = 1, 1 gives i b
⎫⎪ ⎧⎪ ⎫⎪ xD xD ⎪⎧ RA ≅ 1 − ⎨ =1 − ⎨ 10 / 3 ⎬ a ⎬ ⎪⎩θ C10 / (a f FD ) ⎪⎭ ⎪⎩ 4.48 f T f v C10 / (a f FD ) ⎪⎭
[
]
⎧ ⎫ 3. 5 =1 − ⎨ 10 / 3 ⎬ [ ] ( ) ( ) × 4 . 48 1 . 32 12100 / 1 4938 ⎩ ⎭ which is less than 0.995, as expected.
[
]
3/2
(11 - 21)
3/2
= 0.994846
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Example 11-10
Using Eq. (11 - 21) for bearing B with FD = PB = 2654 N ggives ⎧ ⎫ 3.5 RB ≅ 1 − ⎨ 10 / 3 ⎬ ⎩ 4.48(1.32 )[12100 / (1× 2654 )] ⎭ R = RA RB = 0.994616
3/2
= 0.999769 0 999769
Which is greater than the overall reliability goal of 0.99.
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Example 11-11 (1) Consider a constrained housing as depicted in Fig. Fig 11-19 11 19 with two direct-mount tapered roller bearings resisting an external thrust Fae of 8000N 8000N. The shaft speed is 950 rpm, rpm the desired life is 10000h, the expected shaft diameter is approximately 1 in The lubricant is ISO VG 150 (150 cSt at 40 oC) oil with an in. estimated bearing operating temperature of 80 oC. Fig. 11.19
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Example 11-11 (2) The reliability goal is 0.95. The application factor is approximately pp y af =1. (a) Choose a suitable tapered roller bearing for A. (b) Choose a suitable tapered roller bearing B. (c) Find the reliabilities RA, RB, and R. R Solution: (a)The bearing reactions at A are FrA=FrB=0 FaA=Fae= 8000 N Mechanical Design Lab. with Advanced Materials
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Example 11-11 (3) Si Since b i B is bearing i unloaded, l d d we will ill start t t with ith R = R A = 0.95. 0 95 From Table 11 - 6, 00.47 47 FrA 0 47 FrB 0.47 − mFae KB KA N ti that Noting th t Fae to t the th right i ht is i positive, iti with ith direct di t mounting ti m = 1, 1 0.47 FrB 0.47(0) − ( +1)( −8000) KA KB 0 < 8000 N Th top set off Equations The E i i Table in T bl 11 - 6 applies, li so 0.47 FrB 0.47(0) FaA = − mFae = − ( +1)( −8000) = 8000 N KB KB FaB =
0.47(0) =0 KA Mechanical Design Lab. with Advanced Materials
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Example 11-11 (4)
If we set K A = 1, from Table 11 - 6 PA = 0.4 FrA + K A FaA = 0.4(0) + (1)8000 = 8000 N PB = FrB = 0 Th required The i d life lif is i LD = 10000(950) ( )((60)) = 570(10 ( 6 ) rev From Figs.11 - 16 and 11 - 17, f T = 0.76 and f v = 1.12. This gives f T f v = 0.76(1.12) = 0.85.
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Example 11-11 (5) Then, from Eq. (11 − 17), for bearing A ⎡ ⎤ LD C10 = a f P ⎢ 6 ⎥ 2/ 3 ⎣ 4.48 f T f v (1 − RD ) 90 ⋅ 10 ⎦
(LD in revolutions) (11 - 17)
⎡ ⎤ 570(10 6 ) = 16970 N = (1)8000 ⎢ 6 ⎥ 2/3 ⎣ 4.48(0.85)(1 − 0.95) 90 ⋅ 10 ⎦ Fi Figure 11 - 15 presents t one possibili ibility t in i the th 1 - in i bore b (25.4 (25 4 - mm)) size i : cone, HM88630, cup HM88610 with i h a thrust h rating i (C10 ) a = 17200 1 200 N ( K = 1.07) 10 ) (b) Bearing B experiences no load, and the cheapest bearing of this bore size will do, including a ball or roller bearing. Mechanical Design Lab. with Advanced Materials
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Example 11-11 (6) LD 570(10 6 ) (c) x D = = = 6.333 6 L10 90(10 ) b
⎧⎪ ⎫⎪ ⎧⎪ ⎫⎪ xD xD =1 − ⎨ RA ≅ 1 − ⎨ a ⎬ 10 / 3 ⎬ ( ) ( ) / 4 . 48 / θ C a F f f C a F ⎪⎩ ⎪⎭ ⎪⎩ ⎪⎭ f D T v f D 10 10
[
]
[
3/2
]
(11 - 21)
3/2
⎧ ⎫ 6.333 =1 − ⎨ = 0.953 10 / 3 ⎬ ⎩ 4.48(0.85)[17200 / (1 × 8000 )] ⎭ which is greater than 0.95, as one would expect. For bearing B, FD = PB = 0 RB = 1 R = R A RB = 0.953 which is greater than the reliability goal of 0.95, 0 95 as one would expect. expect Mechanical Design Lab. with Advanced Materials
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11-11 Lubrication (1) Elastohydrodynamic lubrication (EHD) is the phenomenon that occurs when a lubricant is introduced between surfaces that are in pure rolling contact contact. The contact of g gear teeth and that found in rollingg bearings g and in cam-and-follower surfaces are typical examples.
When a lubricant is trapped between two surfaces in rolling contact, a tremendous increase in the pressure within the lubricant film occurs. occurs The viscosity is exponentially related to pressure, and so very large increase in viscosity occurs in the lubricant that is trapped between the surfaces. Mechanical Design Lab. with Advanced Materials
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11-11 Lubrication (2) Leibensperger observes that the change in viscosity in and out of contact pressure is equivalent to the difference between cold asphalt and light sewing machine oil. oil
The purposes of an antifriction-bearing lubricant: 1. To provide a film of lubricant between the sliding and rollingg surfaces. 2. To help distribute and dissipate heat. 3. To prevent corrosion of the bearing surfaces. 4 To protect the parts from the entrance of foreign matter. 4. matter Mechanical Design Lab. with Advanced Materials
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11-11 Lubrication (3) GREASE
OIL
1. T