section-10.pdf

7/30/2019 SECTION-10.pdf

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SECTION 10 - BALL AND ROLLER BEARINGS

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( )( )( )( ) 125810150060 610 ==

HRB

hrHR 000,14

(c)

b

P

P

B

B

1

10

10 1ln

1

ln

=

=

9.0

1ln

1ln

10P

mrB 125810 =

mrB 1350=

34.1

1

9.0

1ln

1ln

12581350

=P

891.0=P

602. A certain bearing is to carry a radial load of 500 lb. and a thrust of 300 lb. The

load imposes light shock; the desired 90 % life is 10 hr./day for 5 years at

rpmn 3000= . (a) Select a deep-groove ball bearing. What is its bore? Consider

all bearings that may serve. (b) What is the computed rated 90 % life of the

selected bearing? (c) What is the computed probability of the bearing surviving

the specified life? (d) If the loads were changed to 400 and 240 lb., respectively,determine the probability of the bearing surviving the specified life, and the 90 %

life under the new load.

Solution:

lbFx 500=

lbFz 300=

Assume 1=rC

( )( )6.0

5000.1

300==

xr

z

FC

F

Table 12.2, QFC

F

xr

z>

(a) ztxre FCFCF += 56.0

1=rC

Assume 8.1=tC

( )( ) ( )( ) lbFe 82030093.1500156.0 =+=


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For light shock, service factor ~ 1.1

( )( ) lbFe 9028201.1 ==

( ) ( ) lbFB

BF e

r

r 614,3027701350 313

1

10==

=

( )( )( )( )( )( ) mrB 3285103000601036556

10 ==

( ) ( ) lbFB

BF e

r

r 409,139023285 313

1

10==

=

Table 12.3,

Bearing No. rF , lb sF , lb Bore

217 14,400 12,000 85 mm

312 14,100 10,900 60 mm

Select, Bearing No. 312

lbFr 100,14=

lbFs 900,10=

(b) Table 12.2

0285.0900,10

300==

s

z

F

F

99.1=tC

22.0=Q

ztxre FCFCF += 56.0

( )( ) ( )( ) lbFe 87730099.1500156.0 =+= ( )( ) lbFe 9658771.1 ==

e

r

r FB

BF

3

1

10

=

( )9651

100,143

1

10

=

B

mrB 311910 =

( )( )( )( )( )( ) 31191030006010365 610 ==

YRB

yearsYR 75.4=

(c)

b

P

P

B

B

1

10

10 1ln

1ln

=


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Page 4 of 17

use 125.1=b

mrB 311910 =

mrB 3285=

125.1

1

9.0

1ln

1ln

31193285

=P

8943.0=P

(d) lbFx 400=

lbFz 240=

1=rC

( )( )

6.0

4000.1

240==

xr

z

FC

F

Table 12.2

15.2=tC

6.021.0


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yearsYR 8=

603. The smooth loading on a bearing is 500-lb radial, 100 lb. thrust; rpmn 300= . An

electric motor drives through gears; 8 hr./day, fully utilized. (a) Consideringdeep-groove ball bearings that may serve, choose one end specify its bore. For

the bearing chosen, determine (b) the rated 90 % life and (c) the probability ofsurvival for the design lufe.

Solution:

lbFx 500=

lbFz 100=

Table 12.1, 8 hr./day fully utilized, assume 25,000 hr

( )( )( )( ) mrB 4501030060000,25 610 ==

(a) assume 1=rC

( )( ) 2.05000.1100

==

xr

z

FC

F

considerxr

z

FC

FQ >

( )( ) lbFCF xre 5005000.1 ===

( ) ( ) lbFB

BF e

r

r 3832500450 313

1

10==

=

Table 12.3

Bearing No. rF , lb sF , lb

207 4440 3070306 4850 3340305 3660 2390

Select 305, lbFr 3660= , lbFs 2390=

Bore (Table 12.4) = 25 mm

(a) 0418.02390

100==

s

z

F

F

Table 12.2, 26.022.0 Q<

xr

z

FC

FQ >

( )( ) lbFCF xre 5005000.1 ===

( )5001

36603

1

10

=

B

mrB 39210 =

Rated Life:


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Page 6 of 17

( )( )( )( ) 3921030060 610 ==

HRB

hrHR 000,22

(c)

b

P

P

B

B

1

10

10 1ln

1

ln

=

125.1=b

125.1

1

9.0

1ln

1ln

392

450

=P

884.0=

P

605. A No. 311, single-row, deep-groove ball bearing is used to carry a radial load of

1500 lb. at a speed of 500 rpm. (a) What is the 90 % life of the bearing in hours?

What is the approximate median life? What is the probability of survival if theactual life is desired to be (b) 105 hr., (c) 104 hr.?

Solution:Table 12.3, No. 311

lbFs 9400=

lbFr 12400=

lbFx 1500=

assume 1=rC

( )( ) lbFCF xre 150015001 ===

(a) er

r FB

BF

3

1

10

=

( )15001

124003

1

10

=

B

mrB 56510 =

( )( )( )( ) 5651050060 610 ==

HRB

hrHR 800,18

For median life = 5( 90 % life) = ( ) hr000,94800,185 =


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(b) ( )( )( )( ) mrB 3000105006010 65 ==

b

P

P

B

B

1

10

10 1ln

1ln

=

125.1=b

125.1

1

9.0

1ln

1ln

565

3000

=P

502.0=P

(c) 104

hr

( )( )( )( ) mrB 300105006010 64 ==

b

P

P

B

B

1

10

10 1ln

1ln

=

125.1=b

125.1

1

9.0

1ln

1ln

565

300

=P

950.0=P

606. The load on an electric-motor bearing is 350 lb., radial; 24 hr. service,

rpmn 1200= ; compressor drive; outer race stationary. (a) Decide upon a deep-

groove ball bearing, giving its significant dimensions. Then compute the selectedbearings 90 % life, and the probable percentage of failures that would occur

during the design life. What is the approximate median life of this bearing? (b)

The same as (a), except that a 200 series roller bearing is to be selected.

Solution:

lbFx 350=

xre FCF =

outer race stationary, 1=rC

( )( ) lbFe 3503501 ==


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Page 8 of 17

Table 12.1

90 % Life, hrs = 50,000 hrs

( )( )( )( ) mrB 360010120060000,50 6 ==

(a) ( ) ( ) lbFB

BF e

r

r 53643503600 313

1

10==

=

Table AT 12.3

earing No. rF , lb sF , lb

208 5040 3520

209 5660 4010306 4850 3340

307 5750 4020

Use No. 209 lbFr 5660=

Table 12.4, Dimension

Bore = 45 mm

O.D. = 85 mmWidth of Races = 19 mm

Max. Fillet r = 0.039 mm

90 % Life:

e

r

r FB

BF

3

1

10

=

( )3501

56603

1

10

=

B

mrB 422910 = ( )( )( )( ) 422910120060 610 ==

HRB

hrHR 740,58

Probability.

b

P

P

B

B

1

10

10 1ln

1ln

=

125.1=

b 125.1

1

9.0

1ln

1ln

4229

3600

=P

916.0=P

% failures = 1 0.916 = 0.084 = 8.4 %


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Page 9 of 17

Median Life = 5(58,740) = 293,700 hrs

(b) Table 12.3, lbFr 5364=

use No. 207, lbFr 5900=

Bore = 35 mm

O.D. = 72 mmWidth of Races = 17 mm

90 % life:

e

r

r FB

BF

3

1

10

=

( )3501

59003

1

10

=

B

mrB 479010 =

( )( )( )( ) 479010120060 610 ==

HRB

hrHR 530,66

Probability.

b

P

P

B

B

1

10

10 1ln

1ln

=

125.1=b

125.11

9.0

1ln

1ln

4790

3600

=P

926.0=P % failures = 1 0.926 = 0.074 = 7.4 %

Median Life = 5(66,530) = 332,650 hrs

608. A deep-groove ball bearing on a missile, supporting a radial load of 200 lb., is to

have a design life of 20 hr.; with only a 0.5 % probability of failure while

rpmn 4000= . Using a service factor of 2.1 , choose a bearing. ( A 5- or 6- place

log table is desirable.)

Solution: No need to use log table.

lbFx 200=

assume 1=rC


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Page 10 of 17

( )( ) lbFCF xre 2002000.1 ===

( )( ) lbFe 2402002.1 ==

( )( )( )( ) mrB 8.41040006020 610 ==

995.0005.01 ==P

b

P

P

B

B

1

10

10 1ln

1ln

=

125.1=b

125.1

1

10

9.0

1ln

995.0

1ln

8.4

=

B

mrB 7210 =

( ) ( ) lbFB

BF e

r

r 4.99824072 313

1

10==

=

Table 12.3

Select No. 201, lbFr 1180=

VARIABLE LOADS

610. A certain bearing is to carry a radial load of 10 kip at a speed of 10 rpm for 20 %

of the time, a load of 8 kips at a speed of 50 rpm for 50 % of the time, and a loadof 5 kips at 100 rpm during 30 % of the time, with a desired life of 3000 hr.; no

thrust. (a) What is the cubic mean load? (b) What ball bearings may be used?What roller bearings?

Solution:

(a)3

1

3

3

32

3

21

3

1

+++=

nnFnFnF

FmL

321 nnnn ++= For 1 min.

( )( ) revn 2102.01 ==

( )( ) revn 25505.02 ==

( )( ) revn 301003.03 ==

revn 5730252 =++= kipsF 101 =


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kipsF 82 =

kipsF 53 =

( ) ( ) ( ) ( ) ( ) ( )kipsFm 88.6

57

305258210 31

333

=

++=

(b) lbkipsFx 688088.6 ==

assume 1=rC

( )( ) lbFe 688068800.1 == 1 min = 57 rev

( )( )( )( ) mrB 26.101057603000 610 ==

( ) ( ) lbFB

BF e

r

r 950,14688026.10 313

1

10==

=

Table 12.3, Ball BearingUse Bearing No. 217, lbFr 400,14=

(c) Table 12.3 (Roller Bearing)

Use Bearing No. 213, lbFr 900,14=

612. A deep-groove ball bearing No. 215 is to operate 30 % of the time at 500 rpm

with lbFx 1200= and lbFz 600= , 55 % of the time at 800 rpm with

lbFx 1000= and lbFz 500= , and 15 % of the time at 1200 rpm with

lbFx 800= and lbFz 400= . Determine (a) the cubic mean load; (b) the 90 % life

of this bearing in hours, (c) the average life in hours.

Solution:

Bearing No. 215, lbFr 400,11= , lbFs 250,9=

Table 12.2, sz FF

At 30 % of the time, 500 rpm

065.09250

600==

s

z

F

F

66.1=tC

266.0=Q

( )( )Q

FC

F

xr

z>== 5.0

12001

600

( )( ) ( )( ) lbFCFCF ztcre 166860066.11200156.056.01 =+=+= At 55 % of the time, 800 rpm

054.09250

500==

s

z

F

F


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Page 12 of 17

73.1=tC

257.0=Q

( )( )Q

FC

F

xr

z>== 5.0

10001

500

( )( ) ( )( ) lbFCFCF ztcre 142550073.11000156.056.02=+=+=

At 15 % of the time, 1200 rpm

043.09250

400==

s

z

F

F

84.1=tC

242.0=Q

( )( )Q

FC

F

xr

z>== 5.0

8001

400

( )( ) ( )( ) lbFCFCF ztcre 118440084.1800156.056.01 =+=+=

(a)3

1

3

3

32

3

21

3

1

+++=

nnFnFnF

FmL

321 nnnn ++= lbF 16681 =

lbF 14252 =

lbF 11843 =

For 1 min.

( )( ) revn 1505003.01 ==

( )( ) revn 44080055.02 ==

( )( ) revn 180120015.03 ==

revn 770180440150 =++=

( ) ( ) ( ) ( ) ( ) ( )kipsFm 1434

770

180118444014251501668 31

333

=

++=

(b) lbFF me 1434==

e

r

r FB

BF

31

10

=

( )14341

400,113

1

10

=

B

mrB 50310 =

For 1 min = 770 rev


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Page 13 of 17

( )( )( )( ) 5031077060 610 ==

HRB

hrHR 000,11

(c) Average life = 5(11,000) = 55,000 hrs

MANUFACTURERS CATALOG NEEDED

614. A shaft for the general-purpose gear-reduction unit described in 489 has radial

bearing reactions of lbRC 613= and lbRD 1629= ; rpmn 250= . Assume that

the unit will be fully utilized for at least 8 hr./day, with the likelihood of the sameuses involving minor shock. (a) Select ball bearings for this shaft. (b) Select

roller bearings. (c) What is the probability of both bearings C and D surviving for

the design life?

Solution:

Problem 489, ininD 375.18

31 ==

Ref: Design of Machine Members, Doughtie and Vallance

( ) rtspolac FKKKKKKF =

at C. lbRF Cr 613==

0.1=tK

0.1=pK

0.1=oK

3

c

ars

N

NKK =

rpmNa 250=

rpmNc 500=

5.1=rK

( )( )90856.0

500

2505.13 ==sK

0.1=aK

3

relc

al

KH

HK =

Table 12.1, 8 hr/day, fully utilized, Text

hrHa 000,25=

hrHc 000,10=

assume 0.1=relK for 90 % reliability

3572.1000,10

000,253 ==lK


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( )( )( )( )( )( )( ) lbFc 7566130.190856.00.10.13572.10.1 == Table 9-7, Doughtie and Vallance,

Two-row spherical Type, No. 207

Bore = 1.3780 in, lbFc

880=

At D. lbRF Dr 1629==


( )( )( )( )( )( )( ) lbFc 200916290.190856.00.10.13572.10.1 == Table 9-7, Doughtie and Vallance,

Two-row spherical Type, No. 407

Bore = 1.3780 in, lbFc 2290=

(b) at C, lbFc 756=

Table 9.8, Doughtie and VallanceUse No. 207, Bore = 1.3780 in, lbFc 1540=

at C, lbFc 2009=

Table 9.8, Doughtie and Vallance

Use No. 307, Bore = 1.3780 in, lbFc 2660=

(c) For probability:

(c.1) at C, Bearing No. 207, Two-row spherical bearing, lbFc 880=

( ) ( )( )( )( )( )613190856.0111880 lc KlbF ==

58.1=lK

3

relc

al

KH

HK =

3

000,10

000,2558.1

relK=

634.0=relK

Table 9-3, Reference

Probability = 95.8 %

at D, Bearing No. 407, Deep-groove bearing, lbFc 2290=

( ) ( )( )( )( )( )1627190856.01112290 lc KlbF ==

547.1=lK

3

relc

al

KH

HK =


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Page 15 of 17

3

000,10

000,25547.1

relK=

675.0=relK



(c.2) at C, Roller Bearing No. 207, lbFc 1540=

( ) ( )( )( )( )( )613190856.01111540 lc KlbF ==

765.2=lK

3

relc

al

KH

HK =

3

000,10

000,25765.2

relK=

118.0=relK



at D, Roller Bearing No. 407, lbFc 2660=

( ) ( )( )( )( )( )1627190856.01112660 lc KlbF ==

80.1=lK

3

relc

al

KH

HK =

3

000,10

000,2580.1

relK=

43.0=relK

Table 9-3, ReferenceProbability = 95.7 %

615. A shaft similar to that in 478 has the following radial loads on the bearings, leftto right: 803 lb, 988 lb, 84 lb, and 307 lb; no thrust. The minimum shaft diameter

at the bearings are 1.250 in, 1.125 in, 1.000 in, and 1.0625 in. Assume that the

service will not be particularly gentle; intermittently used, with rpmn 425= . (a)

Select ball bearing for this shaft. (b) Select roller bearings.

Solution:

Ref: Design of Machine Members by Doughtie and Vallance


0.1=aK


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Page 17 of 17

Table 9-7, Ref.Deep-groove type, 106

lbFc 544=

Bore = 1.1811 in

(a.4) 307 lb, inD 0625.1=

( )( )( )( )( )( )( ) lbFc 3333070.10844.10.10.10.10.1 == Table 9-7, Ref.

Deep-groove type, 106

lbFc 544=

Bore = 1.1811 in

(b) Roller Bearing

(b.1) 803 lb, inD 250.1=

lbFc 870= , Bore = 1.3780 in

use No. 207, lbFc 1540=

(b.2) 988 lb, inD 125.1=

lbFc 1071= , Bore = 1.1811 in

use No. 206, lbFc 1320=

(b.3) 84 lb, inD 000.1=

lbFc

91= , Bore = 1.1811 in

use No. 206, lbFc 1320=

(b.4) 307 lb, inD 0625.1=

lbFc 333= , Bore = 1.1811 in

use No. 206, lbFc 1320=

- end -

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