SECTION 10 - BALL AND ROLLER BEARINGS 601.
The radial reaction on a bearing is 1500 lb.; it also carries a thrust of 1000 lb.; shaft rotates 1500 rpm; outer ring stationary; smooth load, 8-hr./day service, say 15,000 hr. (a) Select a deep-groove ball bearing. (b) What is the rated 90 % life of the selected bearing? (c) For b = 1.34 , compute the probability of the selected bearing surviving 15,000 hr.
Solution: Fx = 1500 lb Fy = 1000 lb
( )
B10 = (15,000 )(60 )(1500 ) 10−6 = 1350 mr
F e= 0.56Cr Fx + Ct Fz
Cr = 1 , outer ring stationary assume Ct = 1.8 F e= 0.56(1)(1500) + (1.8)(1000) = 2640 lb 1
1 B 3 Fr = 10 Fe = (1350)3 (2640) = 29,178 lb Br (a) Table 12.3 use 320, Fr = 29,900 lb Fs = 29,900 lb To check: Fz 1000 = = 0.0340 Fs 29,400 Table 12.2, Ct = 1.93 , Q = 0.2286 Fz 1000 = = 0.667 > Q Cr Fx (1.0 )(1500 ) F e= 0.56Cr Fx + Ct Fz F e= 0.56(1)(1500) + (1.93)(1000) = 2770 lb 1
1 B 3 Fr = 10 Fe = (1350)3 (2770) = 30,614 lb Br 2.4 % higher than 29,900 lb. Safe. Therefore use Bearing 320, Deep-Groove Ball Bearing.
(b) Fr = 29,900 lb F e= 2770 lb 1
B 3 29,900 = 10 (2770 ) = 30,614 lb 1 mr B10 = 1258 mr
Page 1 of 17
SECTION 10 - BALL AND ROLLER BEARINGS
( )
B10 = (HR )(60 )(1500 ) 10−6 = 1258 HR ≈ 14,000 hr 1
1 b ln B P (c) = B10 1 ln P10
1 1 ln = ln 0.9 P10 B10 = 1258 mr B = 1350 mr 1
1 1.34 ln 1350 P = 1258 1 ln 0.9 P = 0.891 602.
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 n = 3000 rpm . (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: Fx = 500 lb Fz = 300 lb Assume Cr = 1 Fz 300 = = 0 .6 Cr Fx (1.0 )(500 ) Fz Table 12.2, >Q Cr Fx (a) F e= 0.56Cr Fx + Ct Fz Cr = 1 Assume Ct = 1.8 F e= 0.56(1)(500) + (1.93)(300 ) = 820 lb
Page 2 of 17
SECTION 10 - BALL AND ROLLER BEARINGS For light shock, service factor ~ 1.1 F e= (1.1)(820) = 902 lb 1
1 B 3 Fr = 10 Fe = (1350)3 (2770) = 30,614 lb Br B10 = (5)(365)(10 )(60 )(3000 )(10−6 ) = 3285 mr 1
B 3 Fr = 10 Fe Br Table 12.3, Bearing No. 217 312
1
= (3285)3 (902) = 13,409 lb
Fr , lb 14,400 14,100
Fs , lb 12,000 10,900
Select, Bearing No. 312 Fr = 14,100 lb Fs = 10,900 lb (b) Table 12.2 Fz 300 = = 0.0285 Fs 10,900 Ct = 1.99 Q = 0.22 F e= 0.56Cr Fx + Ct Fz F e= 0.56(1)(500) + (1.99)(300) = 877 lb F e= (1.1)(877 ) = 965 lb 1
B 3 Fr = 10 Fe Br 1
B 3 14,100 = 10 (965) 1 B10 = 3119 mr
( )
B10 = (YR)(365)(10 )(60 )(3000 ) 10 −6 = 3119 YR = 4.75 years 1
1 b ln B P (c) = B10 1 ln P10
Page 3 of 17
Bore 85 mm 60 mm
SECTION 10 - BALL AND ROLLER BEARINGS use b = 1.125 B10 = 3119 mr B = 3285 mr 1
1 1.125 ln 3285 P = 3119 1 ln 0.9 P = 0.8943 (d) Fx = 400 lb Fz = 240 lb Cr = 1 Fz 240 = = 0 .6 Cr Fx (1.0 )(400 ) Table 12.2 Ct = 2.15 Q = 0.21 < 0.6 F e= 0.56Cr Fx + Ct Fz F e= 0.56(1)(400) + (2.15)(240) = 740 lb F e= (1.1)(740) = 814 lb 1
B 3 Fr = 10 Fe Br 1
B 3 14,100 = 10 (814) 1 B10 = 5197 mr 1
1 b ln B P = B10 1 ln P10 1
1 1.125 ln 3285 P = 5197 1 ln 0.9 P = 0.939 Life: B10 = (YR)(365)(10 )(60 )(3000 )(10 −6 ) = 5197
Page 4 of 17
SECTION 10 - BALL AND ROLLER BEARINGS YR = 8 years
603.
The smooth loading on a bearing is 500-lb radial, 100 lb. thrust; n = 300 rpm . An electric motor drives through gears; 8 hr./day, fully utilized. (a) Considering deep-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 of survival for the design lufe.
Solution: Fx = 500 lb Fz = 100 lb Table 12.1, 8 hr./day fully utilized, assume 25,000 hr B10 = (25,000 )(60 )(300 )(10−6 ) = 450 mr (a) assume Cr = 1 Fz 100 = = 0 .2 Cr Fx (1.0 )(500 ) F consider Q > z Cr Fx Fe = Cr Fx = (1.0)(500) = 500 lb 1 3
1 B Fr = 10 Fe = (450)3 (500) = 3832 lb Br Table 12.3 Bearing No. Fr , lb Fs , lb 207 4440 3070 306 4850 3340 305 3660 2390 Select 305, Fr = 3660 lb , Fs = 2390 lb Bore (Table 12.4) = 25 mm
Fz 100 = = 0.0418 Fs 2390 Table 12.2, 0.22 < Q0.26 F Q> z Cr Fx Fe = Cr Fx = (1.0)(500) = 500 lb
(a)
1
B 3 3660 = 10 (500) 1 B10 = 392 mr Rated Life:
Page 5 of 17
SECTION 10 - BALL AND ROLLER BEARINGS
( )
B10 = (HR )(60 )(300 ) 10−6 = 392 HR ≈ 22,000 hr 1
1 b ln B P (c) = B10 1 ln P10 b = 1.125 1
1 1.125 ln 450 P = 392 1 ln 0.9 P = 0.884 605.
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 the actual life is desired to be (b) 105 hr., (c) 104 hr.?
Solution: Table 12.3, No. 311 Fs = 9400 lb Fr = 12400 lb Fx = 1500 lb assume Cr = 1 Fe = Cr Fx = (1)(1500 ) = 1500 lb 1
B 3 (a) Fr = 10 Fe Br 1
B 3 12400 = 10 (1500) 1 B10 = 565 mr
( )
B10 = (HR )(60 )(500 ) 10−6 = 565 HR ≈ 18,800 hr
For median life = 5( 90 % life) = 5(18,800) = 94,000 hr
Page 6 of 17
SECTION 10 - BALL AND ROLLER BEARINGS
( )
( )
(b) B = 105 (60)(500) 10−6 = 3000 mr 1 ln B P = B10 1 ln P10 b = 1.125
1 b
1
1 1.125 ln 3000 P = 565 1 ln 0.9 P = 0.502 (c) 104 hr
( )
( )
B = 104 (60)(500) 10−6 = 300 mr 1 ln B P = B10 1 ln P10 b = 1.125
1 b
1
1 1.125 ln 300 P = 565 1 ln 0.9 P = 0.950 606.
The load on an electric-motor bearing is 350 lb., radial; 24 hr. service, n = 1200 rpm ; compressor drive; outer race stationary. (a) Decide upon a deepgroove ball bearing, giving its significant dimensions. Then compute the selected bearing’s 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: Fx = 350 lb Fe = Cr Fx outer race stationary, Cr = 1 Fe = (1)(350) = 350 lb Page 7 of 17
SECTION 10 - BALL AND ROLLER BEARINGS Table 12.1 90 % Life, hrs = 50,000 hrs B = (50,000)(60)(1200) 10−6 = 3600 mr
( )
1 3
1 B (a) Fr = 10 Fe = (3600)3 (350 ) = 5364 lb Br Table AT 12.3 earing No. Fr , lb Fs , lb 208 5040 3520 209 5660 4010 306 4850 3340 307 5750 4020 Use No. 209 Fr = 5660 lb Table 12.4, Dimension Bore = 45 mm O.D. = 85 mm Width of Races = 19 mm Max. Fillet r = 0.039 mm
90 % Life: 1 3
B Fr = 10 Fe Br 1
B 3 5660 = 10 (350) 1 B10 = 4229 mr
( )
B10 = (HR )(60 )(1200 ) 10−6 = 4229 HR ≈ 58,740 hr Probability.
1 ln B P = B10 1 ln P10 b = 1.125
1 b
1
1 1.125 ln 3600 P = 4229 1 ln 0.9 P = 0.916 % failures = 1 – 0.916 = 0.084 = 8.4 % Page 8 of 17
SECTION 10 - BALL AND ROLLER BEARINGS Median Life = 5(58,740) = 293,700 hrs (b) Table 12.3, Fr = 5364 lb use No. 207, Fr = 5900 lb Bore = 35 mm O.D. = 72 mm Width of Races = 17 mm 90 % life: 1
B 3 Fr = 10 Fe Br 1
B 3 5900 = 10 (350) 1 B10 = 4790 mr
( )
B10 = (HR )(60 )(1200 ) 10−6 = 4790 HR ≈ 66,530 hr Probability. 1
1 b ln B P = B10 1 ln P10 b = 1.125 1
1 1.125 ln 3600 P = 4790 1 ln 0.9 P = 0.926 % 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 n = 4000 rpm . Using a service factor of 1.2 , choose a bearing. ( A 5- or 6- place log table is desirable.)
Solution: No need to use log table. Fx = 200 lb assume Cr = 1
Page 9 of 17
SECTION 10 - BALL AND ROLLER BEARINGS Fe = Cr Fx = (1.0)(200) = 200 lb Fe = (1.2)(200) = 240 lb
( )
B10 = (20 )(60 )(4000 ) 10−6 = 4.8 mr P = 1 − 0.005 = 0.995 1
1 b ln B P = B10 1 ln P10 b = 1.125 1
1 1.125 ln 4.8 0.995 = B10 1 ln 0.9 B10 = 72 mr 1
1 B 3 Fr = 10 Fe = (72)3 (240) = 998.4 lb Br Table 12.3 Select No. 201, Fr = 1180 lb
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 load of 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: 1
F 3n + F23n2 + F33n3 + L 3 (a) Fm = 1 1 ∑n ∑ n = n1 + n2 + n3 For 1 min. n1= (0.2)(10) = 2 rev n 2 = (0.5)(50) = 25 rev n3 = (0.3)(100) = 30 rev
∑ n = 2 + 25 + 30 = 57 rev F1 = 10 kips
Page 10 of 17
SECTION 10 - BALL AND ROLLER BEARINGS F2 = 8 kips F3 = 5 kips 1
(10 )3 (2 ) + (8)3 (25) + (5)3 (30 ) 3 Fm = = 6.88 kips 57
(b) Fx = 6.88 kips = 6880 lb assume Cr = 1 Fe = (1.0)(6880) = 6880 lb 1 min = 57 rev B10 = (3000 )(60 )(57 )(10−6 ) = 10.26 mr 1
1 B 3 Fr = 10 Fe = (10.26)3 (6880) = 14,950 lb Br Table 12.3, Ball Bearing Use Bearing No. 217, Fr = 14,400 lb
(c) Table 12.3 (Roller Bearing) Use Bearing No. 213, Fr = 14,900 lb 612.
A deep-groove ball bearing No. 215 is to operate 30 % of the time at 500 rpm with Fx = 1200 lb and Fz = 600 lb , 55 % of the time at 800 rpm with Fx = 1000 lb and Fz = 500 lb , and 15 % of the time at 1200 rpm with Fx = 800 lb and Fz = 400 lb . 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, Fr = 11,400 lb , Fs = 9,250 lb Table 12.2, Fz Fs At 30 % of the time, 500 rpm Fz 600 = = 0.065 Fs 9250 Ct = 1.66 Q = 0.266 Fz 600 = = 0 .5 > Q Cr Fx (1)(1200 ) Fe1 = 0.56Cr Fc + Ct Fz = 0.56(1)(1200) + (1.66)(600) = 1668 lb At 55 % of the time, 800 rpm Fz 500 = = 0.054 Fs 9250
Page 11 of 17
SECTION 10 - BALL AND ROLLER BEARINGS Ct = 1.73 Q = 0.257 Fz 500 = = 0 .5 > Q Cr Fx (1)(1000 ) Fe 2 = 0.56Cr Fc + Ct Fz = 0.56(1)(1000) + (1.73)(500) = 1425 lb At 15 % of the time, 1200 rpm 400 Fz = = 0.043 Fs 9250 Ct = 1.84 Q = 0.242 Fz 400 = = 0 .5 > Q Cr Fx (1)(800 ) Fe1 = 0.56Cr Fc + Ct Fz = 0.56(1)(800) + (1.84)(400) = 1184 lb 1
F 3n + F23n2 + F33n3 + L 3 (a) Fm = 1 1 ∑n ∑ n = n1 + n2 + n3
F1 = 1668 lb F2 = 1425 lb F3 = 1184 lb For 1 min. n1= (0.3)(500 ) = 150 rev n 2 = (0.55)(800) = 440 rev n3 = (0.15)(1200 ) = 180 rev
∑ n = 150 + 440 + 180 = 770 rev 1 3
(1668) (150 ) + (1425) (440 ) + (1184 ) (180 ) Fm = = 1434 kips 770 3
3
(b) Fe = Fm = 1434 lb 1
B 3 Fr = 10 Fe Br 1
B 3 11,400 = 10 (1434 ) 1 B10 = 503 mr For 1 min = 770 rev
Page 12 of 17
3
SECTION 10 - BALL AND ROLLER BEARINGS B10 = (HR )(60 )(770 )(10−6 ) = 503 HR ≈ 11,000 hr
(c) Average life = 5(11,000) = 55,000 hrs MANUFACTURER’S CATALOG NEEDED 614.
A shaft for the general-purpose gear-reduction unit described in 489 has radial bearing reactions of RC = 613 lb and RD = 1629 lb ; n = 250 rpm . Assume that the unit will be fully utilized for at least 8 hr./day, with the likelihood of the same uses 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: 3 Problem 489, D = 1 in = 1.375 in 8 Ref: Design of Machine Members, Doughtie and Vallance Fc = (K a K l )K o K p K s K t Fr
at C. Fr = RC = 613 lb K t = 1.0 K p = 1 .0 K o = 1.0 Ks = 3
Kr Na Nc
N a = 250 rpm N c = 500 rpm K r = 1.5 Ks = 3
(1.5)(250) = 0.90856 500
K a = 1.0 Ha H c K rel Table 12.1, 8 hr/day, fully utilized, Text H a = 25,000 hr H c = 10,000 hr assume K rel = 1.0 for 90 % reliability Kl = 3
Kl = 3
25,000 = 1.3572 10,000
Page 13 of 17
SECTION 10 - BALL AND ROLLER BEARINGS Fc = (K a K l )K o K p K s K t Fr
Fc = (1.0)(1.3572)(1.0)(1.0 )(0.90856)(1.0)(613) = 756 lb Table 9-7, Doughtie and Vallance, Two-row spherical Type, No. 207 Bore = 1.3780 in, Fc = 880 lb At D. Fr = RD = 1629 lb Fc = (K a K l )K o K p K s K t Fr Fc = (1.0)(1.3572)(1.0)(1.0 )(0.90856)(1.0)(1629) = 2009 lb Table 9-7, Doughtie and Vallance, Two-row spherical Type, No. 407 Bore = 1.3780 in, Fc = 2290 lb (b) at C, Fc = 756 lb Table 9.8, Doughtie and Vallance Use No. 207, Bore = 1.3780 in, Fc = 1540 lb at C, Fc = 2009 lb Table 9.8, Doughtie and Vallance Use No. 307, Bore = 1.3780 in, Fc = 2660 lb (c) For probability: (c.1) at C, Bearing No. 207, Two-row spherical bearing, Fc = 880 lb Fc = 880 lb = (1)K l (1)(1)(0.90856)(1)(613) K l = 1.58 Kl = 3
Ha H c K rel
1.58 = 3
25,000 10,000 K rel
K rel = 0.634 Table 9-3, Reference Probability = 95.8 % at D, Bearing No. 407, Deep-groove bearing, Fc = 2290 lb Fc = 2290 lb = (1)K l (1)(1)(0.90856 )(1)(1627 ) K l = 1.547 Kl = 3
Ha H c K rel
Page 14 of 17
SECTION 10 - BALL AND ROLLER BEARINGS
1.547 = 3
25,000 10,000 K rel
K rel = 0.675 Table 9-3, Reference Probability = 93.3 % (c.2) at C, Roller Bearing No. 207, Fc = 1540 lb Fc = 1540 lb = (1)K l (1)(1)(0.90856)(1)(613) K l = 2.765 Kl = 3
Ha H c K rel
2.765 = 3
25,000 10,000 K rel
K rel = 0.118 Table 9-3, Reference Probability = 98.8 % at D, Roller Bearing No. 407, Fc = 2660 lb Fc = 2660 lb = (1)K l (1)(1)(0.90856)(1)(1627 ) K l = 1.80 Kl = 3
Ha H c K rel
1.80 = 3
25,000 10,000 K rel
K rel = 0.43 Table 9-3, Reference Probability = 95.7 % 615.
A shaft similar to that in 478 has the following radial loads on the bearings, left to 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 n = 425 rpm . (a) Select ball bearing for this shaft. (b) Select roller bearings.
Solution: Ref: Design of Machine Members by Doughtie and Vallance Fc = (K a K l )K o K p K s K t Fr K a = 1.0
Page 15 of 17
SECTION 10 - BALL AND ROLLER BEARINGS
Kl = 3
Ha H c K rel
H c = 10,000 hr Table 12.1, Text, H a = 10,000 hr (intermittent) 90 % reliability, K rel = 1.0 10,000 = 1 .0 10,000 K o = 1.0
Kl = 3
K p = 1 .0
K r = 1.5 assumed Ks = 3
Kr Na Nc
N a = 425 rpm N c = 500 rpm Ks = 3
(1.5)(425) = 1.0844 500
K t = 1.0 (a) Ball Bearing (a.1) 803 lb, D = 1.250 in Fc = (1.0)(1.0)(1.0)(1.0)(1.0844)(1.0)(803) = 870 lb Table 9-7, Ref. Two-row spherical type, 207 Fc = 880 lb Bore = 1.3780 in (a.2) 988 lb, D = 1.125 in Fc = (1.0)(1.0)(1.0)(1.0 )(1.0844)(1.0)(988) = 1071 lb Table 9-7, Ref. Two-row spherical type, 306 Fc = 1050 lb Bore = 1.1811 in (a.3) 84 lb, D = 1.000 in Fc = (1.0)(1.0)(1.0)(1.0)(1.0844)(1.0)(84) = 91 lb
Page 16 of 17
SECTION 10 - BALL AND ROLLER BEARINGS Table 9-7, Ref. Deep-groove type, 106 Fc = 544 lb Bore = 1.1811 in (a.4) 307 lb, D = 1.0625 in Fc = (1.0)(1.0)(1.0)(1.0)(1.0844)(1.0)(307 ) = 333 lb Table 9-7, Ref. Deep-groove type, 106 Fc = 544 lb Bore = 1.1811 in (b) Roller Bearing (b.1) 803 lb, D = 1.250 in Fc = 870 lb , Bore = 1.3780 in use No. 207, Fc = 1540 lb (b.2) 988 lb, D = 1.125 in Fc = 1071 lb , Bore = 1.1811 in use No. 206, Fc = 1320 lb (b.3) 84 lb, D = 1.000 in Fc = 91 lb , Bore = 1.1811 in use No. 206, Fc = 1320 lb (b.4) 307 lb, D = 1.0625 in Fc = 333 lb , Bore = 1.1811 in use No. 206, Fc = 1320 lb - end -
Page 17 of 17
