Volume of a Sphere V sphere ¼ 4 3 pr 3 V sphere ¼ 4 3 pð0:7Þ 3 ¼ 1:43 cm 3 Mass of the Ball m r ¼ rV ¼ 7:87 g=cm 3  1:43 cm 3 ¼ 11:25 g ¼ 0:011 kg The shaft and inner ring angular speed is o ¼ 2pN 60 o ¼ 30;000 rev min  1 min 60 s  2prad 1 rev ¼ 3142 rad=s Angular Speed of Rolling Elements and Cage o C ¼ R in R in þ R out o shaft ¼ 0:025 m 0:025 m þ 0:039 m  3142 rad=s ¼ 1227 rad=s Centifugal Force, F c F c ¼ m r o 2 C R r ¼ 0:011 kg Âð1227 rad=sÞ 2  0:032 m ¼ 530 N Thrust Force Component at Outer Ring Contact S F x ¼ 0 W oa ¼ W ia ¼ 200 N Radial Component of Inner Ring Contact W ir W ia ¼ ctan a i W ir ¼ 200  ctan 40  ¼ 238:4N Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. Radial Component of Outer Ring Contact S F y ¼ 0 W or ¼ W ir þ F c W or ¼ 238:4 þ 530 ¼ 768:4N Outer Ring Contact Angle tan a o ¼ W oa W or tan a o ¼ 200 N 768:4N ¼ 0:26 a o ¼ 14:5  Resultant Force on Outer Ring Contact The resultant force on the outer ring race, W o , is in the direction normal to surface, as shown in Fig. 12-22. It has two components in the axial and radial directions: W oa W o ¼ sin 14:5  W o ¼ 200 N sin 14:5  ¼ 798:8N Normal Contact Force on Inner Ring Race The resultant force on the inner ring race, W i , is in the direction normal to surface, as shown in Fig. 12-22. It has two components in the axial and radial directions: W ia W o ¼ sin 40  W i ¼ 200 N sin 40  ¼ 311:2N Example Problem 12-9 An angular contact ball bearing has a contact angle with the inner ring of a i ¼ 30  . The thrust load of W ¼ 11;500 N is divided evenly on 14 balls. The ball has a diameter of d r ¼ 18 mm. The shaft speed is N ¼ 33;000 RPM. The conformity ratio R r ¼ 0:52. The diameter of the outer race at the contact point with the balls is 118 mm. The balls and rings are made of steel having the Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. following properties: modulus of elasticity E ¼ 2 Â10 11 N=m 2 , Poisson’s ratio n ¼ 0:3, density r ¼ 7870 kg=m 3 . a. Find contact angle, a o , with the outer ring. b. Find the contact force with the inner and outer rings. c. Find the centrifugal force of each rolling element. d. Find the maximum pressure at the contact with the outer ring. e. Find the maximum pressure at the contact with the outer ring, given rings made of steel and balls made of silicone nitride (hybrid bearing). The properties of silicone nitride are: modulus of elasticity E ¼ 3:14  10 11 N=m 2 , Poisson’s ration n ¼ 0:24, density r ¼ 3200 kg=m 3 . Solution a. Contact Angle, a o , with Outer Ring The first step is to find the centrifugal force, F c ¼ m r o 2 C R c , where m r is the mass of a ball, o C is the angular speed of the ball center (or cage), and R c is the radius of the ball center circular orbit. The volume and mass of a ball in the bearing are: V sphere ¼ 4 3 pr 3 ¼ 4 3  p  0:009 3 m 3 ¼ 3:05  10 À6 m 3 m r ¼ rV ¼ 7870 kg=m 3  3:05 Â10 À6 m 3 ¼ 0:024 kg The shaft and inner ring angular speed is o ¼ 2p  33;000 60 ¼ 3456 rad=s The outer diameter is R out ¼ 118 2 mm ¼ 59  10 À3 m In order to find the inner diameter, we assume that a o $ a i and that the difference between the outer and inner radius is d r cos a i . In that case, R in ¼ðR out À d r Þcos a i ¼ð59 mm À 18 mmÞcos 30  ¼ 43:4mm ¼ 43:4 Â10 À3 m Now it is possible to find the angular speed of a rolling-element center (or cage), o c : o c ¼ R in R in þ R out o shaft ¼ 43:4 43:4 þ59  3456 rad=s ¼ 1465 rad=s Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. The distance between the ball center and the bearing center as R c , is required for the calculation of the centrifugal force. R c ¼ R in þ R out 2 ¼ 51:2mm The centrifugal force, F c , of each rolling element is in the radial direction: F c ¼ m r o 2 c R c ¼ 0:024 kg  1465 2  1=s 2  0:0512 m ¼ 2637 N The thrust force component of each ball at the outer ring contact is equal to that of the inner ring: W oa ¼ W ia ¼ 11;500 N 14 ¼ 821 N The radial component of inner ring contact is W ir W ia ¼ ctan a i W ir ¼ 821  ctan 30  ¼ 1422 N The radial component of outer ring contact force is W or ¼ W ir þ F c ¼ 1421 N þ 2637 N ¼ 4059 N The outer ring contact angle is solved as follows: tan a o ¼ W oa W or ¼ 821 N 4058 N ¼ 0:202; a o ¼ 11:43  b. Contact Force with Inner and Outer Rings The resultant (normal) component of the outer ring contact is W oa W o ¼ sin 11:43  W o ¼ 821 N sin 11:43  ¼ 4143 N The resultant (normal) contact force at the inner ring race is W ia W i ¼ sin 30  W i ¼ W ia sin 30  ¼ 1642 N c. Rolling-Element Centrifugal Force From part (a) the centrifugal force is 2637 N, in the radial direction. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. d. Maximum Pressure at Contact of Outer Ring The maximum force is at the race of the outer ring, W o ¼ 4143 N. The radius of the contact curvatures is R 1x ¼ R 1y ¼ 9mm R 2y ¼ R r d r ¼ 18 mm  0:52 ¼ 9:36 mm R 2x ¼ 59 mm The equivalent radius of the outer raceway contact in the y-z plane (referred to as the x plane) is 1 R x ¼ 1 R 1x À 1 R 2x ¼ 1 9mm À 1 59 mm R x ¼ 9  59 mm 43:4 À 9 ¼ 10:62 mm The equivalent inner radius in the y plane is 1 R y ¼ 1 R 1y À 1 R 2y ¼ 1 9mm À 1 9:36 mm R y ¼ R 2y Á R 1y R 2y À R 1y ¼ 9  9:36 mm 9:36 À9 ¼ 234 mm The equivalent curvature of the inner ring and ball contact is 1 R eq ¼ 1 R x þ 1 R y ¼ 1 10:62 mm þ 1 234 mm R eq ¼ R x Á R y R x þ R y ¼ 10:62 Â234 mm 234 þ10:62 ¼ 10:16 mm The ratio a r becomes a r ¼ R y R x ¼ 234 mm 10:62 mm ¼ 22:03 The dimensionless coefficient k is estimated from the ratio a r : k ¼ a 2=p r ¼ 7:16 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. For the calculation of the ellipsoid radii, the following values are required: q a ¼ p 2 À 1 ¼ 0:57 ^ EE % 1 þ q a a r ^ EE % 1 þ 0:57 22:03 ¼ 1:025 The shaft and the bearing are made of identical material, so the equivalent modulus of elasticity is E eq ¼ E 1 À n 2 ¼ 2  10 11 N=m 2 1 À 0:3 2 ¼ 2:2 Â10 11 N=m 2 Now the ellipsoid radii a and b can be determined: a ¼ 6 Á ^ EE Á W max Á R eq p Á k Á E eq ! 1=3 ¼ 6  1:025  4143 N  10:82  10 À3 m p  7:16  2:2 Â10 11 N=m 2  1=3 ¼ 0:37 mm b ¼ 6 Á k 2 Á ^ EE Á W max Á R eq p Á E eq ! 1=3 ¼ 6  7:16 2  1:025  4143 N  10:16 Â10 À3 m p  2:2  10 11 N=m 2  1=3 ¼ 2:68 mm The maximum pressure at the contact with outer ring is p max ¼ 3 2 Á W max p Á ab ¼ 3 2 Á 4134 N p  0:37  2:68  10 À6 m 2 ¼ 1:99 GPa e. Maximum stress of hybrid bearing The rings are made of steel and the balls are made of silicone nitride, which has the following properties: Modulus of elasticity E ¼ 3:14  10 11 N=m 2 Poisson’s ratio n ¼ 0:24 Density r ¼ 3200 kg=m 3 The major advantage of a hybrid bearing is a lower centrifugal force due to lower density of the rolling element. However, the modulus of elasticity of silicone nitride is higher than that of steel, and this can result in a higher maximum pressure. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. For a hybrid bearing, it is necessary to consider the equivalent modulus of elasticity of silicone nitride on steel: 2 E eq ¼ 1 À n 2 21 E 1 þ 1 À n 2 2 E 2 ¼ 1 À 0:24 2 3:14 Â10 11 N=m 2 þ 1 À 0:3 2 2:0 Â10 11 N=m 2 ¼ 0:755 Â10 À11 m 2 =N E eq ¼ 2:65 Â10 11 N=m 2 Due to the low density of the balls, the centrifugal force is lower: m r ¼ rV ¼ 3200 kg m 3  3:05 Â10 À6 m 3 ¼ 0:0098 kg The centrifugal force is F c ¼ m r o 2 c R c ¼ 0:0098 kg  1465 2 1=s 2  0:0512 m ¼ 1077 N The radial component of the inner ring contact is the same as for a steel bearing: W ir ¼ 1422 N The radial component on the outer ring contact becomes W or ¼ W ir þ F c ¼ 1422 N þ 1077 N ¼ 2499 N The outer ring contact angle is tan a o ¼ W oa W or ¼ 821 N 2499 N ¼ 0:329; a o ¼ 18:19  The resultant (normal) component on the outer ring contact becomes W oa W o ¼ sin 18:19  and W o ¼ 821 N sin 18:19  ¼ 2630 N Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. The ellipsoid radii a and b can now be determined by substituting the values already calculated: a ¼ 6 EW max R eq p  k  E eq ! 1=3 ¼ 6  1:025  2630 N  10:16  10 À3 m p  6:86  2:65 Â10 11 N=m 2  1=3 ¼ 0:30 mm b ¼ 6 k 2 EW max R eq p  E eq ! 1=3 ¼ 6  6:86 2  1:028  2630 N  10:82 Â10 À3 m p  2:65  10 11 N=m 2  1=3 ¼ 2:16 mm The maximum pressure at the contact is p max ¼ 3 2 W max pab ¼ 3 2 2630 N p  0:3  2:16  10 À6 m 2 ¼ 1:94 GPa Conclusion In this example of a high-speed bearing, the maximum stress is only marginally lower for the hybrid bearing. The maximum pressure at the contact with outer ring is For an all-steel bearing: p max ¼ 1:99 GPa For a hybrid bearing: p max ¼ 1:94 GPa Example Problem 12-10 For the bearing in Example Problem 12-9, find h min at the contact with the outer race and the inner race for both (a) an all-steel bearing and (b) a hybrid bearing. Use oil SAE 10 at 70  C and a viscosity–pressure coefficient a ¼ 2:2 10 À8 m 2 =N. Solution The analysis of the forces is identical to that of Example Problem 12-9. In this problem, the EHD minimum film thickness is calculated. Bearing Data from Example Problem 12-9: Inner contact diameter, R in ¼ 43:4mm Outer contact diameter, R out ¼ 59 mm Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. Outer ring equivalent radius in the x plane, R x ¼ 10:62 mm (contact at outer race) Shaft speed, o ¼ 3456 rad=s R x ¼ 9 mm, R 2x ¼ 43:4 mm, R y ¼ 234 mm Data for Steel Bearing from Example Problem 12.9: Equivalent modulus of elasticity: E eq ¼ 2:2  10 11 N=m 2 Resultant component of outer ring contact: W o ¼ 4143 N Normal contact force at inner ring race: W i ¼ 1642 N Data for Hybrid Bearing: Equivalent modulus of elasticity: E eq ¼ 2:65 Â10 11 N=m 2 Resultant component of outer ring contact: W o ¼ 2630 N Normal contact force at inner ring race: W i ¼ 1642 N For hard surfaces, such as steel in rolling bearings, the equation for calculating the minimum film thickness, h min , is presented in dimensionless form, as follows: h min R x ¼ 3:63 " UU 0:68 r ða Á E eq Þ 0:49 " WW 0:073 ð1 Àe À0:68k Þ Here, a is the viscosity–pressure coefficient and " UU r and " WW are dimensionless velocity and load, respectively, defined by the following equations: " UU r ¼ m o U r E eq R x and " WW ¼ W E eq R 2 x Here, U r is the rolling velocity, m o is the viscosity of the lubricant at atmospheric pressure and bearing operating temperature, W is the reaction force of one rolling element, and E eq is the equivalent modulus of elasticity. All steel and hybrid bearings have the same rolling velocity, which is calculated from the shaft speed via Eq. (12-34): U rolling ¼ R in R out 2ðR in þ rÞ o ¼ R in R out R in þ R out o U r ¼ 0:0434  0:059 m 2 0:0434 m þ0:059 m  3456 rad=s ¼ 86:42 m=s Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. The first step is to calculate R x and k for the inner and outer contacts. The radius of curvature at the inner contact is 1 R x ¼ 1 R 1x þ 1 R 2x ¼ 1 9mm þ 1 43:4mm R x ¼ 9  43:4mm 43:4 þ9 ¼ 7:45 mm The ratio a r is a r ¼ R y R x ¼ 234 mm 7:45 mm ¼ 31:2 The dimensionless coefficient k is derived directly from the ratio a r : k ¼ a 2=p r ¼ 8:94 For the outer contact, R x and k can be taken from Example Problem 12-9: R x ¼ 10:62 mm and k ¼ 7:16 a. All-Steel Bearing Equivalent modulus of elasticity: E eq ¼ 2:2  10 11 N=m 2 Resultant component of outer ring contact is: W o ¼ 4143 N Normal contact force at inner ring race: W i ¼ 1642 N Inner Race Contact: The dimensionless rolling velocity is " UU r ¼ 0:01 N-s=m 2  86:42 m=s 2:2 Â10 11 N=m 2  7:45  10 À3 m 2 ¼ 52:7 Â10 À11 The dimensionless load at the inner race is " WW ¼ 1642 N 2:2  10 11 N=m 2  7:45 2  10 À6 m 2 ¼ 13:44  10 À5 Substituting these values in the formula for the minimum thickness, we get h min 7:45 Â10 À3 ¼ 3:63  ð52:7  10 À11 Þ 0:68 ð2:2  10 À8  2:2 Â10 11 Þ 0:49 ð13:44 Â10 À5 Þ 0:073 Âð1 Àe À0:68Â8:94Þ ¼ 217:8 Â10 À6 The minimum thickness for the inner race is h min ¼ 217:8 Á 10 À6  7:45 Á10 À3 m ¼ 1:62 Á 10 À6 m h min ¼ 1:62 mm Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. [...]... rolling velocity is 0:01 N-s=m2  86: 42 m=s ¼ 37  10 11 2: 2  101 1 N=m2  10: 62  10 3 m2 The dimensionless load at the outer race is " W ¼ 2: 2  101 1 4134 N ¼ 16:7  10 5 N=m2  10: 622  10 6 m2 Substituting these values in the formula for the minimum thickness, we get hmin ð37  10 11 Þ0:68 2: 2  10 8  2: 2  101 1 Þ0:49 ¼ 3:63  10: 62  10 3 16:7  10 5 Þ0:073  ð1 À eÀ0:68Â7:16 Þ ¼ 167:1  10 6... Static C0 lbs 11 12 13 14 15 17 19 21 23 1 1.5 1.5 1.5 2 2 2 2.5 2. 5 1400 1700 1930 23 20 3000 3800 5000 5700 7350 850 104 0 120 0 1460 1930 25 50 3400 4000 5300 mm 35 37 42 47 52 62 72 80 90 6309 6 310 6311 63 12 6313 6314 6315 6316 6317 6318 6319 6 320 6 321 6 322 6 324 6 326 6 328 6330 6309Z 6310Z 6311Z 6312Z 6313Z 6314Z 6309.2Z 6 310. 2Z 6311.2Z 63 12. 2Z 6313.2Z 6314.2Z Copyright 20 03 by Marcel Dekker, Inc All Rights... 6309 RS 6 310 RS 6309.2RS 6 310. 2RS 45 50 55 60 65 70 75 80 85 90 95 100 105 110 120 130 140 150 100 110 120 130 140 150 160 170 180 190 20 0 21 5 22 5 24 0 26 0 28 0 300 320 25 27 29 31 33 35 37 39 41 43 45 47 49 50 55 58 62 65 2. 5 3 3 3.5 3.5 3.5 3.5 3.5 4 4 4 4 4 4 4 5 5 5 9150 106 00 129 00 14000 16000 18000 19300 19600 21 600 23 200 24 500 28 500 30500 325 00 36000 39000 44000 49000 6700 8150 100 00 108 00 125 00 14000... 100 010 060 100 060 060 100 Load ratings dynamic static C C0 lbs lbs 325 0 6100 4550 8300 7500 11400 3050 4150 4500 22 40 4400 320 0 6300 5600 9150 20 40 29 00 320 0 909 024 909 025 909 026 909 027 909 028 909 029 909030 9090 52 9090 62 909067 909070 1.3 128 8440 1.4065 9379 1.5000 1. 125 0 1. 625 0 1 .28 15 1.3750 75 02 1 .25 00 3.1496 2. 2500 3.1496 2. 8 125 3.7500 3.1875 4.0 625 2. 9630 2. 9630 2. 0800 2. 6500 Copyright 20 03 by Marcel... the inner race contact is hmin ¼ 21 3 :24  10 6  7:45  10 3 m ¼ 1:59  10 6 m hmin ¼ 1:59 mm Copyright 20 03 by Marcel Dekker, Inc All Rights Reserved Outer Race: " Ur ¼ The value of dimensionless velocity is 0:01 N-s=m2  86: 42 m=s m2 ¼ 30:7  10 11 2: 65  101 1 N=m2  10: 62  10 3 The dimensionless load on the outer race is " W ¼ 26 30 N ¼ 8:8  10 5 2: 65  101 1 N=m2  10: 622  10 6 m2 Substituting... 86: 42 m=s ¼ 43:8  10 11 2: 65  101 1 N=m2  7:45  10 3 m2 The dimensionless load on the inner race is " W ¼ 16 42 N ¼ 11:16  10 5 2: 65  101 1 N=m2  7:4 52  10 6 m2 Substituting these values in the equation for the minimum film thickness, we get hmin ð43:8  10 11 Þ0:68 2: 2  10 8  2: 65  101 1 Þ0:49 ¼ 3:63 Á 7:45  10 3 ð11:16  10 5 Þ0:073  ð1 À eÀ0:68Â8:94 Þ ¼ 21 3 :24  10 6 The minimum fluid film thickness... Bearing with two seals d 6300Z 6301Z 6302Z 6303Z 6304Z 6305Z 6306Z 6307Z 6308Z 6300.2Z 6301.2Z 63 02. 2Z 6303.2Z 6304.2Z 6305.2Z 6306.2Z 6307.2Z 6308.2Z 6300 6301 63 02 6303 6304 6305 6306 6307 6308 6300.2RS 6301.2RS 63 02. 2RS 6303.2RS 6304.2RS 6305.2RS 6306.2RS 6307.2RS 6308.2RS 10 12 15 17 20 25 30 35 40 Copyright 20 03 by Marcel Dekker, Inc All Rights Reserved RS RS RS RS RS RS RS RS RS Load ratings... in the equation for the minimum thickness, we get hmin ð30:7  10 11 Þ0:68 2: 2  10 8  2: 65  101 1 Þ0:49 ¼ 3:63  10: 62  10 3 ð8:8  10 5 Þ0:073  ð1 À eÀ0:68Â7:16 Þ ¼ 169:4  10 6 The minimum fluid-film thickness at the outer race is hmin ¼ 169:4  10 6  10: 62  10 3 m ¼ 1:80  10 6 m Conclusion The following is a summary of the results Steel bearing Minimum thickness for inner race: hmin ¼ 1: 62. .. determining the maximum allowed bearing load from basic principles Copyright 20 03 by Marcel Dekker, Inc All Rights Reserved TABLE 13-1 Dimensions and Load Ratings for Deep Ball Bearing Series 6300 (From FAG Bearing Catalogue, with permission) Number Bearing of standard design 6300 6301 63 02 6303 6304 6305 6306 6307 6308 Dimensions Bearing with one shield Bearing with two shields Bearing with one seal Bearing. .. 18600 20 000 22 400 27 000 30000 325 00 38000 43000 50000 60000 TABLE 13 -2 Angular Contact Bearing of Series 909 a ¼ 25  , separable (From FAG Bearing Catalogue, with permission) EQUIVALENT DYNAMIC LOAD F P ¼ Fr when a 0:68 Fr Fa P ¼ 0:41 Fr þ 0:87 Fa when > 0:68 Fr d D 7503 1.1904 8 128 1 .28 15 9379 1.4384 6875 1. 125 0 7503 2. 0800 2. 9630 2. 4370 3.3750 3.0300 3.9300 1.8750 2. 5000 2. 2500 Number 909001 9090 02 909003 . 95 20 0 45 4 24 500 22 400 6 320 100 21 5 47 4 28 500 27 000 6 321 105 22 5 49 4 30500 30000 6 322 110 24 0 50 4 325 00 325 00 6 324 120 26 0 55 4 36000 38000 6 326 130 28 0 58 5 39000 43000 6 328 140 300 62 5. 6309 RS 6309.2RS 45 100 25 2. 5 9150 6700 6 310 6310Z 6 310. 2Z 6 310 RS 6 310. 2RS 50 110 27 3 106 00 8150 6311 6311Z 6311.2Z 55 120 29 3 129 00 100 00 63 12 6312Z 63 12. 2Z 60 130 31 3.5 14000 108 00 6313. ¼ 4134 N 2: 2  10 11 N=m 2  10: 62 2  10 À6 m 2 ¼ 16:7  10 À5 Substituting these values in the formula for the minimum thickness, we get h min 10: 62 10 À3 ¼ 3:63  ð37 10 À11 Þ 0:68 2: 2  10 À8 Â