Bearing Design in Machinery Episode 2 Part 9 doc

wear and optimize the bearing life. In such cases, it is possible to have an optimal design that operates with a full EHD fluid film. For this purpose, the designer can calculate the minimum film thickness, h min , according to the equations discussed in the following sections. For optimal conditions, the minimum film thickness, h min , must be greater than the surface roughness. The equivalent surface roughness at the contact, R s (RMS), is obtained from the roughness of the two individual surfaces in contact, R s1 and R s2 , from the equation (see Hamroch, 1994) R s ¼ðR 2 s1 þ R 2 s2 Þ 1=2 ð12-38Þ The surface roughness is measured by a profilometer, often referred to as stylus measurement. Microscope devices are used when higher-precision measurements are needed. The ratio L of the film thickness to the size of surface asperities, R s , is: L ¼ h min R s ð12-39Þ For a full EHD lubrication of rolling bearings, L is usually between 3 and 5. The desired ratio L is determined according to the expected level of vibrations and FIG. 12-20 Pressure wave and film thickness distribution in elastohydrodynamic lubrication. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. other disturbances in the machine. Although h min is very important for successful bearing operation, calculation of h 0 is often required for determining the viscous shear force, referred to as traction force, which is the resistance to relative sliding. 12.8 ELASTOHYDRODYNAMIC LUBRICATION OF A LINE CON TACT A theoretical line contact is formed between two cylinders, such as in a cylindrical rolling bearing. For this case, Pan and Hamrock (1989) introduced the following empirical equation for the minimum film thickness, h min : h min R x ¼ 1:714 " UU 0:694 r ðaE eq Þ 0:568 " WW 0:128 ð12-40Þ Here, a is the viscosity–pressure coefficient. Although theoretical equations were derived earlier, Eq. (12-40) is more accurate, because it is based on actual measurements. The equation is a function of dimensionless terms. The advantage in using dimensionless terms is that any system of units can be used, as long as the units are consistent and result in dimensionless terms. The dimensionless terms " UU r and " WW are the dimensionless rolling velocity and load per unit of cylinder length, respectively. The dimensionless terms are defined by: " UU r ¼ m 0 E eq R x U r " WW ¼ 1 E eq R x L W ð12-41Þ Here, U r is the rolling velocity, Z 0 is the viscosity of the lubricant at atmospheric pressure, W is the load on one rolling element, E eq is the equivalent modulus of elasticity, and R x is an equivalent radius of contact in the x plane (direction of axis of rotation of rolling element), as defined in Sec. 12.5.2. Jones et al. (1975) measured the viscosity–pressure coefficient a ½m 2 =N for various lubricants. The data is summarized in Table 12-1. The rolling velocity " UU r in Eq. (12-41) is for pure rolling (no relative sliding) (Eq. (12-34)). However, in many other problems, such as in gears and cams, there is a combination of rolling and sliding. If the ratio of the rolling to the sliding is x and U s is the sliding velocity, then the dimensionless rolling velocity " UU r used in Eq. (12-41) is replaced by " UU r ¼ m 0 2E eq R x U s ð1 þxÞð12-42Þ Equation (12-40) indicates that the viscosity and rolling speed have a more significant effect on the minimum film thickness, h min , in comparison to the load Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. and its Poisson ratio is n ¼ 0:3. The bearing is lubricated by oil having an absolute viscosity of m 0 ¼ 0:01 N-s=m 2 at atmospheric pressure and bearing operating temperature. The viscosity–pressure coefficient is a ¼ 2:2  10 À8 m 2 =N. Find the minimum elastohydrodynamic film thickness at the following points: a. At the contact of the rolling elements with the inner raceway b. At the contact of the rolling elements with the outer raceway Solution The radii of the contacting curvatures are: Roller radius: R roller ¼ 0:01 m, Outer ring raceway: R outer raceway ¼ 0:08 m, Inner ring raceway: R inner raceway ¼ 0:06 m The contact between the rolling elements and the inner raceway is convex, and the equivalent contact curvature, R x;in , is derived according to the equation 1 R x;in ¼ 1 R roller þ 1 R inner raceway 1 R x;in ¼ 1 0:01 þ 1 0:06 ) R eq;in ¼ 0:0085 m However, the contact between the rolling elements and the outer raceway is concave, and the equivalent contact curvature, R x;out , is derived according to the equation 1 R x;out ¼ 1 R roller À 1 R outer raceway 1 R x;out ¼ 1 0:01 À 1 0:08 ; R x;out ¼ 0:0114 m If we assume that there is no radial clearance, then the maximum load on one cylindrical rolling element can be calculated from the following formula: W max ¼ 4W shaft n r ¼ 4  11;000 14 ¼ 3142 N where n r is the number of rollers in the bearing. 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 ) E eq ¼ 2:05 Â10 11 1 À0:3 2 ¼ 2:25 Â10 11 N=m 2 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. The dimensionless load on the inner race is " WW ¼ 1 E eq R x;in L W max ¼ 1 2:25 Â10 11  0:0085 Â0:01 3142 ¼ 1:64 Â10 À4 In comparison, the dimensionless load on the outer race is " WW ¼ 1 E eq R x;out L W max ¼ 1 2:25 Â10 11  0:0114 Â0:01 3142 ¼ 1:22 Â10 À4 Comparison of the inner and outer dimensionless loads indicates a higher value for the inner contact. This results in higher contact stresses, including maximum pressure, at the inner contact. The rolling velocity, U r , for a cylindrical roller is calculated via Eq. (12-34). U r ¼ R in R out R out þ R in o where the angular shaft speed, o, is equal to o ¼ 2pN 60 ¼ 2  p  5000 60 ¼ 523 rad=s Hence, the rolling speed is U r ¼ R in R out R out þ R in o ¼ 0:06 Â0:08 0:06 þ0:08 523 ¼ 17:93 m=s The dimensionless rolling velocity of the inner surface is " UU r ¼ m 0 E eq R x;in U r ¼ 0:01 2:25 Â10 11  0:0085 17:93 ¼ 9:375 Â10 À11 In comparison, the dimensionless rolling velocity at the outer ring race is " UU r ¼ m 0 E eq R x;out U r ¼ 0:01 2:25 Â10 11  0:0114 17:93 ¼ 6:99 Â10 À11 The elastohydrodynamic minimum film thickness is derived from Eq. (12.40) for a fully lubricated ball bearing. h min R x ¼ 1:714 " UU 0:694 ðaE eq Þ 0:568 " WW 0:128 The minimum oil film thickness at the contact with the inner raceway is h min R x;in ¼ 1:714ð9:3 Â10 À11 Þ 0:694 ð2:2  10 À8  2:25 Â10 11 Þ 0:568 ð1:64 Â10 À4 Þ 0:128 h min;in ¼ 0:609 mm Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. The minimum oil film thickness at the contact with the outer raceway is h min R x;out ¼ 1:714ð6:99  10 À11 Þ 0:694 ð2:2  10 À8  2:25 Â10 11 Þ 0:568 ð1:22 Â10 À4 Þ 0:128 h min;out ¼ 0:695 mm The minimum film thickness at the contact with the inner ring race is lower because the equivalent radius of convex curvatures is lower. Therefore, it is sufficient to calculate the minimum film thickness at the contact with the inner ring race. However, for a rolling bearing operating at high speed, the centrifugal force of the rolling element is added to the contact force at the contact with the outer raceway. In such cases, h min;out may be lower than h min;in , and the EHD fluid film should be calculated at the inner and outer ring races. Example Problem 12-4 Elastohydrodynamic Fluid Film in a Cam and a Follower A cam and a follower operate in a car engine as shown in Fig. 12-21. The cam radius at the tip of the cam is R ¼ 20 mm, the distance of this radius center from FIG. 12-21 Cam and follower. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. the cam center of rotation is a ¼ 40 mm, and the width of the cam (effective length of contact) is 10 mm. There is a maximum reaction force W ¼ 1200 N between the cam and the follower when the follower reaches its maximum height. The rotation speed of the cam is N ¼ 600 RPM. The cam and the follower are made of steel. The steel modulus of elasticity is E ¼ 2:05 Â10 11 N=m 2 , and its Poisson ratio n is 0.3. The contact between the cam and follower is fully lubricated. The absolute viscosity at atmospheric pressure and engine tempera- ture is m 0 ¼ 0:01 N-s=m 2 , and the viscosity–pressure coefficient is a ¼ 2:2 Â10 À8 m 2 =N. Find the minimum oil film thickness at the contact between the cam and the shaft. Solution The contact is between a plane and a curvature of radius R ¼ 20 mm at the tip of the cam. In that case, the equivalent radius is R. The equation for the equivalent modulus is E eq ¼ E 1 À n 2 ¼ 2:05 Â10 11 1 À 0:3 2 ¼ 2:2 Â10 11 N=m 2 The dimensionless load becomes " WW ¼ 1 E eq RL W max ¼ 1 2:2  10 11  0:020 Â0:01 1200 ¼ 2:727 Â10 À5 The sliding velocity U is equal to the tangential velocity at the tip of the cam: U ¼ oða þ RÞ Here, o is equal to o ¼ 2pN 60 ¼ 2p600 60 ¼ 62:83 rad=s This problem is one of pure sliding, x ¼ 0, and the sliding velocity is U s ¼ð0:04 þ0:02Þ62:83 ¼ 3:76 m=s For pure sliding, x ¼ 0, the dimensionless equivalent rolling velocity is [Eq. (12-42)] " UU r ¼ m 0 2E eq R U s ð1 þ xÞ¼ 0:01 2  2:2 Â10 11  0:02 3:76 ¼ 4:27 Â10 À12 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. For line contact and in the presence of sufficient lubricant, the elastohydro- dynamic minimum film thickness is derived according to the equation h min R ¼ 1:714 " UU 0:694 ðaE eq Þ 0:568 " WW 0:128 The oil film thickness in the inner surface is h min 0:02 ¼ 1:714ð4:27 Â10 À12 Þ 0:694 ð2:2 Â10 À8  2:22 Â10 11 Þ 0:568 ð2:727 Â10 À5 Þ 0:128 h min ¼ 0:21 mm 12.9 ELASTOHYDRODYNAMIC LUBRICATION OF BALL BEARINGS Under load, the theoretical point contact between a rolling ball and raceways becomes an elliptical contact area. The elliptical contact area has radii a and b,as shown in Fig. 12-17. In a similar way to the theoretical line contact, there is a minimum elastohydrodynamic film thickness, h min , near the exit from a uniform film thickness, h 0 . For hard surfaces, such as steel in rolling bearings, and sufficient lubricant, Hamrock and Dowson (1977) obtained the following formula for the minimum film thickness, h min : h min R x ¼ 3:63 " UU 0:68 r ðaE eq Þ 0:49 " WW 0:073 ð1 Àe À0:68k Þð12-44Þ Here, a is the viscosity–pressure coefficient and " UU r and " WW are dimensionless velocity and load, respectively, defined by: " UU r ¼ m o U r E eq R x and " WW ¼ W E eq R 2 x ð12-45Þ Here, U r is the rolling velocity, m 0 is the viscosity of the lubricant at atmospheric pressure and bearing operating temperature, W is the load on one rolling element, and E eq is the equivalent modulus of elasticity. The equations for the equivalent modulus of elasticity and equivalent contact radius are used for calculating the Hertz stresses at a point contact. These equations were discussed in Secs. 12.4 and 12.5. For convenience, these equations are repeated here. Recall that E eq is determined from equation 2 E eq ¼ 1 À n 2 1 E 1 þ 1 À n 2 2 E 2 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. Here, n is poisson’s ratio and E is the modulus of elasticity of the respective two materials. For identical materials, the equation becomes E eq ¼ E 1 À n 2 The equivalent radius of curvature in the plane of rotation, R x , for the contact with the inner ring race is 1 R x ¼ 1 R 1x þ 1 R 2x where R 1x and R 2x are as shown on the right-hand side of Fig. 12-18b for the contact at the inner ring race and R 1x ¼ d=2, where d is the ball diameter. On the left-hand side of Fig. 12-18b, the contact at the inner ring race is concave, and the equivalent contact curvature radius in this plane is 1 R y ¼ 1 R 1y À 1 R 2y The radius ratio, a r (the ratio of the larger radius to the smaller radius, a r > 1) is defined as a r ¼ R y R x The ellipticity parameter, k, is the ratio k ¼ b a The parameter k can be estimated from k % a 2=p r For hard surfaces, sufficient lubricant, and for pure rolling, Hamrock and Dowson (1981) obtained the following formula for the central film thickness, h c (at the center of the fluid film): h c R x ¼ 2:69 " UU 0:67 r ðaE eq Þ 0:53 " WW 0:067 ð1 À0:61e À0:73k Þð12-46Þ This equation is useful for the calculation of the traction force (resistance to relative sliding) where an EHD film is separating the surfaces. For soft surfaces, such as rubber, the contact area is relatively large, resulting in a lower contact pressure. In addition, the viscosity does not increase as much as predicted by the preceding equations. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. Example Problem 12-5 Find the minimum film thickness for a rolling contact of a deep-groove ball bearing having the following dimensions: The bearing has 14 balls of diameter d ¼ 19:04 mm. The radius of curvature of the inner-deep groove (in cross section x-z on the left-hand side of Fig. 12-18b) is 9.9 mm. The inner race diameter, R 2x (at the bottom of the deep groove), is d i ¼ 76:5 mm (cross section y-z on the right-hand side of Fig. 12-18b). The radial load on the bearing is W ¼ 10;500 N, and the bearing speed is N ¼ 5000 RPM. The rolling elements and rings are made of steel. The modulus of elasticity of the steel for rollers and rings is E ¼ 2  10 11 N=m 2 , and Poisson’s ratio is n ¼ 0:3. The properties of the lubricant are: The absolute viscosity at ambient pressure and bearing operating temperature is m 0 ¼ 0:04 N-s=m 2 , and the viscosity–pressure coefficient is a ¼ 2:3 Â10 À8 m 2 =N. Solution Referring to Fig. 12-18b, the radius of curvature in the y-z plane is R 2x ¼ 38:25 mm and R 1x ¼ d 2 ¼ 9:52 mm Equivalent Radius for Inner Raceway Convex Contact in y-z Plane ðx Plane) 1 R x ¼ 1 R 1x þ 1 R 2x ) 1 R x ¼ 1 9:52 þ 1 38:25 ; R x ¼ 7:62 mm Equivalent Inner Radius in y Plane (Concave Contact) The curvatures in this plane are: R 1y ¼ 9:52 mm and R 2y ¼ 9:9 mm. The equivalent inner radius in the y plane is 1 R y ¼ 1 R 1y À 1 R 2y ) 1 R y ¼ 1 9:52 À 1 9:9 ; R y ¼ 248:0mm Equivalent Curvature of Inner Ring and Ball Contact 1 R eq ¼ 1 R x þ 1 R y ) 1 R eq ¼ 1 7:62 þ 1 248 ; R eq ¼ 7:4mm a r ¼ R y R x ¼ 32:55 and k ¼ a 2=p r ¼ 32:54 2=p ¼ 9:18 Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. The shaft and the bearing are made of identical steel, so the equivalent modulus of elasticity is: E eq ¼ E 1 À n 2 ) E eq ¼ 2  10 11 1 À 0:3 2 ¼ 2:2 Â10 11 N=m 2 For a ball bearing without radial clearance, the maximum load at the contact of one rolling element can be estimated with the following equation: W max % 5W shaft n r ¼ 5ð10;500Þ 14 ¼ 3750 N where n r ¼ 14 is the number of balls in the bearing. The angular velocity, o,is o ¼ 2pN 60 ¼ 2p5000 60 ¼ 523:6 rad=s The rolling speed is derived according to 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 or U r ¼ R 2x ðR 2x þ dÞ 2R 2x þ d o U r ¼ 38:25  10 À3 ð38:25 Â10 À3 þ 19:05 Â10 À3 Þ ð2 Â38:25 Â10 À3 Þþ19:04 Â10 À3 523:6 ¼ 12 m=s The dimensionless rolling velocity at the inner ring race is " UU r ¼ m 0 E eq R x U r ¼ 0:01 2:2 Â10 11  0:00762 12 ¼ 7:17  10 À11 The dimensionless load on the inner ring race is " WW ¼ 1 E eq R 2 x W max ¼ 1 2:2 Â10 11  0:00762 2 3750 ¼ 2:94 Â10 À4 Finally, the oil film thickness between the inner race and the roller is h min R x ¼ 3:63 " UU 0:68 r ðaE eq Þ 0:49 " WW 0:073 ð1 Àe À0:68k Þ) h min 0:00762 ¼ 3:63 ð7:17 Â10 À11 Þ 0:68 ð2:3 Â10 À8  2:2 Â10 11 Þ 0:49 ð2:94  10 À4 Þ 0:073 Âð1 À e À0:68Â9:18 Þ Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. [...]... R2x Rx 9: 52 57 : 29 Rx ¼ 11: 42 mm Equivalent Outer Radius in y Plane The curvatures in this plane are R1y ¼ 9: 52 mm and R2y ¼ 9: 9 mm The equivalent inner radius in the y-z plane is 1 1 1 1 1 1 À ; ¼ À ) ¼ Ry R1y R2y Ry 9: 52 9: 9 Ry ¼ 24 8:0 mm The equivalent radius of the curvature of the outer ring and ball contact is 1 1 1 1 1 1 þ ; ¼ þ ) ¼ Req Rx Ry Req 11: 42 248 Ry ar ¼ ¼ 21 : 72 Rx and 2= p k ¼ ar ¼ 21 : 722 =p... 5865 :93  0:010 92 ¼ ¼ 0:43 mm p  7:1  2: 2  1011 !1=3 ^ 6k 2 EWmax Req b¼ pEeq  1=3 6  7: 12  1:03  5865 :93  0:010 92 ¼ ¼ 3:07 mm p  2: 2  1011 The maximum pressure at the outer contact is pmax ¼ 3 Wmax 3 5865 :93 ¼ 21 21:64 N=mm2 ¼ 2: 12 GPa ¼ 2 pab 2 p  0:43  3:07 The maximum pressure is very close to that at the inner ring contact, pmax ¼ 2: 17 GPa (see Example Problem 12. 2) Copyright 20 03... Dekker, Inc All Rights Reserved b Minimum Film Thickness at the Outer Race Contact Rolling occurs only in the x plane, so R2x ðR2x þ dÞ o 2R2x þ d 57 : 29  10À3 ð57 : 29  10À3 þ 19: 04  10À3 Þ Ur ¼ 3141: 59 ¼ 1 02: 81 m=s 2  57 : 29  10À3 Þ þ 19: 04  10À3 Ur ¼ The dimensionless velocity for the outer surface is m 0:01 " Ur ¼ 0 Ur ¼ 1 02: 81 ¼ 4: 09  10À10 2: 2  1011  0:011 42 Eeq Rx where Ur is the rolling... 1 1 1 1 1 1 À ; ¼ À ) ¼ Rx R1x R2x Rx 9: 52 57 : 29 Copyright 20 03 by Marcel Dekker, Inc All Rights Reserved Rx ¼ 11: 42 mm Equivalent Outer Radius in y Plane The curvatures in this plane are R1y ¼ 9: 52 mm and R2y ¼ 9: 9 mm The equivalent inner radius in the y (or x-z) plane is 1 1 1 1 1 1 À ; ¼ À ) ¼ Ry R1y R2y Ry 9: 52 9: 9 Ry ¼ 24 8:0 mm The equivalent radius of the outer ring and ball contact is 1 1 1 1... the shaft, in rad=s The dimensionless load on the outer race is " W ¼ 1 1 W ¼  5865 :93 ¼ 2: 04  10À4 2 max 11  ð0:011 42 2 Eeq Rx 2: 2  10 Finally, the oil film thickness between the outside race and the ball is " Ur0:68 ðaEeq Þ0: 49 hmin ¼ 3:63 ð1 À eÀ0:68k Þ " Rx W 0:073 hmin ð4: 09  10À10 Þ0:68 2: 3  10À8  2: 2  1011 Þ0: 49 ¼ 3:63 0:011 42 2: 04  10À4 Þ0:073 hmin  ð1 À eÀ0:68Â7:1 Þ ¼ 2: 06 mm Example... 11: 42 248 ar ¼ Req ¼ 10: 92 mm Ry ¼ 21 : 72 Rx and 2= p k ¼ ar ¼ 21 : 722 =p ¼ 7:1 ^ It is also necessary to calculate E, which will be used to calculate the ellipsoid radii First qa will be determined p qa ¼ À 1 ¼ 0:57 2 q 0:57 ^ E %1þ a )1þ ¼ 1:03 21 : 72 ar From Example Problem 12- 5, the equivalent modulus of elasticity is Eeq ¼ 2: 2  1011 N=m2 The load on the bearing is divided unevenly between the rolling... pressure and minimum film thickness was calculated in Example Problems 12- 2 and 12- 5 In this problem, the maximum pressure will be compared with that at the outer ring race in the presence of a centrifugal force Contact with Outer Ring Race The contact radii in the x plane (y-z plane) in Fig 12- 18b are: R2x ¼ 57 : 29 mm and R1x ¼ d ¼ 9: 52 mm 2 The equivalent radius for the outer raceway concave contact in the... (steel and silicon nitride), given by 2 1 À n2 1 À n2 21 2 ¼ þ ) Eeq E1 E2 2 1 À ð0 :24 2 1 À ð0:3 2 ¼ þ ) Eeq ¼ 2: 65  1011 Pa Eeq 3:14  1011 2: 0  1011 The maximum load transmitted by one rolling element at the contact with the ring (assuming zero clearance) is estimated by the following equation: Wmax % 5Wshaft 5ð10;500Þ ¼ 3750 N ¼ 14 nr The number of balls in the bearing is nr ¼ 14 The total maximum...Thus, at the inner ring race hmin ¼ 0:4 12 mm Example Problem 12- 6 Centrifugal Force The deep-groove ball bearing in Example Problem 12- 5 is used in a high-speed turbine where the average shaft speed is increased to N ¼ 30;000 RPM The radial bearing load is equal to that in Example Problem 12- 5, W ¼ 10; 500 N The lubricant is also equivalent to that in Example Problem 12- 5 The properties of... equal to o¼ 2pN 2p30;000 ¼ ¼ 3141: 59 rad=s 60 60 The angular velocity of the rolling element center, point C, is given by oC ¼ Rin 38 :25 3141: 59 ¼ 125 7:75 rad=s o¼ 38 :25 þ 57 : 29 Rin þ Rout This angular velocity is used to calculate the centrifugal force, Fc : Fc ¼ mr ðRi þ rÞo2 C The density (silicon nitride) and volume determine the mass of the ball: mr ¼ p 3 p d r )  0:0 190 43  320 0 ¼ 0:0 12 kg 6 6 . according to the equation h min R ¼ 1:714 " UU 0: 694 ðaE eq Þ 0:568 " WW 0: 128 The oil film thickness in the inner surface is h min 0: 02 ¼ 1:714ð4 :27 Â10 À 12 Þ 0: 694 2: 2 Â10 À8  2: 22 Â10 11 Þ 0:568 2: 727 . contact with the inner raceway is h min R x ;in ¼ 1:714 9: 3 Â10 À11 Þ 0: 694 2: 2  10 À8  2: 25 Â10 11 Þ 0:568 ð1:64 Â10 À4 Þ 0: 128 h min ;in ¼ 0:6 09 mm Copyright 20 03 by Marcel Dekker, Inc. All Rights. balls in the bearing. The angular velocity, o,is o ¼ 2pN 60 ¼ 2p5000 60 ¼ 523 :6 rad=s The rolling speed is derived according to Eq. ( 12- 34): U rolling ¼ R in R out 2 R in þ rÞ o ¼ R in R out R in þ