The Tolerance Field Effect On The Angular Contact Ball Bearings Systems Rating Life

80

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THE TOLERANCE FIELD EFFECT ON THE ANGULAR CONTACT BALL BEARINGS SYSTEMS’ RATING LIFE Rezmires Daniel, Racocea Cezar Technical University "Gh. Asachi” of Iasi, [email protected]

ABSTRACT To assure both good dynamic load capacity and higher shaft stiffness, two rolling bearings are usually mounted in pair. The load distribution on the contacts of the two rolling bearings depends on individual stiffness of the each bearing, on the length separator between the bearings, and on the chosen tolerance values. In this work we present a model in five degrees of freedom, which could serve to find the load distribution in bearing arrangements, considering the intermediate elements as rigid bodies. The assembly’ stiffness was determined considering the individual stiffness of each element, and the rating life was expressed as a function of tolerance values of the intermediate elements. This was realised by solving a non-linear system of equations, including the centrifugal effects and some of the friction forces. KEYWORDS: Angular – contact ball bearing, Quasi-dynamic equilibrium, Rating life.

For any rolling bearing pair, (r,j), the distance pieces L1, L2 are considered to have the same initial length and the shaft is considered as rigid. The following co-ordinate systems were considered: • an inertial system OXYZ with its origin on the middle length of the inner ring's curvature centres; • a rolling element frame OX1Y1Z1 for each (r,j) ball. To limit the complexity of the analysis, the following assumptions were admitted: • the bearing was mounted on an elastic shaft and in a rigid housing; • the surfaces in contact have ideal shapes; • the pair bearing system were considered to be rigid except the local contact zones. Related to the ball co-ordinate system (OX1Y1Z1)r,j two degrees of freedom, represented by the translations ux and uz, have been considered for each ball. The external load vector {F}, applied to the entire arrangement, contains 5 components that are further divided to each bearing of the arrangement: {F} = {Fx,Fy,Fz, My, Mz} (1a) {F}r = {Fxr,Fyr,Fzr, Myr, Mzr} (1b) The static equilibrium provides easily the system of equations: Fx1+Fx2=Fx Frz2+Frz1=Fz Fry2+Fry1=Fry Fz.(L+B)=Frz1.B Fy.(L+B)=Fry1.B (2) My1+My2=My Mz1+Mz2=Mz

The displacement vector {δ}r of the “r” inner ring has also five components: (3) {δ}r = {δx, δyr, δzr, γy, γz} L1

B1 r=2

B2

Z1

Z1

Oi

Oi

r=1

X1 Y1

dm/2

1.Analytical Approach

Y1

Ow

Ow

X1

Oe

Oe

Z

αr,1 R

αr,1

C

R

Fz α0

α0

Fa Fy

B1i L

Y L2

14 .2 446

O

X B2i

B

Fig.1. General view

2. Static Equilibrium of the (r,j) ball element To solve the equilibrium system (3), is necessary to find the components of {δ}r vector which are functions of distances OeOw, and OiOw respectively. The following notations were introduced: loe= OeOw= Ro-Dw/2-Sd/4; (4a) loi= OiOw =Ri-Dw/2-Sd/4; (4b) Lie=loi+loe (4c)

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Considering identical bearings in the arrangement pair, the D1 and D2 values presented in Figures 2 and 3, are : D1= L1+(Bi,1+Bi,2)/2 (5a) D2= L2+(Bo,1+Bo,2 )/2 (5b) D1= D2 (5c) Considering further the tolerances TBi,o,r corresponding to B1, B2 distances and the tolerances TL1,2 corresponding to L1 and L2 distances, the values D1 and D2 become: D2p=L2+TBo,1+TL2+ TBo,2+(Bo,1+Bo,2)/2 (6) D1p=L1+TBi,1+TL1+ TBi,1+(Bi,1+Bi,2)/2 (7) The initial contact angle α0 and Lie parameters depend also on D2p and D1p values, so that new values α0,r and Lie(r) have to be considered. Also, α1,r and R parameters are different versus the initial values. The supplementary axial clearance introduced by the effective values for D2p and D1p parameters is: ja =D2p-D1p (8) r=1

r=2 D1 Oi

Oi

Owp

Ow

Ow

+ δ yr . sin( ψ( r , j )) + R.[cos( α 1 ( r , j )) − cos( α r ,1 )]

D2p

Fig. 2. The α0 and Lie when L2+(Bo,1+Bo,2)/2> L1+(Bi,1+Bi,2)/2 r=1

D 2p D1

Oi

O ip

Oi Ow

Ow α0 Oe

(11)

where: z( r , j ) = Lie ( r ). cos( α 0 ,r ) + δ zr .cos( ψ( r , j )) +

D2

r=2

and • αr,1=arctan{[ D2p]/[2.[ dm/2+loi(r).cos(α0,r)]]} • R=[ D2p]/[2.sin(αr,1)]; If the inner ring is misaligned around OY and Oz axes with γy and γz angles respectively, then the initial angle αr,1 become a function of α1(r,j): α1(r,j)=αr,1+sgn(r).γy.cos(ψ(r,j))+ sgn(r).γz.sin(ψ(r,j)) (9) where: • ψ(r,j) defines the angular position of the ball element in the inertial system; • sgn(r) defines "r" row:  1, r = 1 sgn( r ) =  (10) •  − 1, r = 2 Because in the static load case the inner and outer contact angles are equal for any individual ball element, but different for every ball, the total deformation that acts on the (r,j) ball can be written as:

δ( r, j ) = x( r, j )2 + z( r, j )2 −loi −loe

Oe

Oe

for r=1 α0,1 =arctan((Lie.sin(α0)+ja)/(Lie.cos(α0)); sd1(1)=[(Lie.cos(α0,1))2+(Lie.sin(α0,1)+ja)2)0.5-Lie; Lie(1)=loi+loe+sd1(1); • for r=2 • α0,2=α0; sd1(2):=0; Lie(2):=Lie

α0

α0p

Oep

• • • •

81

Owp α 0p

Oe D2

Figure 3. The α0 and Lie when L2+(Bo,1+Bo,2)/2< L1+(Bi,1+Bi,2)/2 Assuming “ja” as decision criteria results: if ja>0 • for r=2 • α0,2 =arctan((Lie.sin(α0)+ja)/(Lie.cos(α0)); • sd1(2)=[(Lie.cos(α0,2))2+(Lie.sin(α0,2)+ja)2] 0.5-Lie; • Lie(2) =loi+loe+sd1(2); • for r=1 • α0,1=α0; sd1(1):=0; Lie(1):=Lie and • αr,1=arctan{[ D2p]/[2.[ dm/2+loi(r).cos(α0,r)]]} • R=[ D2p]/[2.sin(αr,1)]; if ja<0

x( r , j ) = Lie ( r ). sin( α 0 ,r ) + δ x + R [sin( α r ,1 ) − − sin( α 1 ( r , j ))] The contact angle for the (r,j) roller element is:  x( r , j )   α s ( r , j ) = α i ( r , j ) = α e ( r , j ) = arctan  z( r , j )  (12) The normal load and contact angle are given by: (13a) Q(r,j) = Kech.δ(r,j)n αi(r,j) = αe(r,j) (13b) The {δ}r displacement vector results by solving the equilibrium equation system for the inner ring. Using the previous relations, the equilibrium of forces and moments are: Fz =

∑∑ Q( r , j ).cos( α ( r , j )) cos( ψ( r , j )) i

r

=

∑∑ F

zr ( r ,

r

Fy =

∑∑ Q( r , j ).cos( α ( r , j )) sin( ψ( r , j )) i

j

∑∑ F r

(15a) j)

j

r

=

j

j

yr ( r ,

(15b) j)

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82

Fx =

∑∑ Q( r , j ) sin( α ( r , j )) = i

r

=

∑∑ r

My =

j

(15c)

j

x

r

j ).bx ( r , j ).

x

{δ}

z

j

∂{δ}r

j

   D  by ( r , j ) = C +  δi ( r , j ) + loi − w .cos(α s ( r , j )). 2     . sin( ψ( r , j ))    D  bz ( r , j ) = C +  δi ( r , j ) + loi − w .cos(α s ( r , j )). 2     .cos(ψ( r , j )) δi(r,j)=δ(r,j).(Kech/Ki)1/n (16) The {δ}r components represent the solution of the Eq. (15a-15e) and it was found by an Newton-Raphson algorithm. To solve the equilibrium system (15) is necessary to write the Jacobian matrix for the two bearings. The rigidity matrix Mr is: ∂Fxr ∂δy ∂Fry ∂δy ∂Frz ∂δy ∂My ∂δy ∂Mz ∂δy

∂Fxr ∂δz ∂Fry ∂δz ∂Frz ∂δz ∂My ∂δz ∂Mz ∂δz

∂Fxr ∂γy ∂Fry ∂γy ∂Frz ∂γy ∂My ∂γy ∂Mz ∂γy

∂Fxr  ∂γz   ∂Fry  ∂γz  ∂Frz   ∂γz  ∂My   ∂γz  ∂Mz  ∂γz  r

∂M z {δ}

∑ F ( r , j ).b ( r , j ) + ∂∑ x

y

j

∂{δ}r

r

(15e)

where: • Q(r,j) represents the load acting on the (r,j) ball; • Fzr(r,j), Fyr(r,j) represent the radial forces which act in “r,j” ball; • bx,y,z(r,j), represents the distance from the point of inner raceway - ball contact to the center of the inertial system. D  B  bx ( r , j ) = +  δi ( r , j ) + loi − w . sin( α s ( r , j )) 2  2 

 ∂Fxr  ∂δx   ∂Fry  ∂δx  ∂Frz Mr =   ∂δx  ∂My   ∂δx  ∂Mz  ∂δx

(18b) ∂ [ K ech .δ( r , j ) . cos( α i ( r , j )). cos( ψ( r , j ))] n

∑

=

Fyr ( r , j ).bx ( r , j ).

j

∂ =

x

z

j

∂{δ}r

r

(18c) Fz ( r , j ).bx ( r , j )

j

.

∑ F ( r , j ).b ( r , j ) + ∂∑

(18d) Fy ( r , j ).bx ( r , j )

j

. (18e)

δx

Oi

δz

loi

Z

O' i

ux

Ow

uz

α 0,r

αi

O* w

loe αe

X

Oe

Fig. 4. The center of mass displacement (ux, uz) for the (r,j) ball. Fmo Qo Ffo

Fc αe

(17) Ffi α Qi

where: ∂Fxr = ∂{δ} r

∂{δ}r

j

∂

∂M y

∑∑ F ( r , j ).b ( r , j ) +

∑∑

∑

(15d)

j

r

+

y

j

zr ( r ,

Mz =

∂Frz = ∂{δ} r

∑∑ F ( r , j ).b ( r , j ) +

∑∑ F r

∂ [ K ech .δ( r , j )n . cos( α i ( r , j )). sin( ψ( r , j ))]

Fx ( r , j )

r

+

∂Fry = ∂{δ} r

i

Fmi

Fig. 5 The forces on (r,j) ball

∑ j

∂ [ K ech .δ( r , j )n . sin( α i ( r , j ))] ∂{δ}r

(18a)

3. Quasi-dynamic effects Due to centrifugal force, both the load and the contact angle are modified versus the static values.

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Considering the existence of the centrifugal force the final position for the mass centre of the (r,j) ball is presented in Fig. 4 as function of “ux” and “uz“ parameters. The new position of Owe point is found also with the Newton Raphson algorithm applied this time to all balls. The loads that act on the (r,j) ball are presented in Fig.5. Considering the guiding ball assumption [1], the equilibrium equations for the (j) ball are: EFCA(r,j)=Qi(r,j).sin(αi(r,j))-Qo(r,j).sin(αe(r,j))[Fmi.cos(αi(r,j))-Fmo.cos(αe(r,j))]=0 (19a) EFCR(r,j)=Qi(r,j).cos(αi(r,j))Qo(r,j).cos(αe(r,j))+[Fmi.sin(αi(r,j))Fmo.sin(αe(r,j))]+Fc=0 (19b) where: Fmi=2.(1-λ).Mg/Dw; Fmo=2.λ.Mg/Dw; Ffo=µ.Qo; Ffi=µ.Qi Mg=Dw2.m.10-7.ωc .ωw . sin(β) β - atitude angle, [rad]  sin( α e ( r , j ))   and λ=1, for outer race β = arctan   cos( α e ( r , j )) + γ  guiding  sin( α i ( r , j ))   and λ=0, for inner race β = arctan   cos( α i ( r , j )) − γ  guiding The following must be considered: • Ffo and Ffi act like blocking forces; • If Fmi>Ffi then Fmi=Fmi-Ffi else Fmi=0; • If Fmo>Ffo then Fmo=Fmo-Ffo else Fmo=0; In these conditions the rigidity matrix for (r,j) element  ∂EFCA( r , j ) ∂EFCA( r , j )    ∂ux ∂uz is: MFC( r , j ) =  ∂EFCR( r , j ) ∂EFCA( r , j )    ∂ux ∂uz   (20)

83

From these notations results:   − 1.5.XI 2 −  rri.( rri + loi ).ZI . Tdi  ∂EFCAi = Tqi .  ∂ux 1 XI 2  − + 3  ZI . Tdi ZI .Tdi 1.5 

  1.5.XO2 +  rro.( rro+ loe).ZO. Tdo  ∂EFCAo = Tqo.  ∂ux 1 XO2 +  − 3  Zo. Tdo ZO .Tdo1.5  ∂EFCA ∂EFCAi ∂EFCAo (23) = − ∂ux ∂ux ∂ux − 1.5.XI .ZI   −  + rri .( rri loi ). ZI . Tdi ∂EFCAi  = Tqi .  ∂uz XI XI 3  −  + 4 1.5 2  ZI . Tdi ZI .Tdi 

1.5.XO.ZO   +  + rro .( rro loe ). ZO . Tdi ∂EFCAo  = Tqo.  ∂uz XO XO3  +  + 4 1.5 2  ZO . Tdo ZO .Tdo 

∂EFCA ∂EFCAi ∂EFCAo = − ∂uz ∂uz ∂uz

(24)

− 1.5.XI   +  ∂EFCRi rri .( rri loi ). Tdi +  = Tqi .   ∂ux XI + 2  1.5  ZI .Tdi  1.5.XO   −  ∂EFCRo rro.( rro + loe ). Tdo  = Tqo.   ∂ux XO −   ZO 2 .Tdo 1.5 

The following notations were introduced to simplify the MFC(r,j) components:

∂EFCR ∂EFCRi ∂EFCRo − = ∂ux ∂ux ∂ux

dto=dto(r,j) - contact deformation for static load case at outer contact level; dti=dti(r,j) - contact deformation for static load case at inner contact level;

− 1.5.ZI   −  rri .( rri + loi ). Tdi  ∂EFCRi = Tqi .   ∂uz XI 2  _ 3   ZI .Tdi 1.5 1.5.ZO   +  rro .( rro loe ). Tdo + ∂EFCRo  = Tqo. 2   ∂uz XO  + 3 1 . 5   ZO .Tdo

αs=αs(r,j) - contact angle in the static case; ZO=(loe+dto).cos(αs)+uz; XO=(loe+dto).sin(αs)+ux; ZI=(loi+dti).cos(αs)-uz; XI=(loi+dti).sin(αs)-ux; rri=(ZI2+XI2)0.5-loi; Tqi=Ki.rri1.5 Tdi=1+(XI/ZI)2

rro=(ZO2+XO2)0.5-loe Tqo=Ko.rro1.5 Tdo=1+(XO/ZO)2

(21a) (21b) (22a) (22b)

∂EFCR ∂EFCRi ∂EFCRo − = ∂uz ∂uz ∂uz

(25)

……..(26)

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84

4. The rigidity matrix of the “r” bearing

1

The displacements ux and uz are obtained solving the eq. (20) to (26). The rigidity matrix (17) has the following elements: ∂ [ K i .δ i ( r , j ,ux ,uz )n . sin( α i ( r , j ,ux ,uz ))] ∂Fx = ∂{δ}r ∂{δ}r j

∑

∂Fry = ∂{δ}r ∂Frz = ∂{δ}r

∂M y

{δ}r

∑

∂[ Ki .δi ( r , j ,ux,uz )n .cos(αi ( r , j ,ux,uz )).sin(ψ( r , j ))] ∂{δ}r

j

∑

∂[ Ki .δi ( r , j ,ux,uz )n .cos(αi ( r , j ,ux,uz )).cos(ψ( r , j ))]

∂M z = {δ}r

.

∑F ( r, j,ux,uz).b ( r, j ) + ∂∑F ( r, j,ux,uz).b ( r, j )

∂ =

∂{δ}r

j

x

y

z

j

∂{δ}r

x

j

.

∑

x

z

j

y

∂{δ}r

.

where:

∑K .δ ( r, j,ux,uz ).cos(α ( r, j,ux,uz ))cos(ψ( r, j ))) F ( r , j ,ux,uz ) = ∑K .δ ( r , j ,ux,uz ).cos(α ( r , j ,ux,uz )) sin(ψ( r , j )) ) F ( r , j , ux , uz ) = ∑ K .δ ( r , j , ux , uz ). sin( α ( r , j , ux , uz ))

Fz ( r , j ,ux,uz ) =

i i

i

i i

i

r

y

r

x

i

i

(29)

6. Numerical examples

x

j

∑

The (r) bearing life in terms of millions of revolutions can be calculated as: L10 ( r ) = ( L−r e ( r ) + L−s e ( r ))−1 / e (30) Results that the pair bearings life in million of revolutions is: L10 = ( L−10e ( 1 ) + L−10e ( 2 ))−1 / e (31) Bearing life in term of hour can be calculated as: L .10 6 (32) B10 = 10 60.RPM

∑F ( r, j,ux,uz ).b ( j ) + ∂∑F ( r, j,ux,uz ).b ( r, j )

∂

1

 Nr −1  Nr −1 p  pe   Q( r , j ) p  Q( r , j ) pe       j =0  j =0   Qer ( r ) =  Q ( r ) = ; es    Z Z                

i

r

5. The (r) pair bearings life

The following geometrical values have been considered: B1=Bi1 =Bi2 =10 [mm]; B2=Bo1 =Bo2 =10 [mm] L2=L1=10 [mm]; Nr=12 balls / bearing; Dw=9.525 [mm]; dm=46 [mm]; αo=15 [deg] Ro/Dw=0.52; Ri/Dw=0.53; L=50 [mm] The effect of the “ja” parameter versus Fx1 and Fx2 evolution is related in Fig. 6. The external forces are: Fz=10 [N], Fy=10 [N], Fx=200 [N], L=50 [mm] To point out the influence of the length tolerances on load distribution, the axial displacement is related in Fig. 7 as function of the “ja” parameter. 180

   

4

Q Ls ( r ) =  cs  Qes

   

4

(28)

120

Fx2, ni=8000 [rpm]

100

Fx2, ni=10000 [rpm] Fx2, ni=12000 [rpm]

80

Fx1, ni=8000 [rpm] Fx1, ni=10000 [rpm]

60

Fx1, ni=12000 [rpm]

40

(1 ! γ )1.39  Dw 0.3 .D1.8 .N −1 / 3 (27)   (1 ± γ )1.3  dm  w r

where: A=98; λ=1; The calculation of the inner and outer race lives are given in Eq. (28), for the case of the inner race rotating with respect to the load. The race lives are calculated per Eq. (29), for the case when the inner race is stationary with respect to the load. The combination of the race lives to give the bearing life is shown by Eqs. (30) and (31). If the inner race rotates with respect to load, the lives can be calculated as: Q Lr ( r ) =  cr  Qer where:

140

20 0 -5

-3

-1

1

3

5

"ja" parameter, [mm]*100

Fig. 6. “ja” parameter versus axial displacement Fz=10 [N], Fy=10[N], Fx=200 [N], L=50 [mm] 10

Axial displacement, [mm]*100

0.41

 2. f   Qc = A.λ.  2. f − 1 

160

Fx1, Fx2, [N]

Using the Lundberg-Palmgreen life rating method applied to the (r,j) pair bearings, the basic dynamic element capacity, Qc, is defined as the ball load which will result in a life of a million revolutions of the raceway with 90 percent probability of survival. For a ball with diameter 25mm, basic dynamic ball capacity can be calculated as:

9.5 9 8.5 8 7.5

ni=8000 [rpm]

7

ni=10000 [rpm]

6.5

ni=12000 [rpm]

6 -6

-4

-2

0

2

4

6

"ja" parameter, [mm]*100

Fig. 7. Axial displacement versus “ja” parameter for the “r” bearing

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The influence of the length tolerances on the bearing life is related in Fig. 8 and 9, as function of the “ja” parameter. The external conditions are: Fz=5[N], Fy=0 [N], Fx=500 [N]; RPM=25000 [rpm]., L=100 [mm]. 1160

B=10, [mm] B=12 [mm]

1140

B=30 [mm] B=20 [mm]

B10, [hours]

1120 1100 1080 1060 1040 1020 -6

-4

-2

0

2

4

6

ja, [mm]*1000

Fig.8. Pair bearing's life versus “ja” parameter

B10, [hours]

1160 1140

ja=0

1120

ja=5

1100

ja=3

1080 1060

Oi,e L1,L2 TL1,2 B1, B2

centers of curvature lengths of intermediate parts tolerances of the L1 and L2 lengths, [mm] “r” bearing width, [mm]

Bi,r, Bo,r

inner and outer ring width, [mm]

R, B, C, L D1, D2 D1p, D2p Ro,i TBi,o,r OXYZ OX1Y1Z1 ux,uz x,y,z loe loi Lie, Lie(r) Sd, Sd1(r) Q(r,j) Qi,o(r,j) Fmi, Fmo, Ffi, Ffo ja {}

lengths, [mm] initial distances between curvature centers final distances between curvature centers outer and inner raceway radius, [mm] length tolerances of the inner and outer rings, inertial system rolling element frame for the (r,j) ball displacement of the mass center index to describe the axes distance between Oe and Ow points, [mm] distance between Oi and Ow points, [mm] distance between Oi and Oe points, [mm] diametrical clearance of the bearing [mm] normal load, [N] normal load –inner and outher racevay tangential forces axial clearance, [mm] vector index

Forces and moments {F}, {F}r {M} Mg Fx, Fxr Fy, Fyr Fz, Fzr My, Mz, Myr, Mzr

force vectors moment vector gyroscopic moment axial load along OX axes; radial loads along OY axes radial load along OZ axes external moments which act around of the OY and OZ axis respectively

1040

Life parametrs

1020 1000 0

10

20

30

40

50

60

70

B, [mm]

Fig.9. Pair bearing's life versus “B” parameter

6. Conclusions The parameters "ja" and "B" influnce the components of the load vector {F}r, and also the load distribution in the "r" rolling bearings' arrangement. - From the viewpoint of the maximum rating life, an optimum distance between the curvature centres of the races of the two bearings has been found (Fig.8 and 9). - In order to obtain a maximum rating life for a tandem mounted bearings (Fig.1), the parameter "ja" must approach to zero. Notations Indexes, distances and coordinate systems r j i o n Nr m Kech, Ki,, Ke dm Dw Ow

bearing index ball number in the “r” bearing inner ring outer ring point contact constant, n=1.5 the number of balls for the “r” bearing ball weight ,[kg] equivalent, inner, and outer, rigidity factors pitch diameter, [mm] ball diameter, [mm] mass center of the (r,j) ball;

L10 B10 Qc Qer Qes RPM

bearing life, mr, 90% prob. survival bearing life, hours, 90% prob. survival dynamic capacity dynamic equivalent load of the rotating race dynamic equivalent load of the stationary race revolutions per minute

Greek notations α0, α0,r, α0p αr,1 α1(r,j) αs(r,j) αi(r,j) αe(r,j) αi(r,j,ux,uz) αe(r,j,ux,uz) δi(r,j,ux,uz) δi(r,j,ux,uz) γ µ λ ωc,w

initial contact angle initial contact angle affected by “ja” parameter initial angle between “R” and OZ axis final angle between “R” and OZ axis contact angle obtained from static equilibrium inner contact angle obtained from the static equilibrium outer contact angle obtained from the static equilibrium outer contact angle inner contact angle local contact deformation at the inner ring level for the (r,j) index local contact deformation at the outer ring level for the (r,j) index γ=Dw/dm, dimensionless parameter friction coefficient John’s coefficient angular speed of the cage and ball,

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THE ANNALS OF UNIVERSITY “DUNĂREA DE JOS” OF GALAŢI FASCICLE VIII, 2002, ISSN 1221-4590 TRIBOLOGY

REFERENCES 1.T.Harris, „Rolling Bearing Analysis”. John Wiley & Sons, New York, London, Sydney, 1966 2.M.D.Gafitanu, D.N. Olaru, M.C. Cocea, „Die Veluste wegen der Reibung in Radial-Axial-Kugellagern bei hohen Drehzahlen“, Weqr, 160 (1993), 51-60 3.M.D. Gafitanu, Cretu, Sp, Olaru D – „Rulmenti, Proiectare si tehnologie“, vol 1, Ed. tehnica, Bucuresti, 1985 4.P.K. Gupta, „Dynamics of Rolling-Element Bearings, Part III: Ball Bearing Analysis”, Transactions of the ASME, vol. 101, July 1979 5.Rumbarger, J.H. Poplawski, J., V, „Correlating Computerized Rolling Bearing Analysis” Techniques to the ISO Standards on Load Rating Life, Tribology Transactions, vol.37 (1994), 793