The Load Distribution in Double Row Spherical Roller Bearings and Spherical Roller Bearings Systems in the Static Case

8/9/2019 The Load Distribution in Double Row Spherical Roller Bearings and Spherical Roller Bearings Systems in the Static C

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The Load Distribution in Double Row Spherical

Roller Bearings and Spherical Roller Bearings

Systems in the Static Case

D. Rezmires, I Bercea, Sp. Cretu, D. Olaru

Technical University GH Asachi of Iasi, Department of Machine Design & Tribology, Bvd.

D Mangeron, 61-63, 6600 Iasi, Romania

Abstract

This paper proposes an analytic formulation to simulate the mechanical interactions in double

row spherical roller bearing and double row spherical roller bearings systems taking into account

the type of the contact developed on the contact area and the contact angle modification as effect

of the external load application.. In this paper the relations who describe, the centre of the massof the roller element displacement are presented. Correlating the geometrical particularities of

distinct bearings, in this paper, using an analytical method, the geometrical interactions were

founded. The interdependence between the external load and the bearing system response was

takes into account.

1. Introduction

Spherical roller bearings are optimally suited for applications that require bearings with a high

degree of angular aligning capability and a very high load carrying capacity. The analytical

studies presented in literature for the spherical roller bearings consider that a single type ofcontact is developed, point contact or line contact and not as a function of the external load

magnitude [1]. It is know that when the bearing is high externally loaded some conjunctions can

be of line contact type and another of point contact type. For simplicity, double row spherical

roller bearings were abbreviated as SRB. The SRB has a special geometry that assures a high

combined radial and axial load. Because the SRB has an outer race that is a portion of a sphere,

result that they are internally self-aligning. Considering that property result that the external

moment with a minimum energy is supported, but in the same time the initial contact angle is

modified and as the effect, the contact load magnitude also is modified. The load distribution in

the double row spherical roller bearings system it is a function of the contact type developed on

the contact area. For a simply reading and writing the spherical roller bearing systems was noted

with SSRB.

2. Analytical ApproachTo describe the SRB geometry some specific functions and index was introduced. The index r =

1 or r=2 denoted bearing row 1 or 2 respectively. To describe the roller element position on the

"r" row an "j" index was introduced. To perform the analysis, the following coordinate systems

were used:

an inertial system OeXYZ (which has the origin in the geometrical centre of the outer racenoted Oe),

a rolling element frame OXrYrZr.

The relation between inertial system and the centre of mass of the rolling element with {r M(r,j)}vector were described. The inner ring has five degrees of freedom: it can be displaced in the axial

VAREHD 10, Suceava, 20-21 octombrie 2000


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direction along OeX axis and in radial directions OeY, OeZ, respectively and can be also tilt

around the radial axis OeY and OeZ respectively. These directions in Fig 1 were illustrated.

dm

LwD

w

r = 1r = 2

=0 =0

Z

Y

X

X2

Z2

X1

Z1

M(1,j)M(2,j) rr

j = 1

Oe

Outer ring

Inner ring

Roller element

Bearing

axis

Fig.1. The coordinate system attached to SRB

For each rolling element was considered two degrees of freedom: two translations x, z that are

related with respect to the roller coordinate system (OXrYrZr)r. To limit the complexity of the

analysis, the following major assumptions are made:

the bearing is mounted on a sturdy shafts and in a rigid housing;

the pressure distributions at the contacts have been assumed according to Hertzian theory;

the surfaces in contact had ideal shapes; the bearing components are considered to be rigid excepting the local contact zones.

2.1 Static Equilibrium of Rolling ElementFor the (j,r) parameter which describes the centre of the mass position of the (r,j) roller element a

sign function sgn(r), was introduced :

sgn( )rr

r=

=

=

1 1

1 2(1)

Also a (j,r) function which describe the initial contact angle modification as effect of the inner

ring misalignments, was introduced. If the inner ring was misalign with y and angle z angle

respectively, then the result that the initial contact angle became (j,r) :

(j,r)=0+sgn(r).(y.cos((j,r))+ z.sin((j,r)) (2)


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where:

( , )

( , )

( , )r j

r j for r

r j for r=

=

+ =

1

2(3)

When the phase difference between the bearings rows is zero then =0.The bearing is considered externally load by the force vector {F} as:

{F} = {Fx ,Fy ,Fz}

T

(4)where:

Fx represents the axial load,

Fy and Fz are the radial loads along to Y and Z axis respectively.

As effect of external force application, the inner ring can be displaced along the X, Y and Z axis

with x, y and z respectively. In these conditions the (r,j) roller element will be loaded by{Q(j,r)} force. In Fig. 2 the nominal and final point position for the curvature centre of the inner

and outer raceway was presented. Also the centre of the mass and curvature centres of roller

element was listed.

X

Z

A

B

A

B

Oi

Oi

x

zx

z

z

x

x

M

M

iO

iO

B

A

Z

X

.

.

.

0

e

i

Oe Oe

'

'

'

''

'

'

z

+

cos( )+ysin( )loi=BOi

loe=AOe

l'oi

l'oe

l'oe=OeA'

l'oi=B'Oi'

Fig.2. The initial and final position of curvature centres for rollers and raceways

The points presented in Fig 2 are:

Oe - outer race curvature centre

Oi - inner race curvature centre

M - centre of mass of the roller element

A,B - curvature centres of the roller element() - final position index for Oe, Oi, A,B,M points

Because in the static load case the inner and outer contact angle are equal for any individual

roller element but different for every (r,j) roller element, result that an generic translation for M

and B points in A point that is possible to do. Also a generic translation for the M, B points in

A point is possible to do. The generic point which result from the preview translation with P and

P respectively was noted. In Fig.3 the contact angle evolution in the plane described by the

(r,j) angle was presented.


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Oe

Oi

iO

eOeO

eO

Oi

Oi Oi

iO

iO

iO

P

PP

P

P

P

PP

(i)(ii)

(iii) (iv)

'

'

'

'

'

'

''

Fig.3. The contact angle evolution as function of (r,j) parameters.In the basis of Fig. 2 and 3 respectively, the following expression for the roller-raceway total

deformation and contact angle was founded:

L)]sin(.)cos(.)cos(.L[(])sin(.L[)j,r( 2yz2

x ++++= (5)

++

+=

)sin(.)cos(.)cos(.L

)sin(.Larctan)j,r(

yz

x

i (6)

were:

loe=Ro-R-Sd/2

loi=Ri-R-Sd/2

L=loi+loeSd represents the diametrical clearance of the bearing

=(r,j) is give by Equation.2.Using Eq.5 result that the normal load and the contact angle will be:

Q(r,j)= Kech.(r,j)n

(7a)

i(r,j)= e(r,j) (7b)where

Kech represents the equivalent rigidity factor

Because in the case of the SRB it is possible to obtain different contact type, result that Kech

factor for any roller element is a function of the external load. If the major axis of elliptical

contact noted ai(r,j) is greater like lw/2 than result that Kech is give by line contact type expression

and contact angle is equal with the nominal contact angle else Kech is give by point contact type


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expression and contact angle is different versus the initial contact angle. In these conditions the

equivalent contact rigidity became:

K

Ki Ko

ech

n n

n

=

+

1

1 11 1 , (8)

As a criterion to select the contact type is used the following relation:

contact type =p po contact if a r j l

l line contact if a r j l

i w

i w

( int ), ( , ) /

( ), ( , ) /

2

2(9)

In the case of point contact the rigidity factors Ki and Ko are defined by:

Ki o i o i o, ,.

,

* .. *= 2 15 105 0 5 1 5 , n=1.5, i(r,j)=o(r,j)In the case of line contact the rigidity parameters are equal and this parameters was noted with

"K". In these conditions the relation between rigidity terms is:

Ki = Ko = K,

K lw= 8 05 104 8 9. * / , n=1.11, i(r,j)=o(r,j)=

where:

i,o represent the curvature sum for inner and outer contact respectively, [mm]

*i,o represent a non-dimensional parameter;E is the equivalent elastic modulus;

lw represent the roller length [mm]

ai(r,j) is the major axis of the elliptical contact,

a r jQ r j

E.

i

i

( , ). ( , )

/

=

31 3

. (10)

According to these assumption to obtain the x, y and z displacement of inner the ring it isnecessary to solve the bearing equilibrium equations.

2.2. Bearing Equilibrium EquationsThe equations of bearing equilibrium in axial and radial directions are:

F Q r j r j r jz ijr

i= ( , ) cos( ( , )) cos( ( , )) (11a)

F Q r j r j r jy ijr

i= ( , ) cos( ( , )) sin( ( , )) (11b)

F Q r j r j sng rx ijr

i= ( , ) sin( ( , )) ( ) (11c)

2.3. The centre of the mass displacement for (r,j) roller elementIn Fig. 4 the nominal and final position of the roller centre of mass was listed. Thus, the point Oi

and P represents the displaced position of the Oi and P points respectively.


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O'

eO

0

X

i

iOx

i

.

z

Z

x

z

P

P'

cos( ) sin( )

y+

Fig.4. The centre of the mass displacement as function of inner ring displacement and (r,j)angle

In basis of Fig 4 result that the centre of mass displacement following components will had :

xx oe

oi oe

n

oe

oi oe

i

l

l l

l

l l=

++

+

. K

K. ( ). sin ( )

ec h

o

/1

(12a)

zz y oe

oi oe

n

oe

oi oe

i

l

l l

l

l l=

+

++

+

[ .cos( ) .sin( )]. K

K.( ).cos( )

ech

o

/

1

(12b)

were =(r,j) and it is give by Eq.5.

2.4. Numerical example to describe the contact type developed in SRBThe model developed in the previous section to 22212C SRB was applied. In Table 1 the internal

geometry of 22212C and 22308C SRB was listed.

SRB type 22212 C 22308 C

Roller diameter [mm] 12.5 13

Initial contact angle [o] 9.21 14.33

Number of roller per row, Z 18 13

Roller effective length, [mm] 10.56 12

Roller contour radius [mm] 48.8 39.5

Inner raceway contour radius, [mm] 50 40.35

Outer raceway contour radius, [mm] 50 40

Bearing pitch diameter, [mm] 85.077 66Diametrical play, Sd [mm] 0.005 0.005

Table 1. Some geometric characteristics of the 22212C and 22308C SRBs.


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To illustrate the contact type algorithm detection applies to 22212C SRB, in Table 2 some cases

of external load was presented.The contact type detected by algorithm for different external loaded applied to 22212C,

l=line contact type, p=point contact type, d=discharged

Index Fx=1000, Fz=4000 Fx=5000, Fz=40000 Fx=5000, Fz=40000 Fx=1000, Fz=40000

roller Fy=3000, =0 Fy=30000, =0 Fy=30000, =10 Fy=100, =0

j r=1 r=2 r=1 r=2 r=1 r=2 r=1 r=2

1 p p l l l l l l

2 p p l l l l l l

3 p p l l l l l l

4 p p l l l l l p

5 p p l l l p p p

6 p d l p l p d d

7 p d p d p d d d

8 p d p d p d d d

9 p d d d d d d d

10 d d d d d d d d

11 d d d d d d d d

12 d d d d d d d d13 d d d d d d d d

14 d d d d d d d d

15 p d d d d d p p

16 p d p d p d l p

17 p d l d l p l l

18 p p l p l p l l

Table.2. The algorithm detection of the contact type on the 22212C SRB for different external

loaded

The listing data presents in Tables 2 denote that at high load is possible to have linear contacts

and point contacts in the same time but for different (r,j) roller element. At the low external load(Fx=1000 N, Fz=4000 N, Fy=3000 N), the load vector is transferred by point contact for any

roller elements If the external load is increasing (Fx=5000 N, Fz=40000 N, Fy=30000 N), for

the 5th

roller element on the 2nd

row have a line contact in the case when =0 and the same

roller element described by (r,j) parameters will have a point contact for =10o. In Fig. 5, the

load distribution on the two-spherical roller bearing in the case =0 and =10o respectivelywas presented.

0

1000

2000

3000

4000

5000

6000

7000

8000

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Roller element

Load,

[N]

r=1, w ithout dephasage

r=2 w ithout dephasage

r=1, w ith dephasage

r=2, w ith dephasage

Fig 5. The load distribution on 22212C SRB considering algorithm contact type for =0 and

=10o respectively


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In Fig 6 is evidenced the influence of the contact type on the load distribution. It is presented the

report between the load distribution for (r,j) roller element considering the point contact type

versus algorithm contact type.

0.7

0.8

0.9

1

1.1

1.2

1.3

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Roller element, j

Qp/Qalg

r=1, Qp/Qalg

r=2, Qp/Qalg

Fig 6. The rapport between the load distribution considering only the point contact type versus

the load distribution obtained for algorithm contact type

3. Coordinate systems attached to SSRB

The load analysis was made keeping the outer race of the bearings fixed in space and letting the

inner rings to be displaced under external load. For any individual SRB described with

(r,j)indexes an index idx was attached. The idx index is 1 or 2 were idx=1 represent the

first SRB and idx=2 represent the second SRB. For the any SRB's from the system, the

following coordinate systems was attached:

an inertial system OeXYZidx (this system has the origin in the geometrical centre of the outerrace denoted by Oe),

a rolling element frame OXYZr,j,idx.

The relation between the "idx" inertial system and the centre of mass of rolling element with the

{rM(r,j,}idx vector was described. The inner ring of "idx" SRB has five degrees of freedom: it can

be displaced in the axial direction along OeXidx axis and in radial directions OeYidx, OeZidx,

respectively and can be also misalign around the radial axis OeYidx and OeZidx. In the Fig. 7, the

SSRB components was listed.


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Lw

Dw

r = 1r = 2

=0 =0

Z

Y

X

X2

Z2

X1

Z1

M(1,j)M(2,j) rr

j = 1

Oe

Outer ring

Inner ring

Roller element

Bearing

axis

1

Inner ring

Roller element

Lw

Dw

Outer ring

eO

M(2,j)

r = 2

r

Z

2X

Y

Bearing

axis

X

r = 1

M(1,j)r

X

=02Zj = 1

=01Z

Shaft

idx=1 idx=2

.L

Shaft initial axis

Fig.7. The coordinate system attached to SSRB system

In this analysis for each rolling elements two degrees of freedom was considered.

3.1. External load in idx SRBFor the (r,j)idx parameter which describe the centre of mass of the roller it is necessary to

implement a sign function noted sgn(r,idx), with the following expression:

sgn( , )r idxr

r=

=

=

1 1

1 2(13)

Also, to describe the SSRB a general function which describe the initial contact angle wasintroduced. That function was noted (r,j)ix and describe the initial contact evolution of initial

contact angle 0,idx under the effect of the "idx" inner ring misalignment. Considering that the

y,idx and z,idx are the misalignment angle of the idx inner ring result that the new initial freecontact angle became:

(r,j,idx)=0,idx+sgn(r,idx).(y,idx.cos((r,j,idx))+ z,idx.sin((r,j,idx)) (14)where:

( , , )

( , , )

( , , )r j idx

r j idx for r

r j idx for ridx=

=

+ =

1

2(15)

The SSRB is externally loading by the force vector {F}. In Eq 16. the {F}vector were presented.

{F} = {Fx,Fry,Frz}T (16)where:

Fx is the axial load

Fry and Frz are the radial loads along to Y and Z axis respectively.

As effect, the "idx" inner ring of the "idx" SRB can be displaced along the X, Y and Z axis with

x, yidx and zidx respectively. As effect of these displacements, some roller elements will beloaded. For the (r,j,idx) roller element the load will be {Q(r,j)}idx . Generally, the shaft is

relatively rigid and bearing misalignment due to the shaft is inconsequential magnitude. The load

distribution on idx SRB can be obtained if is know the effect of external load on any SRB. In

Fig 8 and Fig 9, two hypothesis cases of external load applied to a SRB system are shown.


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Frz

Fry

Fa

La

SRB 1 SRB 2

F

F

F

F

F

FF

F

rz1

ry1

rz2

a1 a2

ry2

Fig.8. The reactions in a simple two-bearing-shaft system

Fig.9. The reactions into a simple two-bearing shaft system, overhung load

The reactions acting along the Y and Z axis can determined from the simple equations of static

equilibrium. The reaction along the X axis is a function of the axial load distribution between the

idx SRB which respect the sum of effect.

for the case presented in Fig 8 the simple equations of static equilibrium are:

Frz1+Frz2=Frz Frz2.L-Frz.a=0Fry1+Fry2=Fry Fry2.L-Fry.a=0 (17)

Fa1+Fa2=Fa

for the case presented in Fig 9 the simple equations of static equilibrium are:Frz1+Frz2=Frz Frz2.(L-a)-Frz.L=0

Fry1+Fry2=Fry Fry2.(L-a)-Fry.L=0 (18)

Fa1+Fa2=Fa

The external load acting in the two-bearing-shaft system produce a misalignment effect on the

idx SRB, which modify the contact angle and also the load distribution in the idx SRB.

Thus, the misalignment angle for idx SRB are given by:

rz,idx = Frz/(6.E.Iz).f(a,L)idx

ry,idx = Fry/(6.E.Iz).f(a,L)idx (19)From Eq.5, result that normal load and contact angle are given by:

Q(r,j)idx= Kech,idx.(r,j)nidx (20a)

i(r,j)idx= e(r,j)idx (20b)where

Kech,idx represent the equivalent rigidity factor for the "idx" SRB

To find the x, y,idx and z,idx displacement of the "idx" inner ring it is necessary to solve thebearing equilibrium equations which was described by the Eqs 17 and Eqs 18 respectively.

3.2. System Equilibrium Equations

The equations of system equilibrium in axial and radial directions are:F Q r j r j r jrz1

jri= 1( , ) cos( ( , )) cos( ( , ))


11/12

F Q r j r j r jrzjr

i2 2= ( , ) cos( ( , )) cos( ( , ))

F Q r j r j r jryjr

i1 1= ( , ) cos( ( , )).sin( ( , ))

F Q r j r j r jryjr

i2 2= ( , ) cos( ( , )).sin( ( , ))

F Q r j r j sng r Q r j r j sng rajr

ijr

i= + 1 2( , ) sin( ( , )) ( ) ( , ) sin( ( , )) ( ) (21)

where: the Q1,2(r,j) terms are the load acting on (r,j) roller element and the idx SRB.

3.3. Numerical results for SSRB static analysisIn Fig 10 the load distribution for 22308 C SSRB was presented, considering that the external

load is apply like in the case listed in Fig.2. The external load applied to shaft is: Frz=2000 [N],

Fry=3000 [N], Fx=Fa=500 [N]. The geometric parameters a and L are: a=200 [mm], L=500

[mm] respectively.

The load distribution on SRB system

0

100

200

300

400

500

600

1 2 3 4 5 6 7 8 9 10 11 12 13

Roller index, j

Load[N]

r=1, idx=2

r=2, idx=2

r=1, idx=1

r=2, idx=1

Fig .10. The load distribution on two-spherical roller bearing system for the case presented in

Fig.8

Considering the case presented in Fig 11, in Fig 5 is listing the load distribution for an 22308C

SSRB was presented. For this case, the external load applied to shaft is: Frz=2000 [N], Fry=3000

[N], Fx=Fa=500 [N]. The geometric parameters a and L are: a=200 [mm], L=700 [mm]

respectively.


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The load distribution on SRB system

0

200

400

600

800

1000

1200

1 2 3 4 5 6 7 8 9 10 11 12 13

Roller index, j

Load

[N]

r=1, idx=1

r=2, idx=1

r=1, idx=2

r=2, idx=2

Fig .11. The load distribution on two-spherical roller bearing system for the case presented in

Fig.9

Notations and abbreviations:

SRB spherical roller bearing

SSRB spherical roller bearings system

r row index

j roller element index

idx bearing index

i,e inner and outer index parameter

Sd diametrical clearance

R roller radius curvatureRo outer race roller radius

Ri inner race roller radius

Oe outer race curvature centre

Oi inner race curvature centre

M centre mass of rolling element

A,B curvature centres of the rolling

element

P generic point

nominal contact angle

inner ring displacement

inner ring tilt angle

centre roller displacement

anglular position of roller element

phase differenceX,Y,Z index to describe the bearing axis

F external load

Q roller load

{ } vector

final position index

p punctual index

l linear indexalg algorithm index

References

1. Harris T.A. Rolling bearing analysis,3

rdedition, 1991

Documents

The Load Distribution in Double Row Spherical Roller Bearings and Spherical Roller Bearings Systems in the Static Case