High Speed Tapered Roller Bearing Optimization
High Speed Tapered Roller Bearing Optimization by Brady Walker A Project Submitted to the Graduate Faculty of Rensselaer Polytechnic Institute in Partial Fulfillment of the Requirements for the degree of MASTER OF MECHANICAL ENGINEERING
Approved: _________________________________________ Ernesto Gutierrez-Miravete, Project Adviser
Rensselaer Polytechnic Institute Hartford, Connecticut April, 2008
© Copyright 2008 by Brady Walker All Rights Reserved
ii
CONTENTS LIST OF TABLES............................................................................................................ iv LIST OF FIGURES ........................................................................................................... v LIST OF SYMBOLS........................................................................................................ vi ACKNOWLEDGMENT .................................................................................................. ix 1.0 ABSTRACT ................................................................................................................ x 2.0 INTRODUCTION ....................................................................................................... 1 3.0 THEORY ..................................................................................................................... 5 3.1 INTERNAL LOADING ....................................................................................... 5 3.2 CENTRIFUGAL LOADING ............................................................................... 8 3.3 CAGE SPEED ...................................................................................................... 8 3.4 POINT CONTACT STRESS ............................................................................... 9 3.5 LINE CONTACT STRESS ................................................................................ 13 4.0 RESULTS AND DISCUSSION............................................................................... 16 4.1 EXAMPLE CALCULATION ............................................................................ 16 4.2 SENSITIVITY STUDY...................................................................................... 22 5.0 CONCLUSION......................................................................................................... 28 6.0 REFERENCES ......................................................................................................... 30 7.0 APPENDIX A........................................................................................................... 31
iii
LIST OF TABLES No.
Page
1.
Bearing Characteristics
2
2.
Cup Angle Sensitivity Study Results
22
iv
LIST OF FIGURES No.
Page
1.
Common Types of Ball Bearings
1
2.
Common Types of Roller Bearings
2
3.
3D Tapered Roller Bearing Cutaway View
3
4.
Tapered Roller Bearing Terminology
4
5.
Tapered Roller Bearing Internal Geometry
5
6.
Tapered Roller Bearing Loading
6
7.
Cup Free Body Diagram
6
8.
Roller Free Body Diagram (x- direction)
7
9.
Roller Free Body Diagram (y-direction)
7
10.
Tapered Roller Bearing Kinematics
8
11.
General Case of 2 Point Contact
10
12.
Two Bodies in Point Contact, Resulting Elliptical Contact Area
10
13.
Tapered Roller Bearing Rib- Roller End Contact Shape
11
14.
Radius of Curvature
13
15.
Line Contact Stress Profile
14
16.
Hertzian Contact Shape for Cup and Cone
19
17
3D View of Rib-Roller End Contact Ellipse
21
18.
2D View of Rib-Roller End Contact Ellipse
22
19.
Example TRB at 0.25 Million DN
23
20.
Example TRB at 0.50 Million DN
23
21.
Example TRB at 0.75 Million DN
24
22.
Example TRB at 1.00 Million DN
24
23.
Example TRB at 1.50 Million DN
25
24.
Example TRB at 2.00 Million DN
25
25.
Example TRB at 3.00 Million DN
26
v
LIST OF SYMBOLS a = Semimajor Axis of the Projected Contact
mm (in)
a* = Dimensionless (a) Parameter b = Semiminor Axis of the Projected Contact
mm (in)
bi = Cone Semiminor Axis of the Contact Ellipse
in (mm)
bo = Cup Semiminor Axis of the Contact Ellipse
in (mm)
b* = Dimensionless (b) Parameter DN = Product of Shaft Speed in RPM Times Bearing Bore in millimeters df = Diameter to Rib – Roller End Contact Point
mm (in)
dm = Bearing Pitch Diameter
mm (in)
D1 = Diameter to Cone Large End Raceway
mm (in)
Dmax = Roller Large End Diameter
mm (in)
Dmean = Roller Mean Diameter
mm (in)
Dmin = Roller Small End Diameter
mm (in)
EI = Young’s Modulus (body I)
N ( psi ) mm 2
EII = Young’s Modulus (body II)
N ( psi ) mm 2
Fc= Roller Cnetrifugal Load
N (lbf)
h = height of Rib-Roller End Contact Point from Cone Raceway
mm (in)
L2 = Roller Apex Length
mm (in)
lt = Roller Length
mm (in)
leff = Effective Roller Length
mm (in)
m = Roller Mass
⎛ sec 2 kg ⎜⎜ lb in ⎝
Q = Roller Load
N (lbf)
Qf = Rib Normal Load
N (lbf)
Qi = Cone Normal Load
N (lbf)
Qo = Cup Normal Load
N (lbf)
Qfa = Rib Axial Load
N (lbf)
Qia = Cone Axial Load
N (lbf) vi
⎞ ⎟⎟ ⎠
Qfr = Rib Radial Load
N (lbf)
Qoa = Cup Axial Load
N (lbf)
Qir = Cone Radial Load
N (lbf)
Qor = Cup Radial Load
N (lbf)
r = radius
mm (in)
rI1 = Body I Radius of Curvature in Direction 1
mm (in)
rI2 = Body I Radius of Curvature in Direction 2
mm (in)
rII1 = Body II Radius of Curvature in Direction 1
mm (in)
rII2 = Body II Radius of Curvature in Direction 2
mm (in)
Rx = Directional Equivalent Radii (in the x direction)
mm (in)
Ry = Directional Equivalent Radii (in the y direction)
mm (in)
v = Velocity
rad sec
vc = Cage Velocity
rad sec
vi = Cone Velocity
rad sec
vie = Rolling Element Velocity at Cone Contact Point
rad sec
vo = Cup Velocity
rad sec
voe = Rolling Element Velocity at Cup Contact Point
rad sec
x = Principal Direction Distance
mm (in)
y = Principal Direction Distance
mm (in)
αdm = Pitch Angle
rad
αf = Rib Angle
rad
αi = Cone Angle
rad
αo = Cup Angle
rad
αR = Roller Included Angle
rad
γi = Cone Geometry Term γo = Cup Geometry Term vii
k = Elliptical Eccentricity Parameter θ = Angle From Cone Raceway to Rib Face
rad kg mm 3
ρ = Roller Density
⎛ sec 2 ⎜⎜ lb 4 ⎝ in
⎞ ⎟⎟ ⎠
( ) (in ) (in )
∑ ρ = Curvature Sum (point contact)
mm −1 in −1
∑ρ
i
= Curvature Sum (line contact-Cone)
mm −1
∑ρ
o
= Curvature Sum (line contact-Cup)
mm −1
−1
−1
σ = Hertzian Contact Stress
N ( psi ) mm 2
σmax = Maximum Hertzian Contact Stress
N ( psi ) mm 2
ω = Speed
rpm
ωc = Cage Speed
rpm
ωe = Rolling Element Speed
rpm
ωi = Cone Speed
rpm
ωo = Cup Speed
rpm
ξI= Poisson’s Ratio (body I) ξII= Poisson’s Ratio (body II) ζ = Complete Elliptical Integral of Second Kind
viii
ACKNOWLEDGMENT
ix
1.0 ABSTRACT Tapered roller bearings historically have been used in low speed applications. There has been an interest in evaluating the usage of tapered roller bearings at high speeds where ball bearings are usually used. The major benefit of using a tapered roller bearing is reduced size and weight.
Tapered roller bearings have a significantly larger load
capacity than ball bearings so a much smaller bearing could be used to carry the same load.
The purpose of this project was to determine the effects of cup and cone angles on tapered bearing stresses at high speeds. A FORTRAN program was created to evaluate the stress distributions within the bearing. The program takes the kinematics of the bearing motion into account but ignores gyroscopic motions. The program was run for a given bearing size while varying the cup angle and DN level. The results show that at high speeds > 1.0 million DN there is a significant amount of centrifugal loading that occurs. The centrifugal loading is generated by the rolling element speed and weight. The centrifugal loading increases the load on the rib and decreases the load on the cone. The cup load remains unchanged over all DN levels. To minimize these effects the cup angle should be optimized at each DN level. It was found that at low speeds the optimum cup angle is 40 degrees, as the DN level increases the optimum cup angle decreases down to 10 degrees for 3.0 million DN.
When designing tapered roller bearings for high speed applications there is an optimum cup angle to reduce stresses within the bearing. The rib stress is especially sensitive to the speed and care should be given to reduce the rib stress as low as possible. It can be clearly seen in the results that there is a “trough” shape for the rib stress, which indicates that there is a optimal angle for reducing rib stress.
Additionally, it was determined that there is a limiting speed for tapered roller bearings. For a given design there exists a speed that completely unloads the cone and all the load is transmitted through the rib. This condition is referred to as “declutching” and must
x
avoided by either limiting the speed or modifying the bearing geometry (i.e shallower cup and cone angles).
xi
2. Introduction Bearings are devices that allow relative motion between two objects. Typically, bearings are attached to a shaft (through the bore) and a housing (around the outer diameter). The types of rotation that can occur are shaft rotation, housing rotation, corotation or counter rotation. In addition to allowing relative motion, bearings also carry load. Bearings are designed to tolerate differing loading conditions depending upon the type of bearing used. The most common types of bearings are ball bearings, and roller bearings. Ball bearings contain balls (spheres) as rolling elements an can be found in a variety of configurations (fig 1.) [ref 1]. Roller bearings contain “rollers” as the rolling elements, the term “roller” refers to a cylindrical shaped object. Roller bearings may contain rolling elements that have a large aspect ratio (length/diameter), small aspect ratio, spherical shape, or a tapered shape (fig 2) [ref 1].
Figure No. 1: Common Types of Ball Bearings
1
Figure No. 2: Common Types of Roller Bearings Table No. 1 summarizes the general differences of the various types of bearings shown in figures 1 and 2. Roller bearings are stiffer and have higher load capacity than comparably sized ball bearings. These increases are due to the type of contact present in roller bearings (line contact for roller bearings, point contact for ball bearings). Generally, ball bearings operate at higher speeds than roller bearings. The only bearings that can take thrust loads are ball bearings, spherical roller and tapered roller bearings.
Table No. 1: Bearing Characteristics 2
Tapered roller bearings have a higher thrust load capacity than a ball bearing of similar size. There has been an interest in using tapered roller bearings in the place of ball bearings at high speeds due the benefit of size and weight reduction. Typically, tapered roller bearings have been used in lower speed applications. The speed limitation is primarily due to the rib/roller-end contact, which requires careful lubrication and stress considerations. Speeds have been limited to 0.5 million DN unless special design consideration is given to the rib/roller-end contact stress and lubrication. A special lubrication method was devised to provide an ample supply of lubrication to the surfaces in contact. Subsequent testing proved that high speed tapered bearings were viable and warranted continued research. The focus now turns to decreasing the stresses within the bearing to ensure an optimum design. Figure no. 3 illustrates a 3D view of a tapered roller bearing [ref 1]. The “tapered” shape of the cup, cone and roller can clearly be seen.
Figure No. 3: 3D Tapered Roller Bearing Cutaway View Figure no. 4 defines the terminology for tapered roller bearings. The outer ring is referred to as a “cup”, the inner ring is referred to as a “cone” and the rib is the part of the cone that contacts the face of the large end of the roller. The bearing also contains a cage (or separator) to keep the rolling elements equally spaced.
3
Figure No 4: Tapered Bearing Terminology The intent of this project is to develop a method for optimizing tapered roller bearing stress fields at high speeds. The areas of interest are the cup, cone and rib-roller end contact zones. The goal is to understand the effects of centrifugal loading, force balance, bearing kinematics and stress calculations.
4
3.
Theory
Tapered roller bearing terminology is different from traditional ball and roller bearing. The roller included angle, cup angle, and cone angle all meet at the apex point along the bearing centerline. The pitch diameter is measured to the center of the rolling element, and the roller length is measured from roller small face to roller large face along the pitch angle (cone angle + ½ roller angle). The flange angle is nearly perpendicular to the cone angle, however it is common to adjust the flange angle small amounts to ensure that the contact ellipse is fully contained (see section 4.0 for further discussion). The following figure illustrates the terms for tapered roller bearing internal geometry:
Figure No. 5: Tapered Roller Bearing Internal Geometry 3.1
Force Balance
Figure 6 [ref 1] illustrates a free body diagram for a tapered bearing under combined loading (including centrifugal). The forces Qo, Qi, and Qf all act perpendicular to their respective surfaces. The centrifugal force Fc acts vertically at the C.G. of the roller. The force balance is affected by the magnitude of the centrifugal force generated by the rolling element speed. A large portion of the centrifugal load is transmitted to the rib contact at high speeds. The following is a derivation of the contact loading for a tapered roller bearing, including centrifugal forces.
5
Point Contact
Line contact
Figure No. 6: Tapered Roller Bearing Loading For a tapered roller bearing roller centrifugal forces alter the distribution of load between the outer raceway and central guide flange. For equilibrium to exist, Qia + Q fa − Qoa = 0 Qir − Q fr + Fc − Qor = 0
and, By taking the sum of the forces on the cup equal to the applied load in the x direction:
Qo =
AppliedThrust z sin (α o )
3.1.0
Applied Thrust
Qoa
Figure No. 7: Cup Free Body Diagram Taking the sum of the roller forces in the x direction equal to zero:
Qi sin (α i ) − Qo sin (α o ) + Q f sin (α f ) = 0
or, Qi =
Qo sin (α o ) − Q f sin (α f
sin (α i )
)
3.1.1
6
Qoa
Qfa Qia
Figure No. 8 : Roller Free Body Diagram (x- direction) Taking the sum of the roller forces in the y direction equal to zero: sin (α f ) ⎞ ⎛ Q sin (α o ) ⎟⎟ − Qo cos(α o ) + Fc + o =0 Q f ⎜⎜ − cos(α f ) − tan (α i ) ⎠ tan (α i ) ⎝
or, Qo sin (α o ) tan (α i ) sin (α f ) ⎞ ⎛ ⎜⎜ − cos(α f ) − ⎟ tan (α i ) ⎟⎠ ⎝
Qo cos(α o ) − Fc − Qf =
3.1.2
Qor
Fc Qfr
Qir Figure No. 9 : Roller Free Body Diagram (x- direction) Equations 3.1.0, 3.1.1 and 3.1.2 are the solutions to the roller reaction forces for a bearing under pure thrust, and including roller centrifugal forces. As shown in figure 6, applied thrust loads are transmitted into forces normal to the contact surfaces. These normal forces can be several times larger than the applied thrust load due to the large radial component of the force that is generated by shallow raceway angles. As described above, the centrifugal loading of the rolling elements will effect the load distribution within the bearing. The most significant effect will be upon the flange contact which tends to carry the majority of the centrifugal loading.
7
3.2
Roller Centrifugal Force
To determine the centrifugal force of the rolling element, the following method was derived: 1 2 m(d m )ωc 2 where, 1 2 m = ρπ (Dmean ) lt 4 Fc =
By substitution, Fc =
1⎛1 2 ⎞ 2 ⎜ ρπ (Dmean ) lt ⎟(d m )ωc 2⎝4 ⎠
Which simplifies to:
Fc =
1 ρπ (Dmean )2 lt d mω c 2 8
2.3
3.2.0
Rolling Speeds and Velocities
To determine ω c (cage speed) the following method was derived:
lt Q
vo vc
Dmean
y
dm/2
P
r z
vi
αr αdm
αo
x Figure No 10: Tapered Roller Bearing Kinematics
The velocity of any point on a rotating body can be given as a function of radius and speed: v = rω
8
Similarly, the velocity of the cone at the roller mid length can be written as: ⎛d ⎞ ⎛D ⎞ vi = ω i ⎜⎜ m − ⎜ mean ⎟ cos(α dm )⎟⎟ (Velocity of inner ring at point P) ⎝ 2 ⎝ 2 ⎠ ⎠
3.3.0
The cage velocity can also be written using this same method (at roller center) ⎛d ⎞ vc = ω c ⎜ m ⎟ (Velocity of cage at roller midpoint) ⎝ 2 ⎠
3.3.1
Since there is no outer ring rotation, the velocity at point Q is equal to zero. vo = 0
3.3.2
The respective rolling element velocities are given below: ⎛d ⎞ ⎛D ⎞ vie = ω c ⎜ m ⎟ − ω e ⎜ mean ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠
3.3.3
⎛d ⎞ ⎛D ⎞ voe = ω c ⎜ m ⎟ + ω e ⎜ mean ⎟ = 0 ⎝ 2 ⎠ ⎝ 2 ⎠
3.3.4
Setting Eqn.3.3.3 equal to Eqn. 3.3.0 (inner ring speed equal to element speed) ⎛ d m ⎛ Dmean ⎞ ⎞ ⎞ ⎛d ⎞ ⎛D −⎜ ⎟ cos(α dm )⎟⎟ = ω c ⎜ m ⎟ − ω e ⎜ mean ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎝ 2 ⎠ ⎠
ω i ⎜⎜
3.3.5
Again assuming no slip occurs at midspan, solve equation 3.3.5 in terms of element speed ⎛ dm ⎞ ⎟⎟ D ⎝ mean ⎠
ω e = −ω c ⎜⎜
3.3.6
Now we have solution for element speed as a function of cage speed.
Plugging
Eqn.3.3.6 into Eqn.3.3.5, results in a solution for cage speed as a function of inner ring speed. ⎞ ω 1 ⎛⎜ ⎛ Dmean ⎞ ⎜⎜ ⎟⎟ cos(α dm )⎟ = c 1 − ⎟ ω 2 ⎜⎝ ⎝ d m ⎠ i ⎠ 3.4
3.3.7
Point Contact Stress Calculations
9
To develop the stress fields, for point contact the hertz contact theory was used. (ref 1). It is assumed that the hertz point contact method will be representative for roller end to rib contact stresses. The method is defined below: The contact zone between the roller-end and rib creates an ellipsoidal surface compressive stress distribution, see figure no 4 [ref 1]. The roller end has a spherical radius, and the rib face is flat at an angle αf. The maximum compressive hertzian contact stress is given as:
σ max =
3Q 2πab
3.4.0
The elliptical contact area is shown in figure number 11, the semimajor and semiminor axes are also shown.
Figure No. 11: General Case of 2 Point Contact
Figure 12: Two bodies in point contact, resulting elliptical contact area. 10
The normal stress at other points within the contact area is given by the following equation: 1
2 2 3Q ⎡ ⎛ x ⎞ ⎛ y ⎞ ⎤ 2 σ= ⎢1 − ⎜ ⎟ − ⎜ ⎟ ⎥ 2πab ⎣⎢ ⎝ a ⎠ ⎝ b ⎠ ⎦⎥
3.4.1
Where, x and y are principle direction distances In terms of a tapered roller bearing, the contact area is shown in figure 12, it can be seen that the contact shape is an ellipse because of the conformity of the rib in the x direction. The physical location and orientation of the rib to roller end contact area is shown in figure no 13. It can clearly be seen that the contact area is an elliptical shape, and the assumption is that the contact can be modeled using hertz point contact theory.
Figure No 13: Tapered Roller Bearing Rib- Roller End Contact Shape To solve the hertz contact stress equations the “a” and “b” terms must be solved. The values of a and b are determined from the following equations: ⎡ 3Q ⎛ 1 − ξ I 2 1 − ξ II 2 ⎜ + a = a* ⎢ ⎜ E2 ⎢⎣ 2∑ ρ ⎝ E I
1
⎞⎤ 3 ⎟⎥ ⎟ ⎠⎥⎦
3.4.2
and, ⎡ 3Q ⎛ 1 − ξ I 1 − ξ II ⎜ + b = b* ⎢ ⎜ E2 ⎢⎣ 2∑ ρ ⎝ E I 2
2
⎞⎤ ⎟⎥ ⎟ ⎠⎥⎦
1 3
3.4.3
11
For steel bodies in contact equations 3.4.2 and 3.4.3 reduce to; 1
⎡ Q ⎤3 a = 0.0236a * ⎢ ⎥ ⎢⎣ ∑ ρ ⎥⎦
3.4.4
1
⎡ Q ⎤3 b = 0.0236b * ⎢ ⎥ ⎣⎢ ∑ ρ ⎦⎥
ρ xI =
1 rI 1
ρ xII =
1 rII 1
ρ yI =
1 rI 2
ρ yII =
1 rII 2 1
∑ρ = r
+
I1
3.4.5
1 1 1 + + rI 2 rII 1 rII 2
rI 1 = rI 2 = Roller end Spherical Radius rII 1 = ∞ (i.e. flat surface) rII 2 =
df
2 sin (α f
)
The radius of curvature is positive in value if the body is convex, and negative if the body is concave. Where, ⎛ sin (θ − α i ) ⎞ ⎟⎟ d f = Dl + 2h⎜⎜ ⎝ cos(θ ) ⎠ and D1 = 2 L2 sin (α i ) Dmax ⎛α ⎞ 2 sin ⎜ R ⎟ ⎝ 2 ⎠ h = rI 1 (sin (φ ) + cos(θ )) L2 =
⎛ Dmax ⎝ 2rI 1
φ = a sin ⎜⎜
⎞ αR ⎟⎟ − ⎠ 2
θ = 90 − (α f − α i )
12
h L2 rII2
df/2 D1/2 90−αf
αf
Figure No. 14: Radius of Curvature for Rib – Roller End Contact (direction 2) a * and b * are defined as : ⎛ 2k 2ζ a* = ⎜⎜ ⎝ π
1
⎞3 ⎟⎟ ⎠
1
⎛ 2ζ ⎞ 3 b* = ⎜ ⎟ ⎝ kπ ⎠
Brewe and Hamrock [6], using a least squares method of linear regression, obtained simplified approximations for k, and ζ. The simplifications are listed in equations 3.4.6 and 3.4.7: ⎛R k = 1.0339⎜ x ⎜R ⎝ y
ζ = 1.0003 +
⎞ ⎟ ⎟ ⎠
0.636
3.4.6
0.5968 ⎛ Rx ⎞ ⎜ ⎟ ⎜R ⎟ ⎝ y⎠
3.4.7
and Rx Ry 3.5
−1
= ρ xI + ρ xII
−1
= ρ yI + ρ yII
Line contact Stress Calculations:
For the roller body- raceway contacts under ideal line contact the following equations are used:
13
σ max = i ,o
2Qi , o πleff bi , o
3.5.0
l
σ max
σ
Y y
2b
Figure No. 15: Line Contact Stress Profile (ideal) The stress at any point along the roller can be defined as:
σ i ,o =
2Qi ,o ⎡ ⎛ y ⎢1 − ⎜ πl eff bi ,o ⎢ ⎜⎝ bi ,o ⎣
⎡ 4Qi ,o bi ,o = ⎢ ⎢⎣ πl eff ∑ ρ i ,o
⎞ ⎟ ⎟ ⎠
2
⎤ ⎥ ⎥ ⎦
1 2
3.5.1
⎛1− ξI 1 − ξ II ⎜ + ⎜ E E2 I ⎝ 2
2
⎞⎤ ⎟⎥ ⎟ ⎠⎥⎦
1 2
3.5.2
For steel contacting bodies: ⎛ Qi ,o bi ,o = 3.35 x10 −3 ⎜ ⎜l ∑ρ i ,o ⎝ eff
⎞ ⎟ ⎟ ⎠
1 2
3.5.3
Where,
14
∑ρ
i
=
2 Dmean (1 − γ i )
assumes no crown on roller or raceways
∑ρ
o
=
2 Dmean (1 + γ o )
assumes no crown on roller or raceways
γi = γo =
Dmean cos(α i ) dm
Dmean cos(α o ) dm
The calculation methods listed in sections 4 and 5 will be used to determine the stress distributions and contact shapes for the raceways and rib contacts.
15
4.
4.1
Results And Discussion
EXAMPLE CALCULATION
For this example the following are known parameters: Inner Ring Speed = 8000 RPM; Thrust Load = 35,000 lbf; Pitch Diameter = 9.7”; Cup Angle = 18 degrees; Included Roller Angle=4.5 degrees; Cone Angle=13.5 degrees; Rib angle = 13.5 degrees; Roller Large End Diameter = 1.5”; Roller Length = 1.9”; Roller Large End Crown = 18”; Roller Qty = 17; Roller Mean Diameter = 1.425”, Pitch Angle = 15.75 (all angles must be converted to radians by multiplying by π/180). Calculate the maximum hertzian contact stress at the cup, cone, and rib contacts.
The calculation procedure is as follows:
Step 1: Calculate the contact forces. Begin with calculating the Cup contact force from equation 3.1.0, as shown below: Qo =
AppliedThrust z sin (α o )
Qo =
35,000 = 6662.5 lbf (29650 N) 17 sin (18)
Step 2: Calculate the cage speed from equation 3.3.7, as shown below:
1 ⎛⎜ ⎛ Dmean 1− ⎜ 2 ⎜⎝ ⎜⎝ d m
⎞ ω ⎞ ⎟⎟ cos(α dm )⎟ = c ⎟ ω i ⎠ ⎠
1⎛
⎛D
1⎛
⎛ 1.425 ⎞
⎞
⎞
ω c = ⎜⎜1 − ⎜⎜ mean ⎟⎟ cos(α dm )⎟⎟ω i 2 ⎝ ⎝ dm ⎠ ⎠ ⎞
ω c = ⎜⎜1 − ⎜ ⎟ cos(15.75)⎟⎟8000 2 ⎝ ⎝ 9.7 ⎠ ⎠ ω c = 3434.4 RPM Step 3: Calculate the centrifugal force generated by the rolling element from equation 3.2.0 as shown below:
16
1 ρπDmean 2 l t d mω c 2 8 1 2 2 Fc = (0.3)π (1.425) (1.9 )(9.7 )(3434.4) 8 Fc = 1399.3 lbf (6227 N) Fc =
Step 4: Now that the centrifugal forces have been calculated the rib force can be determined. Qo sin (α o ) tan (α i ) sin (α f ) ⎞ ⎛ ⎜⎜ − cos(α f ) − ⎟ tan (α i ) ⎟⎠ ⎝
Qo cos(α o ) − Fc − Qf =
6662.5 sin (18) tan (13.5) Qf = sin (13.5) − cos(13.5) − tan (13.5) Q f = 849.4 lbf (3780 N) 6662.5 cos(18) − 1399.3 −
Step 5: The cone reaction force can be calculated. Qi =
Qo sin (α o ) − Q f sin (α f sin (α i )
)
6662.5 sin (18) − 849.4 sin (13.5) sin (13.5) Qi = 5281.3 lbf (23502 N) Qi =
Now that the reaction forces have been calculated, the stresses can be determined. We will begin with the cup and cone contact stresses. Step 6: The cup contact stress is given by equation 3.5.0 as shown below:
σ max o =
2Qo πl eff bo
The term bo is given by equation 3.5.3 (English unit conversion). ⎛ Qo bo = 2.78 x10 − 4 ⎜ ⎜l ∑ρ o ⎝ eff
1
⎞2 ⎟ ⎟ ⎠
First we must solve for γo
17
Dmean cos(α o ) dm
γo =
1.425 cos(18) 9.7 γ o = 0.1397
γo =
Now, solve for the curvature sum
∑ρ
o
=
2 Dmean (1 + γ o )
∑ρ
o
=
2 1.425(1 + 0.1397 )
∑ρ
o
= 1.2315
1 ⎛ 1 ⎞ ⎜ 31.28 ⎟ in ⎝ mm ⎠
Substituting back into the equation for bo 1
⎛ 6662.5 ⎞ 2 ⎟⎟ bo = 2.78 x10 ⎜⎜ ⎝ 1.9(1.2315) ⎠ bo = 0.01483 in (0.3767 mm ) −4
And finally, substituting the back into the equation for max hertz stress
σ max o =
2Qo πl eff bo
2(6662.5) π (1.9)(0.01483) = 150483 psi (1036828 kPa )
σ max o = σ max o
The cone contact stress shall be determined similar to the cup contact stress Step 7: The difference in the calculations lie in the bo term.
⎛ Qi bi = 2.78 x10 − 4 ⎜ ⎜l ∑ρ i ⎝ eff
1
⎞2 ⎟ ⎟ ⎠
First we must solve for γi
18
Dmean cos(α i ) dm
γi =
1.425cos(13.5) 9.7 γ i = 0.1428
γi =
Now, solve for the curvature sum
∑ρ
i
=
2 Dmean (1 − γ i )
∑ρ
i
=
2 1.425(1 − 0.1428)
∑ρ
i
= 1.6374
1 ⎛ 1 ⎞ ⎜ 51.59 ⎟ in ⎝ mm ⎠
Substituting back into the equation for bi ⎛ 5281.3 ⎞ ⎟⎟ bi = 2.78 x10 − 4 ⎜⎜ ⎝ 1.9(1.6374 ) ⎠ bi = 0.01145 in (0.2908 mm )
1 2
And finally, substituting the back into the equation for max hertz stress
σ max i =
2Qi πl eff bi
2(5281.3) π (1.9 )(0.01145) = 154492 psi (1064450 kPa )
σ max i = σ max i
_____ Cone Contact Stress _____ Cup Contact Stress
σ (psi)
y (in) Figure No. 16: Hertzian Contact Stress Profile for Cup and Cone 19
Step 8: Determine the geometry of the rib to roller end contact:
θ = 90 − (α f − α i ) = 90 − (13.5 − 13.5) = 90 deg ⎞ αr ⎛ 1.5 ⎞ 13.5 ⎟⎟ − ⎟⎟ − = a sin ⎜⎜ = .0024 rad 2 ⎝ 2(18) ⎠ ⎠ 2 ⎛ 90π ⎞ h = rI 1 (sin (φ ) + cos(θ )) = 18(sin (.0024 )) + cos⎜ ⎟ = 0.0434 in (1.102 mm ) ⎝ 180 ⎠ Dmax 1.5 L2 = = = 19.1 in (485.14 mm ) ⎛αr ⎞ ⎛ 4.5 ⎞ 2 sin ⎜ ⎟ 2 sin ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ D1 = 2 L2 sin (α i ) = 2(19.1)sin (13.5) = 8.919 in (226.54mm ) ⎛ Dmax ⎝ 2rI 1
φ = a sin ⎜⎜
⎛ sin (θ − α i ) ⎞ ⎛ sin (90 − 13.5) ⎞ ⎟⎟ = 8.919 + 2(0.0434)⎜⎜ ⎟⎟ = 9.003 in (228.6 mm ) d f = D1 + 2h⎜⎜ ⎝ cos(θ ) ⎠ ⎝ cos(90) ⎠ To determine the radius of curvature of body II in the x direction. The following geometry must be used, and since the radius is concave the value must be negative. rI 1 = rI 2 = 18 rII 1 = ∞ (i.e. flat surface) rII 2 =
−df
2 sin (α f 1
∑ρ = r
I1
ρ xI =
+
)
=
− 9.003 = −19.28 2 sin (13.5)
1 1 1 1 1 1 1 ⎛ 1 ⎞ + + = + +0− = 0.00592 ⎜ 0.1504 ⎟ 19.28 in ⎝ mm ⎠ rI 2 rII 1 rII 2 18 18
1 1⎛ 1 ⎞ = .05556 ⎜1.411 ⎟ 18 in ⎝ mm ⎠
ρ xII = −
1 1 ⎛ 1 ⎞ = −.058562 ⎜1.487 ⎟ 19.28 in ⎝ mm ⎠
1 ⎞ 1 1 ⎛ = .05556 ⎜1.411 ⎟ 18 in ⎝ mm ⎠ =0
ρ yI = ρ yII
Rx =
1 1 1⎛ 1 ⎞ = = 270.3 ⎜ 6865.6 ⎟ ρ xI + ρ xII .05556 + (− .058562) in ⎝ mm ⎠
Ry =
1 1 1⎛ 1 ⎞ = = 18.0 ⎜ 457.2 ⎟ ρ yI + ρ yII .05556 + 0 in ⎝ mm ⎠
Step 9: Determine the dimensionless parameters and max stress.
20
0.636
0.636 ⎛ Rx ⎞ ⎛ 270.3 ⎞ ⎜ ⎟ k = 1.0339 = 1.0339⎜ = 5.7913 ⎟ ⎜R ⎟ ⎝ 18 ⎠ ⎝ y⎠ 0.5968 0.5968 ζ = 1.0003 + = 1.0003 + = 1.04004 ⎛ Rx ⎞ ⎛ 270.3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜R ⎟ ⎝ 18 ⎠ y ⎝ ⎠
⎛ 2k 2ζ a* = ⎜⎜ ⎝ π
1
1
⎞ 3 ⎛ 2(5.7913)2 (1.04004 ) ⎞ 3 ⎟ = 2.811 ⎟⎟ = ⎜ ⎜ ⎟ π ⎠ ⎝ ⎠
1
1
⎛ 2ζ ⎞ 3 ⎛ 2(1.04004 ) ⎞ 3 b* = ⎜ ⎟ =⎜ ⎟ = 0.4853 ⎝ kπ ⎠ ⎝ 5.7913π ⎠ 1
1
⎡ Q ⎤3 ⎛ 849.6 ⎞ 3 a = 0.0045a * ⎢ ⎟ = 0.3074 in (7.808mm ) ⎥ = 0.0045(2.811)⎜ ⎝ .05919 ⎠ ⎣⎢ ∑ ρ ⎦⎥ 1 3
1
⎡ Q ⎤ ⎛ 849.6 ⎞ 3 b = 0.0045b * ⎢ ⎟ = .05308 in (1.348mm ) ⎥ = .0045(.4853)⎜ ⎝ .05919 ⎠ ⎢⎣ ∑ ρ ⎥⎦ 3Q f 3(849.6 ) σ max = = = 24863 psi (171306kPa ) 2πab 2π (0.3074 )(0.05308)
σ (psi)
x (in)
y (in)
Figure No. 17: 3D View of Rib- Roller End Contact Ellipse
21
y (in)
x (in)
Figure No. 18: 2D View Rib-Roller End Contact Ellipse
4.2
SENSITIVITY STUDY
The following figures, illustrate the effects of cup angle on bearing contact stresses for a given speed (DN). The geometry used in section 4.1 was used for this study, with the exception of the cup angle, cone angle and rib angle. The cone angle was varied from 5 degrees to 40 degrees, the cone angle equals the cup angle – roller included angle, and the rib angle is assumed to be equal to the cone angle. Table No. 2 summarizes the optimal cup angles for each DN level of the sample bearing. DN (million)
Cup Angle (deg)
0.25
40
0.50
40
0.75
30
1.00
25
1.50
20
2.00
15
3.00
10
Table No. 2: Sample Bearing Optimized Cup Angle Based on DN leve. Figures 19-25 illustrate the cup, cone and rib stresses for various DN levels for a range of cup angles. 22
Tapered Bearing Stress Distribution (0.25 million DN) 350000
50000 45000 40000
250000
35000 30000
200000
25000 150000
20000 15000
100000
Rib Stress (psi)
Cup, Cone Stress (psi)
300000
Cup Stress Cone Stress Rib Stress
10000 50000
5000
0
0 5
10
15
20
25
30
35
40
Cup Angle (deg)
Figure No 19: Example TRB at 0.25 million DN Tapered Bearing Stress Distribution (0.50 million DN) 350000
50000 45000 40000
250000
35000 30000
200000
25000 150000
20000 15000
100000
Rib Stress (psi)
Cup, Cone Stress (psi)
300000
10000 50000
5000
0
0 5
10
15
20
25
30
35
40
Cup Angle (deg)
Figure No 20: Example TRB at 0.50 million DN
23
Cup Stress Cone Stress Rib Stress
Tapered Bearing Stress Distribution (0.75 million DN) 350000
50000 45000 40000
250000
35000 30000
200000
25000 150000
20000 15000
100000
Rib Stress (psi)
Cup, Cone Stress (psi)
300000
Cup Stress Cone Stress Rib Stress
10000 50000
5000
0
0 5
10
15
20
25
30
35
40
Cup Angle (deg)
Figure No 21: Example TRB at 0.75 million DN
Tapered Bearing Stress Distribution (1.00 million DN) 350000
50000 45000 40000
250000
35000 30000
200000
25000 150000
20000 15000
100000
Rib Stress (psi)
Cup, Cone Stress (psi)
300000
10000 50000
5000
0
0 5
10
15
20
25
30
35
40
Cup Angle (deg)
Figure No 22: Example TRB at 1.00 million DN
24
Cup Stress Cone Stress Rib Stress
Tapered Bearing Stress Distribution (1.50 million DN) 350000
50000 45000 40000
250000
35000 30000
200000
25000 150000
20000 15000
100000
Rib Stress (psi)
Cup, Cone Stress (psi)
300000
Cup Stress Cone Stress Rib Stress
10000 50000
5000
0
0 5
10
15
20
25
30
35
40
Cup Angle (deg)
Figure No 23: Example TRB at 1.50 million DN
Tapered Bearing Stress Distribution (2.00 million DN) 350000
50000 45000 40000
250000
35000 30000
200000
25000 150000
20000 15000
100000
Rib Stress (psi)
Cup, Cone Stress (psi)
300000
10000 50000
5000
0
0 5
10
15
20
25
30
35
40
Cup Angle (deg)
Figure No 24: Example TRB at 2.00 million DN
25
Cup Stress Cone Stress Rib Stress
Tapered Bearing Stress Distribution (3.00 million DN) 300000
50000 45000 40000 35000
200000
30000 150000
25000 20000
100000
15000
Rib Stress (psi)
Cup, Cone Stress (psi)
250000
Cup Stress Cone Stress Rib Stress
10000
50000
5000 0
0 5
10
15
20
25
30
35
40
Cup Angle (deg)
Figure No 25: Example TRB at 3.00 million DN
Several things can be ascertained from this sensitivity study. The first and most important is that there is an optimum cup angle, for every speed. At low speeds (
Approved: _________________________________________ Ernesto Gutierrez-Miravete, Project Adviser
Rensselaer Polytechnic Institute Hartford, Connecticut April, 2008
© Copyright 2008 by Brady Walker All Rights Reserved
ii
CONTENTS LIST OF TABLES............................................................................................................ iv LIST OF FIGURES ........................................................................................................... v LIST OF SYMBOLS........................................................................................................ vi ACKNOWLEDGMENT .................................................................................................. ix 1.0 ABSTRACT ................................................................................................................ x 2.0 INTRODUCTION ....................................................................................................... 1 3.0 THEORY ..................................................................................................................... 5 3.1 INTERNAL LOADING ....................................................................................... 5 3.2 CENTRIFUGAL LOADING ............................................................................... 8 3.3 CAGE SPEED ...................................................................................................... 8 3.4 POINT CONTACT STRESS ............................................................................... 9 3.5 LINE CONTACT STRESS ................................................................................ 13 4.0 RESULTS AND DISCUSSION............................................................................... 16 4.1 EXAMPLE CALCULATION ............................................................................ 16 4.2 SENSITIVITY STUDY...................................................................................... 22 5.0 CONCLUSION......................................................................................................... 28 6.0 REFERENCES ......................................................................................................... 30 7.0 APPENDIX A........................................................................................................... 31
iii
LIST OF TABLES No.
Page
1.
Bearing Characteristics
2
2.
Cup Angle Sensitivity Study Results
22
iv
LIST OF FIGURES No.
Page
1.
Common Types of Ball Bearings
1
2.
Common Types of Roller Bearings
2
3.
3D Tapered Roller Bearing Cutaway View
3
4.
Tapered Roller Bearing Terminology
4
5.
Tapered Roller Bearing Internal Geometry
5
6.
Tapered Roller Bearing Loading
6
7.
Cup Free Body Diagram
6
8.
Roller Free Body Diagram (x- direction)
7
9.
Roller Free Body Diagram (y-direction)
7
10.
Tapered Roller Bearing Kinematics
8
11.
General Case of 2 Point Contact
10
12.
Two Bodies in Point Contact, Resulting Elliptical Contact Area
10
13.
Tapered Roller Bearing Rib- Roller End Contact Shape
11
14.
Radius of Curvature
13
15.
Line Contact Stress Profile
14
16.
Hertzian Contact Shape for Cup and Cone
19
17
3D View of Rib-Roller End Contact Ellipse
21
18.
2D View of Rib-Roller End Contact Ellipse
22
19.
Example TRB at 0.25 Million DN
23
20.
Example TRB at 0.50 Million DN
23
21.
Example TRB at 0.75 Million DN
24
22.
Example TRB at 1.00 Million DN
24
23.
Example TRB at 1.50 Million DN
25
24.
Example TRB at 2.00 Million DN
25
25.
Example TRB at 3.00 Million DN
26
v
LIST OF SYMBOLS a = Semimajor Axis of the Projected Contact
mm (in)
a* = Dimensionless (a) Parameter b = Semiminor Axis of the Projected Contact
mm (in)
bi = Cone Semiminor Axis of the Contact Ellipse
in (mm)
bo = Cup Semiminor Axis of the Contact Ellipse
in (mm)
b* = Dimensionless (b) Parameter DN = Product of Shaft Speed in RPM Times Bearing Bore in millimeters df = Diameter to Rib – Roller End Contact Point
mm (in)
dm = Bearing Pitch Diameter
mm (in)
D1 = Diameter to Cone Large End Raceway
mm (in)
Dmax = Roller Large End Diameter
mm (in)
Dmean = Roller Mean Diameter
mm (in)
Dmin = Roller Small End Diameter
mm (in)
EI = Young’s Modulus (body I)
N ( psi ) mm 2
EII = Young’s Modulus (body II)
N ( psi ) mm 2
Fc= Roller Cnetrifugal Load
N (lbf)
h = height of Rib-Roller End Contact Point from Cone Raceway
mm (in)
L2 = Roller Apex Length
mm (in)
lt = Roller Length
mm (in)
leff = Effective Roller Length
mm (in)
m = Roller Mass
⎛ sec 2 kg ⎜⎜ lb in ⎝
Q = Roller Load
N (lbf)
Qf = Rib Normal Load
N (lbf)
Qi = Cone Normal Load
N (lbf)
Qo = Cup Normal Load
N (lbf)
Qfa = Rib Axial Load
N (lbf)
Qia = Cone Axial Load
N (lbf) vi
⎞ ⎟⎟ ⎠
Qfr = Rib Radial Load
N (lbf)
Qoa = Cup Axial Load
N (lbf)
Qir = Cone Radial Load
N (lbf)
Qor = Cup Radial Load
N (lbf)
r = radius
mm (in)
rI1 = Body I Radius of Curvature in Direction 1
mm (in)
rI2 = Body I Radius of Curvature in Direction 2
mm (in)
rII1 = Body II Radius of Curvature in Direction 1
mm (in)
rII2 = Body II Radius of Curvature in Direction 2
mm (in)
Rx = Directional Equivalent Radii (in the x direction)
mm (in)
Ry = Directional Equivalent Radii (in the y direction)
mm (in)
v = Velocity
rad sec
vc = Cage Velocity
rad sec
vi = Cone Velocity
rad sec
vie = Rolling Element Velocity at Cone Contact Point
rad sec
vo = Cup Velocity
rad sec
voe = Rolling Element Velocity at Cup Contact Point
rad sec
x = Principal Direction Distance
mm (in)
y = Principal Direction Distance
mm (in)
αdm = Pitch Angle
rad
αf = Rib Angle
rad
αi = Cone Angle
rad
αo = Cup Angle
rad
αR = Roller Included Angle
rad
γi = Cone Geometry Term γo = Cup Geometry Term vii
k = Elliptical Eccentricity Parameter θ = Angle From Cone Raceway to Rib Face
rad kg mm 3
ρ = Roller Density
⎛ sec 2 ⎜⎜ lb 4 ⎝ in
⎞ ⎟⎟ ⎠
( ) (in ) (in )
∑ ρ = Curvature Sum (point contact)
mm −1 in −1
∑ρ
i
= Curvature Sum (line contact-Cone)
mm −1
∑ρ
o
= Curvature Sum (line contact-Cup)
mm −1
−1
−1
σ = Hertzian Contact Stress
N ( psi ) mm 2
σmax = Maximum Hertzian Contact Stress
N ( psi ) mm 2
ω = Speed
rpm
ωc = Cage Speed
rpm
ωe = Rolling Element Speed
rpm
ωi = Cone Speed
rpm
ωo = Cup Speed
rpm
ξI= Poisson’s Ratio (body I) ξII= Poisson’s Ratio (body II) ζ = Complete Elliptical Integral of Second Kind
viii
ACKNOWLEDGMENT
ix
1.0 ABSTRACT Tapered roller bearings historically have been used in low speed applications. There has been an interest in evaluating the usage of tapered roller bearings at high speeds where ball bearings are usually used. The major benefit of using a tapered roller bearing is reduced size and weight.
Tapered roller bearings have a significantly larger load
capacity than ball bearings so a much smaller bearing could be used to carry the same load.
The purpose of this project was to determine the effects of cup and cone angles on tapered bearing stresses at high speeds. A FORTRAN program was created to evaluate the stress distributions within the bearing. The program takes the kinematics of the bearing motion into account but ignores gyroscopic motions. The program was run for a given bearing size while varying the cup angle and DN level. The results show that at high speeds > 1.0 million DN there is a significant amount of centrifugal loading that occurs. The centrifugal loading is generated by the rolling element speed and weight. The centrifugal loading increases the load on the rib and decreases the load on the cone. The cup load remains unchanged over all DN levels. To minimize these effects the cup angle should be optimized at each DN level. It was found that at low speeds the optimum cup angle is 40 degrees, as the DN level increases the optimum cup angle decreases down to 10 degrees for 3.0 million DN.
When designing tapered roller bearings for high speed applications there is an optimum cup angle to reduce stresses within the bearing. The rib stress is especially sensitive to the speed and care should be given to reduce the rib stress as low as possible. It can be clearly seen in the results that there is a “trough” shape for the rib stress, which indicates that there is a optimal angle for reducing rib stress.
Additionally, it was determined that there is a limiting speed for tapered roller bearings. For a given design there exists a speed that completely unloads the cone and all the load is transmitted through the rib. This condition is referred to as “declutching” and must
x
avoided by either limiting the speed or modifying the bearing geometry (i.e shallower cup and cone angles).
xi
2. Introduction Bearings are devices that allow relative motion between two objects. Typically, bearings are attached to a shaft (through the bore) and a housing (around the outer diameter). The types of rotation that can occur are shaft rotation, housing rotation, corotation or counter rotation. In addition to allowing relative motion, bearings also carry load. Bearings are designed to tolerate differing loading conditions depending upon the type of bearing used. The most common types of bearings are ball bearings, and roller bearings. Ball bearings contain balls (spheres) as rolling elements an can be found in a variety of configurations (fig 1.) [ref 1]. Roller bearings contain “rollers” as the rolling elements, the term “roller” refers to a cylindrical shaped object. Roller bearings may contain rolling elements that have a large aspect ratio (length/diameter), small aspect ratio, spherical shape, or a tapered shape (fig 2) [ref 1].
Figure No. 1: Common Types of Ball Bearings
1
Figure No. 2: Common Types of Roller Bearings Table No. 1 summarizes the general differences of the various types of bearings shown in figures 1 and 2. Roller bearings are stiffer and have higher load capacity than comparably sized ball bearings. These increases are due to the type of contact present in roller bearings (line contact for roller bearings, point contact for ball bearings). Generally, ball bearings operate at higher speeds than roller bearings. The only bearings that can take thrust loads are ball bearings, spherical roller and tapered roller bearings.
Table No. 1: Bearing Characteristics 2
Tapered roller bearings have a higher thrust load capacity than a ball bearing of similar size. There has been an interest in using tapered roller bearings in the place of ball bearings at high speeds due the benefit of size and weight reduction. Typically, tapered roller bearings have been used in lower speed applications. The speed limitation is primarily due to the rib/roller-end contact, which requires careful lubrication and stress considerations. Speeds have been limited to 0.5 million DN unless special design consideration is given to the rib/roller-end contact stress and lubrication. A special lubrication method was devised to provide an ample supply of lubrication to the surfaces in contact. Subsequent testing proved that high speed tapered bearings were viable and warranted continued research. The focus now turns to decreasing the stresses within the bearing to ensure an optimum design. Figure no. 3 illustrates a 3D view of a tapered roller bearing [ref 1]. The “tapered” shape of the cup, cone and roller can clearly be seen.
Figure No. 3: 3D Tapered Roller Bearing Cutaway View Figure no. 4 defines the terminology for tapered roller bearings. The outer ring is referred to as a “cup”, the inner ring is referred to as a “cone” and the rib is the part of the cone that contacts the face of the large end of the roller. The bearing also contains a cage (or separator) to keep the rolling elements equally spaced.
3
Figure No 4: Tapered Bearing Terminology The intent of this project is to develop a method for optimizing tapered roller bearing stress fields at high speeds. The areas of interest are the cup, cone and rib-roller end contact zones. The goal is to understand the effects of centrifugal loading, force balance, bearing kinematics and stress calculations.
4
3.
Theory
Tapered roller bearing terminology is different from traditional ball and roller bearing. The roller included angle, cup angle, and cone angle all meet at the apex point along the bearing centerline. The pitch diameter is measured to the center of the rolling element, and the roller length is measured from roller small face to roller large face along the pitch angle (cone angle + ½ roller angle). The flange angle is nearly perpendicular to the cone angle, however it is common to adjust the flange angle small amounts to ensure that the contact ellipse is fully contained (see section 4.0 for further discussion). The following figure illustrates the terms for tapered roller bearing internal geometry:
Figure No. 5: Tapered Roller Bearing Internal Geometry 3.1
Force Balance
Figure 6 [ref 1] illustrates a free body diagram for a tapered bearing under combined loading (including centrifugal). The forces Qo, Qi, and Qf all act perpendicular to their respective surfaces. The centrifugal force Fc acts vertically at the C.G. of the roller. The force balance is affected by the magnitude of the centrifugal force generated by the rolling element speed. A large portion of the centrifugal load is transmitted to the rib contact at high speeds. The following is a derivation of the contact loading for a tapered roller bearing, including centrifugal forces.
5
Point Contact
Line contact
Figure No. 6: Tapered Roller Bearing Loading For a tapered roller bearing roller centrifugal forces alter the distribution of load between the outer raceway and central guide flange. For equilibrium to exist, Qia + Q fa − Qoa = 0 Qir − Q fr + Fc − Qor = 0
and, By taking the sum of the forces on the cup equal to the applied load in the x direction:
Qo =
AppliedThrust z sin (α o )
3.1.0
Applied Thrust
Qoa
Figure No. 7: Cup Free Body Diagram Taking the sum of the roller forces in the x direction equal to zero:
Qi sin (α i ) − Qo sin (α o ) + Q f sin (α f ) = 0
or, Qi =
Qo sin (α o ) − Q f sin (α f
sin (α i )
)
3.1.1
6
Qoa
Qfa Qia
Figure No. 8 : Roller Free Body Diagram (x- direction) Taking the sum of the roller forces in the y direction equal to zero: sin (α f ) ⎞ ⎛ Q sin (α o ) ⎟⎟ − Qo cos(α o ) + Fc + o =0 Q f ⎜⎜ − cos(α f ) − tan (α i ) ⎠ tan (α i ) ⎝
or, Qo sin (α o ) tan (α i ) sin (α f ) ⎞ ⎛ ⎜⎜ − cos(α f ) − ⎟ tan (α i ) ⎟⎠ ⎝
Qo cos(α o ) − Fc − Qf =
3.1.2
Qor
Fc Qfr
Qir Figure No. 9 : Roller Free Body Diagram (x- direction) Equations 3.1.0, 3.1.1 and 3.1.2 are the solutions to the roller reaction forces for a bearing under pure thrust, and including roller centrifugal forces. As shown in figure 6, applied thrust loads are transmitted into forces normal to the contact surfaces. These normal forces can be several times larger than the applied thrust load due to the large radial component of the force that is generated by shallow raceway angles. As described above, the centrifugal loading of the rolling elements will effect the load distribution within the bearing. The most significant effect will be upon the flange contact which tends to carry the majority of the centrifugal loading.
7
3.2
Roller Centrifugal Force
To determine the centrifugal force of the rolling element, the following method was derived: 1 2 m(d m )ωc 2 where, 1 2 m = ρπ (Dmean ) lt 4 Fc =
By substitution, Fc =
1⎛1 2 ⎞ 2 ⎜ ρπ (Dmean ) lt ⎟(d m )ωc 2⎝4 ⎠
Which simplifies to:
Fc =
1 ρπ (Dmean )2 lt d mω c 2 8
2.3
3.2.0
Rolling Speeds and Velocities
To determine ω c (cage speed) the following method was derived:
lt Q
vo vc
Dmean
y
dm/2
P
r z
vi
αr αdm
αo
x Figure No 10: Tapered Roller Bearing Kinematics
The velocity of any point on a rotating body can be given as a function of radius and speed: v = rω
8
Similarly, the velocity of the cone at the roller mid length can be written as: ⎛d ⎞ ⎛D ⎞ vi = ω i ⎜⎜ m − ⎜ mean ⎟ cos(α dm )⎟⎟ (Velocity of inner ring at point P) ⎝ 2 ⎝ 2 ⎠ ⎠
3.3.0
The cage velocity can also be written using this same method (at roller center) ⎛d ⎞ vc = ω c ⎜ m ⎟ (Velocity of cage at roller midpoint) ⎝ 2 ⎠
3.3.1
Since there is no outer ring rotation, the velocity at point Q is equal to zero. vo = 0
3.3.2
The respective rolling element velocities are given below: ⎛d ⎞ ⎛D ⎞ vie = ω c ⎜ m ⎟ − ω e ⎜ mean ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠
3.3.3
⎛d ⎞ ⎛D ⎞ voe = ω c ⎜ m ⎟ + ω e ⎜ mean ⎟ = 0 ⎝ 2 ⎠ ⎝ 2 ⎠
3.3.4
Setting Eqn.3.3.3 equal to Eqn. 3.3.0 (inner ring speed equal to element speed) ⎛ d m ⎛ Dmean ⎞ ⎞ ⎞ ⎛d ⎞ ⎛D −⎜ ⎟ cos(α dm )⎟⎟ = ω c ⎜ m ⎟ − ω e ⎜ mean ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎝ 2 ⎠ ⎠
ω i ⎜⎜
3.3.5
Again assuming no slip occurs at midspan, solve equation 3.3.5 in terms of element speed ⎛ dm ⎞ ⎟⎟ D ⎝ mean ⎠
ω e = −ω c ⎜⎜
3.3.6
Now we have solution for element speed as a function of cage speed.
Plugging
Eqn.3.3.6 into Eqn.3.3.5, results in a solution for cage speed as a function of inner ring speed. ⎞ ω 1 ⎛⎜ ⎛ Dmean ⎞ ⎜⎜ ⎟⎟ cos(α dm )⎟ = c 1 − ⎟ ω 2 ⎜⎝ ⎝ d m ⎠ i ⎠ 3.4
3.3.7
Point Contact Stress Calculations
9
To develop the stress fields, for point contact the hertz contact theory was used. (ref 1). It is assumed that the hertz point contact method will be representative for roller end to rib contact stresses. The method is defined below: The contact zone between the roller-end and rib creates an ellipsoidal surface compressive stress distribution, see figure no 4 [ref 1]. The roller end has a spherical radius, and the rib face is flat at an angle αf. The maximum compressive hertzian contact stress is given as:
σ max =
3Q 2πab
3.4.0
The elliptical contact area is shown in figure number 11, the semimajor and semiminor axes are also shown.
Figure No. 11: General Case of 2 Point Contact
Figure 12: Two bodies in point contact, resulting elliptical contact area. 10
The normal stress at other points within the contact area is given by the following equation: 1
2 2 3Q ⎡ ⎛ x ⎞ ⎛ y ⎞ ⎤ 2 σ= ⎢1 − ⎜ ⎟ − ⎜ ⎟ ⎥ 2πab ⎣⎢ ⎝ a ⎠ ⎝ b ⎠ ⎦⎥
3.4.1
Where, x and y are principle direction distances In terms of a tapered roller bearing, the contact area is shown in figure 12, it can be seen that the contact shape is an ellipse because of the conformity of the rib in the x direction. The physical location and orientation of the rib to roller end contact area is shown in figure no 13. It can clearly be seen that the contact area is an elliptical shape, and the assumption is that the contact can be modeled using hertz point contact theory.
Figure No 13: Tapered Roller Bearing Rib- Roller End Contact Shape To solve the hertz contact stress equations the “a” and “b” terms must be solved. The values of a and b are determined from the following equations: ⎡ 3Q ⎛ 1 − ξ I 2 1 − ξ II 2 ⎜ + a = a* ⎢ ⎜ E2 ⎢⎣ 2∑ ρ ⎝ E I
1
⎞⎤ 3 ⎟⎥ ⎟ ⎠⎥⎦
3.4.2
and, ⎡ 3Q ⎛ 1 − ξ I 1 − ξ II ⎜ + b = b* ⎢ ⎜ E2 ⎢⎣ 2∑ ρ ⎝ E I 2
2
⎞⎤ ⎟⎥ ⎟ ⎠⎥⎦
1 3
3.4.3
11
For steel bodies in contact equations 3.4.2 and 3.4.3 reduce to; 1
⎡ Q ⎤3 a = 0.0236a * ⎢ ⎥ ⎢⎣ ∑ ρ ⎥⎦
3.4.4
1
⎡ Q ⎤3 b = 0.0236b * ⎢ ⎥ ⎣⎢ ∑ ρ ⎦⎥
ρ xI =
1 rI 1
ρ xII =
1 rII 1
ρ yI =
1 rI 2
ρ yII =
1 rII 2 1
∑ρ = r
+
I1
3.4.5
1 1 1 + + rI 2 rII 1 rII 2
rI 1 = rI 2 = Roller end Spherical Radius rII 1 = ∞ (i.e. flat surface) rII 2 =
df
2 sin (α f
)
The radius of curvature is positive in value if the body is convex, and negative if the body is concave. Where, ⎛ sin (θ − α i ) ⎞ ⎟⎟ d f = Dl + 2h⎜⎜ ⎝ cos(θ ) ⎠ and D1 = 2 L2 sin (α i ) Dmax ⎛α ⎞ 2 sin ⎜ R ⎟ ⎝ 2 ⎠ h = rI 1 (sin (φ ) + cos(θ )) L2 =
⎛ Dmax ⎝ 2rI 1
φ = a sin ⎜⎜
⎞ αR ⎟⎟ − ⎠ 2
θ = 90 − (α f − α i )
12
h L2 rII2
df/2 D1/2 90−αf
αf
Figure No. 14: Radius of Curvature for Rib – Roller End Contact (direction 2) a * and b * are defined as : ⎛ 2k 2ζ a* = ⎜⎜ ⎝ π
1
⎞3 ⎟⎟ ⎠
1
⎛ 2ζ ⎞ 3 b* = ⎜ ⎟ ⎝ kπ ⎠
Brewe and Hamrock [6], using a least squares method of linear regression, obtained simplified approximations for k, and ζ. The simplifications are listed in equations 3.4.6 and 3.4.7: ⎛R k = 1.0339⎜ x ⎜R ⎝ y
ζ = 1.0003 +
⎞ ⎟ ⎟ ⎠
0.636
3.4.6
0.5968 ⎛ Rx ⎞ ⎜ ⎟ ⎜R ⎟ ⎝ y⎠
3.4.7
and Rx Ry 3.5
−1
= ρ xI + ρ xII
−1
= ρ yI + ρ yII
Line contact Stress Calculations:
For the roller body- raceway contacts under ideal line contact the following equations are used:
13
σ max = i ,o
2Qi , o πleff bi , o
3.5.0
l
σ max
σ
Y y
2b
Figure No. 15: Line Contact Stress Profile (ideal) The stress at any point along the roller can be defined as:
σ i ,o =
2Qi ,o ⎡ ⎛ y ⎢1 − ⎜ πl eff bi ,o ⎢ ⎜⎝ bi ,o ⎣
⎡ 4Qi ,o bi ,o = ⎢ ⎢⎣ πl eff ∑ ρ i ,o
⎞ ⎟ ⎟ ⎠
2
⎤ ⎥ ⎥ ⎦
1 2
3.5.1
⎛1− ξI 1 − ξ II ⎜ + ⎜ E E2 I ⎝ 2
2
⎞⎤ ⎟⎥ ⎟ ⎠⎥⎦
1 2
3.5.2
For steel contacting bodies: ⎛ Qi ,o bi ,o = 3.35 x10 −3 ⎜ ⎜l ∑ρ i ,o ⎝ eff
⎞ ⎟ ⎟ ⎠
1 2
3.5.3
Where,
14
∑ρ
i
=
2 Dmean (1 − γ i )
assumes no crown on roller or raceways
∑ρ
o
=
2 Dmean (1 + γ o )
assumes no crown on roller or raceways
γi = γo =
Dmean cos(α i ) dm
Dmean cos(α o ) dm
The calculation methods listed in sections 4 and 5 will be used to determine the stress distributions and contact shapes for the raceways and rib contacts.
15
4.
4.1
Results And Discussion
EXAMPLE CALCULATION
For this example the following are known parameters: Inner Ring Speed = 8000 RPM; Thrust Load = 35,000 lbf; Pitch Diameter = 9.7”; Cup Angle = 18 degrees; Included Roller Angle=4.5 degrees; Cone Angle=13.5 degrees; Rib angle = 13.5 degrees; Roller Large End Diameter = 1.5”; Roller Length = 1.9”; Roller Large End Crown = 18”; Roller Qty = 17; Roller Mean Diameter = 1.425”, Pitch Angle = 15.75 (all angles must be converted to radians by multiplying by π/180). Calculate the maximum hertzian contact stress at the cup, cone, and rib contacts.
The calculation procedure is as follows:
Step 1: Calculate the contact forces. Begin with calculating the Cup contact force from equation 3.1.0, as shown below: Qo =
AppliedThrust z sin (α o )
Qo =
35,000 = 6662.5 lbf (29650 N) 17 sin (18)
Step 2: Calculate the cage speed from equation 3.3.7, as shown below:
1 ⎛⎜ ⎛ Dmean 1− ⎜ 2 ⎜⎝ ⎜⎝ d m
⎞ ω ⎞ ⎟⎟ cos(α dm )⎟ = c ⎟ ω i ⎠ ⎠
1⎛
⎛D
1⎛
⎛ 1.425 ⎞
⎞
⎞
ω c = ⎜⎜1 − ⎜⎜ mean ⎟⎟ cos(α dm )⎟⎟ω i 2 ⎝ ⎝ dm ⎠ ⎠ ⎞
ω c = ⎜⎜1 − ⎜ ⎟ cos(15.75)⎟⎟8000 2 ⎝ ⎝ 9.7 ⎠ ⎠ ω c = 3434.4 RPM Step 3: Calculate the centrifugal force generated by the rolling element from equation 3.2.0 as shown below:
16
1 ρπDmean 2 l t d mω c 2 8 1 2 2 Fc = (0.3)π (1.425) (1.9 )(9.7 )(3434.4) 8 Fc = 1399.3 lbf (6227 N) Fc =
Step 4: Now that the centrifugal forces have been calculated the rib force can be determined. Qo sin (α o ) tan (α i ) sin (α f ) ⎞ ⎛ ⎜⎜ − cos(α f ) − ⎟ tan (α i ) ⎟⎠ ⎝
Qo cos(α o ) − Fc − Qf =
6662.5 sin (18) tan (13.5) Qf = sin (13.5) − cos(13.5) − tan (13.5) Q f = 849.4 lbf (3780 N) 6662.5 cos(18) − 1399.3 −
Step 5: The cone reaction force can be calculated. Qi =
Qo sin (α o ) − Q f sin (α f sin (α i )
)
6662.5 sin (18) − 849.4 sin (13.5) sin (13.5) Qi = 5281.3 lbf (23502 N) Qi =
Now that the reaction forces have been calculated, the stresses can be determined. We will begin with the cup and cone contact stresses. Step 6: The cup contact stress is given by equation 3.5.0 as shown below:
σ max o =
2Qo πl eff bo
The term bo is given by equation 3.5.3 (English unit conversion). ⎛ Qo bo = 2.78 x10 − 4 ⎜ ⎜l ∑ρ o ⎝ eff
1
⎞2 ⎟ ⎟ ⎠
First we must solve for γo
17
Dmean cos(α o ) dm
γo =
1.425 cos(18) 9.7 γ o = 0.1397
γo =
Now, solve for the curvature sum
∑ρ
o
=
2 Dmean (1 + γ o )
∑ρ
o
=
2 1.425(1 + 0.1397 )
∑ρ
o
= 1.2315
1 ⎛ 1 ⎞ ⎜ 31.28 ⎟ in ⎝ mm ⎠
Substituting back into the equation for bo 1
⎛ 6662.5 ⎞ 2 ⎟⎟ bo = 2.78 x10 ⎜⎜ ⎝ 1.9(1.2315) ⎠ bo = 0.01483 in (0.3767 mm ) −4
And finally, substituting the back into the equation for max hertz stress
σ max o =
2Qo πl eff bo
2(6662.5) π (1.9)(0.01483) = 150483 psi (1036828 kPa )
σ max o = σ max o
The cone contact stress shall be determined similar to the cup contact stress Step 7: The difference in the calculations lie in the bo term.
⎛ Qi bi = 2.78 x10 − 4 ⎜ ⎜l ∑ρ i ⎝ eff
1
⎞2 ⎟ ⎟ ⎠
First we must solve for γi
18
Dmean cos(α i ) dm
γi =
1.425cos(13.5) 9.7 γ i = 0.1428
γi =
Now, solve for the curvature sum
∑ρ
i
=
2 Dmean (1 − γ i )
∑ρ
i
=
2 1.425(1 − 0.1428)
∑ρ
i
= 1.6374
1 ⎛ 1 ⎞ ⎜ 51.59 ⎟ in ⎝ mm ⎠
Substituting back into the equation for bi ⎛ 5281.3 ⎞ ⎟⎟ bi = 2.78 x10 − 4 ⎜⎜ ⎝ 1.9(1.6374 ) ⎠ bi = 0.01145 in (0.2908 mm )
1 2
And finally, substituting the back into the equation for max hertz stress
σ max i =
2Qi πl eff bi
2(5281.3) π (1.9 )(0.01145) = 154492 psi (1064450 kPa )
σ max i = σ max i
_____ Cone Contact Stress _____ Cup Contact Stress
σ (psi)
y (in) Figure No. 16: Hertzian Contact Stress Profile for Cup and Cone 19
Step 8: Determine the geometry of the rib to roller end contact:
θ = 90 − (α f − α i ) = 90 − (13.5 − 13.5) = 90 deg ⎞ αr ⎛ 1.5 ⎞ 13.5 ⎟⎟ − ⎟⎟ − = a sin ⎜⎜ = .0024 rad 2 ⎝ 2(18) ⎠ ⎠ 2 ⎛ 90π ⎞ h = rI 1 (sin (φ ) + cos(θ )) = 18(sin (.0024 )) + cos⎜ ⎟ = 0.0434 in (1.102 mm ) ⎝ 180 ⎠ Dmax 1.5 L2 = = = 19.1 in (485.14 mm ) ⎛αr ⎞ ⎛ 4.5 ⎞ 2 sin ⎜ ⎟ 2 sin ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ D1 = 2 L2 sin (α i ) = 2(19.1)sin (13.5) = 8.919 in (226.54mm ) ⎛ Dmax ⎝ 2rI 1
φ = a sin ⎜⎜
⎛ sin (θ − α i ) ⎞ ⎛ sin (90 − 13.5) ⎞ ⎟⎟ = 8.919 + 2(0.0434)⎜⎜ ⎟⎟ = 9.003 in (228.6 mm ) d f = D1 + 2h⎜⎜ ⎝ cos(θ ) ⎠ ⎝ cos(90) ⎠ To determine the radius of curvature of body II in the x direction. The following geometry must be used, and since the radius is concave the value must be negative. rI 1 = rI 2 = 18 rII 1 = ∞ (i.e. flat surface) rII 2 =
−df
2 sin (α f 1
∑ρ = r
I1
ρ xI =
+
)
=
− 9.003 = −19.28 2 sin (13.5)
1 1 1 1 1 1 1 ⎛ 1 ⎞ + + = + +0− = 0.00592 ⎜ 0.1504 ⎟ 19.28 in ⎝ mm ⎠ rI 2 rII 1 rII 2 18 18
1 1⎛ 1 ⎞ = .05556 ⎜1.411 ⎟ 18 in ⎝ mm ⎠
ρ xII = −
1 1 ⎛ 1 ⎞ = −.058562 ⎜1.487 ⎟ 19.28 in ⎝ mm ⎠
1 ⎞ 1 1 ⎛ = .05556 ⎜1.411 ⎟ 18 in ⎝ mm ⎠ =0
ρ yI = ρ yII
Rx =
1 1 1⎛ 1 ⎞ = = 270.3 ⎜ 6865.6 ⎟ ρ xI + ρ xII .05556 + (− .058562) in ⎝ mm ⎠
Ry =
1 1 1⎛ 1 ⎞ = = 18.0 ⎜ 457.2 ⎟ ρ yI + ρ yII .05556 + 0 in ⎝ mm ⎠
Step 9: Determine the dimensionless parameters and max stress.
20
0.636
0.636 ⎛ Rx ⎞ ⎛ 270.3 ⎞ ⎜ ⎟ k = 1.0339 = 1.0339⎜ = 5.7913 ⎟ ⎜R ⎟ ⎝ 18 ⎠ ⎝ y⎠ 0.5968 0.5968 ζ = 1.0003 + = 1.0003 + = 1.04004 ⎛ Rx ⎞ ⎛ 270.3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜R ⎟ ⎝ 18 ⎠ y ⎝ ⎠
⎛ 2k 2ζ a* = ⎜⎜ ⎝ π
1
1
⎞ 3 ⎛ 2(5.7913)2 (1.04004 ) ⎞ 3 ⎟ = 2.811 ⎟⎟ = ⎜ ⎜ ⎟ π ⎠ ⎝ ⎠
1
1
⎛ 2ζ ⎞ 3 ⎛ 2(1.04004 ) ⎞ 3 b* = ⎜ ⎟ =⎜ ⎟ = 0.4853 ⎝ kπ ⎠ ⎝ 5.7913π ⎠ 1
1
⎡ Q ⎤3 ⎛ 849.6 ⎞ 3 a = 0.0045a * ⎢ ⎟ = 0.3074 in (7.808mm ) ⎥ = 0.0045(2.811)⎜ ⎝ .05919 ⎠ ⎣⎢ ∑ ρ ⎦⎥ 1 3
1
⎡ Q ⎤ ⎛ 849.6 ⎞ 3 b = 0.0045b * ⎢ ⎟ = .05308 in (1.348mm ) ⎥ = .0045(.4853)⎜ ⎝ .05919 ⎠ ⎢⎣ ∑ ρ ⎥⎦ 3Q f 3(849.6 ) σ max = = = 24863 psi (171306kPa ) 2πab 2π (0.3074 )(0.05308)
σ (psi)
x (in)
y (in)
Figure No. 17: 3D View of Rib- Roller End Contact Ellipse
21
y (in)
x (in)
Figure No. 18: 2D View Rib-Roller End Contact Ellipse
4.2
SENSITIVITY STUDY
The following figures, illustrate the effects of cup angle on bearing contact stresses for a given speed (DN). The geometry used in section 4.1 was used for this study, with the exception of the cup angle, cone angle and rib angle. The cone angle was varied from 5 degrees to 40 degrees, the cone angle equals the cup angle – roller included angle, and the rib angle is assumed to be equal to the cone angle. Table No. 2 summarizes the optimal cup angles for each DN level of the sample bearing. DN (million)
Cup Angle (deg)
0.25
40
0.50
40
0.75
30
1.00
25
1.50
20
2.00
15
3.00
10
Table No. 2: Sample Bearing Optimized Cup Angle Based on DN leve. Figures 19-25 illustrate the cup, cone and rib stresses for various DN levels for a range of cup angles. 22
Tapered Bearing Stress Distribution (0.25 million DN) 350000
50000 45000 40000
250000
35000 30000
200000
25000 150000
20000 15000
100000
Rib Stress (psi)
Cup, Cone Stress (psi)
300000
Cup Stress Cone Stress Rib Stress
10000 50000
5000
0
0 5
10
15
20
25
30
35
40
Cup Angle (deg)
Figure No 19: Example TRB at 0.25 million DN Tapered Bearing Stress Distribution (0.50 million DN) 350000
50000 45000 40000
250000
35000 30000
200000
25000 150000
20000 15000
100000
Rib Stress (psi)
Cup, Cone Stress (psi)
300000
10000 50000
5000
0
0 5
10
15
20
25
30
35
40
Cup Angle (deg)
Figure No 20: Example TRB at 0.50 million DN
23
Cup Stress Cone Stress Rib Stress
Tapered Bearing Stress Distribution (0.75 million DN) 350000
50000 45000 40000
250000
35000 30000
200000
25000 150000
20000 15000
100000
Rib Stress (psi)
Cup, Cone Stress (psi)
300000
Cup Stress Cone Stress Rib Stress
10000 50000
5000
0
0 5
10
15
20
25
30
35
40
Cup Angle (deg)
Figure No 21: Example TRB at 0.75 million DN
Tapered Bearing Stress Distribution (1.00 million DN) 350000
50000 45000 40000
250000
35000 30000
200000
25000 150000
20000 15000
100000
Rib Stress (psi)
Cup, Cone Stress (psi)
300000
10000 50000
5000
0
0 5
10
15
20
25
30
35
40
Cup Angle (deg)
Figure No 22: Example TRB at 1.00 million DN
24
Cup Stress Cone Stress Rib Stress
Tapered Bearing Stress Distribution (1.50 million DN) 350000
50000 45000 40000
250000
35000 30000
200000
25000 150000
20000 15000
100000
Rib Stress (psi)
Cup, Cone Stress (psi)
300000
Cup Stress Cone Stress Rib Stress
10000 50000
5000
0
0 5
10
15
20
25
30
35
40
Cup Angle (deg)
Figure No 23: Example TRB at 1.50 million DN
Tapered Bearing Stress Distribution (2.00 million DN) 350000
50000 45000 40000
250000
35000 30000
200000
25000 150000
20000 15000
100000
Rib Stress (psi)
Cup, Cone Stress (psi)
300000
10000 50000
5000
0
0 5
10
15
20
25
30
35
40
Cup Angle (deg)
Figure No 24: Example TRB at 2.00 million DN
25
Cup Stress Cone Stress Rib Stress
Tapered Bearing Stress Distribution (3.00 million DN) 300000
50000 45000 40000 35000
200000
30000 150000
25000 20000
100000
15000
Rib Stress (psi)
Cup, Cone Stress (psi)
250000
Cup Stress Cone Stress Rib Stress
10000
50000
5000 0
0 5
10
15
20
25
30
35
40
Cup Angle (deg)
Figure No 25: Example TRB at 3.00 million DN
Several things can be ascertained from this sensitivity study. The first and most important is that there is an optimum cup angle, for every speed. At low speeds (
