n11_bearings
page 1 of 6
=?
hydrodynamic journal bearing
1800 rpm
SAE 30 oil @ 180 F°
80 lbf radial load
ON = 25
cd / d = 0.002
d = 0.5 in
ON = 25
(prefer ON < 30)
cd / d = 0.002
(prefer 0.001 < cd / d < 0.003)
 p avg  d   c d 
   
O N  
  n '     d 
2
p avg 
P
d
n = 1800 rpm
3 
Eq 11.12c Norton
Eq 11.6d Norton
 P d  cd 

O N  

3 
  n '   d 
 = 2 reyn
2
2
 P d  c d 
 
  
  n ' O N  d 
2
3
Figure 11-1 Norton
n’ = 30 rev/sec
80 lbf 0.5 in   sec 0.0022  reyn.in 2  10 6 reyn   0.10667 in 3
 lbf . sec  reyn 
2 reyn 25  30 rev 



 = 0.474 in
use  = 0.5 in
 = 0.21394 + 0.38517 log ON – 0.0008 (ON – 60) = 0.78038
ON

cd / d = 0.002
  e / cr
20
0.74706
cd = 0.001 in
e = 3.901 x 10-4 in
25
0.78038
30
0.80688
cr = cd / 2 = 0.0005 in
Eq 11.3 Norton
Eq 11.13b Norton
35
0.82867
40
0.84701
n11_bearings
hmin = cr (1 - ) = 109.8 in
OK
Ts 
page 2 of 6
Eq 11.4b Norton
Rq < 30 to 40 in for precision milled/ground surface
 d 3  n'  2
cd 1  
Eq 11.9c Norton
2
3

2 reyn 0.5 in  0.5 in  2  30 rev  lbf . sec  reyn 

 6
Ts 


2 
0.001 in  1  0.780382  sec  reyn.in  10 reyn 
tan  
 1  2
4
 = 32.18°
Tr = Ts + P e sin = 0.07867 in.lbf
Eq 11.8a Norton
Eq 11.9a
PLOSS = Tr  = 2  Tr n’ = 14.83 in.lbf/sec = 0.00225 HP
 = 2 Tr / P d = 0.00336
Eq 11.11
= 0.0592 in.lbf
Eq 11.10 Norton
n11_bearings
page 3 of 6
Hydrodynamic journal bearing
given
radial load P = 54 lbf
diameter d = 0.591 inch (15 mm)
speed = 1725 rpm
clearance ratio cd / d = 0.0017
length ratio  / d = 0.75
Ocvirk number ON = 20
select lubricant at 190° F
n = 1725 rpm
n' = 28.75 rev/sec
cd / d = 0.0017
cd = 0.0010 inch
cr = cd / 2 = 0.0005 inch = 500 in
 = 0.21394 + 0.38517 log ON – 0.0008 (ON – 60) = 0.74706
hmin = cr (1 - ) = 126.5 in
 / d = 0.75
p avg 
OK
 = 0.443 inch
54 lbf
P
= 206 psi

d  0.591 in 0.443 in 
 p avg  d   c d 
   
O N  
  n '     d 
2
2
2
2
 p avg  d   c d 
   
  
 O N n '     d 
2
 206 lbf  1  sec  1 
2  reyn .in 
-6






0
.
0017





2
 28.75 rev  0.75
 lbf . sec  = 1.841 x 10 reyn = 1.84 reyn
20
in


 



2
Fig 11-1 at 190° F
AGMA 3, SAE 30
n11_bearings
In-line Six Cylinder Engine
page 4 of 6
n11_bearings
page 5 of 6
ball bearing selection
Fr = 1449 lbf
P = X V Fr + Y Fa
V=1
try
desire L10 = 100 for 100x106 rev
Fa = 1310 lbf
inner race rotating
Eq. 11.22a
Fig. 11-24
X=1
L10 = (C / P)3
Y=0
P = Fr = 1449 lbf
Eq. 11.20a
CDESIRED = P (L10)1/3 = 1449 lbf (100)1/3 = 6726 lbf
try
6308 bearing
Fa / C0 = 0.2472
C = 7350 lbf
C0 = 5300 lbf
interpolate e = 0.3681
Fa / (V Fr) = 0.9041 > e
X = 0.56
P = X V Fr + Y Fa = 2381 lbf
Fig. 11-23
Fig. 11-24
interpolate Y = 1.198
Fig. 11-24
Eq. 11.22a
CDESIRED = P (L10)1/3 = 2381 lbf (100)1/3 = 11,051 lbf
try
6311 bearing
Fa / C0 = 0.131
C = 12,900 lbf
interpolate e = 0.314
Fa / (V Fr) = 0.9041 > e
P = X V Fr + Y Fa = 2647 lbf
X = 0.56
C0 = 10,000 lbf
Fig. 11-24
interpolate Y = 1.401
Eq. 11.22a
CDESIRED = P (L10)1/3 = 2647 lbf (100)1/3 = 12,286 lbf
OK
6311 bearing
Fig. 11-23
C = 12,900 lbf > CDESIRED
Fig. 11-24
n11_bearings
page 6 of 6
helical gear with  = 25° pressure angle and  = 20° helix angle
WT = 2000 lbf
WR = WR tan  = 932.6 lbf
WA = WT tan  = 727.9 lbf
force on bearing
Fr =
WT2  WR2 = 2206 lbf
Fa = WA = 727.9 lbf
desire L10 = 500x106 rev
P = X V Fr + Y Fa
V=1
try
inner race rotating
Eq. 11.22a
Fig. 11-24
X=1
L10 = (C / P)3
Y=0
P = Fr = 2206 lbf
Eq. 11.20a
CDESIRED = P (L10)1/3 = 2206 lbf (500)1/3 = 17,509 lbf
try
6314 bearing
Fa / C0 = 0.0520
C = 18,000 lbf
interpolate e = 0.2543
Fa / (V Fr) = 0.3230 > e
X = 0.56
P = X V Fr + Y Fa = 2509 lbf
C0 = 14,000 lbf
Fig. 11-23
Fig. 11-24
interpolate Y = 1.750
Fig. 11-24
Eq. 11.22a
CDESIRED = P (L10)1/3 = 2509 lbf (500)1/3 = 19,914 lbf
try
6316 bearing
Fa / C0 = 0.04044
C = 21,200 lbf
interpolate e = 0.2378
Fa / (V Fr) = 0.3230 > e
P = X V Fr + Y Fa = 2594 lbf
X = 0.56
C0 = 18,000 lbf
Fig. 11-24
interpolate Y = 1.866
Eq. 11.22a
CDESIRED = P (L10)1/3 = 2594 lbf (500)1/3 = 20,589 lbf
OK
6316 bearing
Fig. 11-23
C = 21,200 lbf > CDESIRED
Fig. 11-24