Page 232 - Rollingbearings

Bearing size, locating
Given:
P = 18,6 kN
u
P = F = 3,29 kN (Loads, page 509) support
r
then η P /P = 0,2 × 18,6/3,29 = 1,13
c u
3. Life modiication factor, a SKF
Given: The given operating conditions, and the effects of rolling contact
κ = 2,8 fatigue, indicate that bearing size should be determined using the
η P /P = 1,13 basic rating life and SKF rating life.
c
u
NU 311 ECP is an SKF Explorer bearing Product data for 7312 BECBP is on page 414
then, using diagram 10, page 97, a SKF = 50 Basic rating life

Given: p
⎛ 10 6 w ⎛ C w
L 10h > 1 000 000 h L 10h = J J J
⎝ 60 n ⎠ ⎝ P ⎠
then L 10mh > 50 × 1 000 000 h
then L 10mh >> 1 000 000 h indicating that the bearing is oversized From Loads, page 398:

for the operating conditions.
C = 1,62 C single bearing = 1,62 × 104 = 168,5 kN
Minimum load
From Loads, page 398, for bearing pairs arranged back-to-back:
The fact that the basic rating life and SKF rating life are both very
high and above the required bearing life indicates that the bearing F /F = 11,5/1,45 > 1,14
r
a
may be too lightly loaded.
Using the minimum load equation from Loads, page 509, the So use:
minimum radial load, F , required to avoid skidding and roller slip
rm
for cylindrical roller bearings is given by: P = 0,57 F + 0,93 F = (0,57 × 1,45) + (0,93 × 11,5) = 11,52 kN
r
a
q 4 n w q d w 2 Therefore, the load ratio C/P = 168,5/11,52 = 14,6
m
rm r < n z < 100 z
F = k 6 + —— ——
r ⎛ 10 6 w ⎛ 168,5 w 3
L 10h = J J J J J J = 17 400 h
⎝ 60 × 3 000 ⎠ ⎝ 11,52 ⎠
Given:
d = 87,5 mm
m
k = 0,15 SKF rating life
r
n = 3 000 r/min
L
n = 6 000 r/min L 10mh = a SKF 10h
r
then F = 0,94 kN < F = 3,29 kN 1. Lubrication condition – the viscosity ratio, κ
r
rm
Conclusion κ = ν/ν 1

The bearing is oversized / lightly loaded. Options are: Given:
oil viscosity grade = ISO VG 68
• Continue to use the current bearing. There is no risk that the bearing operating temperature = 85 °C (185 °F)
will be damaged due to being too lightly loaded.
• Downsize the bearing, and in so doing reduce cost. Consider one of then, using diagram 13, page 100, ν = 13 mm 2 /s
the following:
– Keep the shaft diameter the same, but use the smaller NU 2 Given:
C.3 Centrifugal pump However, both of these downsizing actions require design modiica- then, using diagram 14, page 101, ν = 7 mm 2 /s
n = 3 000 r/min
series bearing NU 211 ECP (refer to the product section).
d = 0,5 (60 + 130) = 95 mm
– Reduce the shaft diameter one step, provided the shaft design
m
permits (strength and stiffness), and use the smaller NU 2 series
bearing NU 210 ECP (refer to the product section).
1
Therefore, κ = 13/7 = 1,8
tions to the adjacent components.

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