SME
Machine Design · Problem type
Ball and roller bearing life
Use this for “A ball bearing rated C = 30 kN carries 5 kN at 1800 rpm. Find its L₁₀ life in revolutions and in hours.”
Diagram
Balls
touch
the
races
at
points,
so
a
=
3:
L₁₀
=
(
C
P
)
3
=
6
3
=
216
million
revolutions.
At
1,800
rpm,
Lₕ
=
L₁₀
×
10
6
60N
=
2,000
h.
90% of such bearings reach this life.
This is an example. Type your own values to start a new problem.
L₁₀
=
C (kN)
P (kN)
Lₕ (h)
=
×
10
6
60
×
N (rpm)
B
Bearing type
example
Formulas
a
=
3
3 = 3
L₁₀
=
(
C
P
)
a
216
=
(
30
5
)
3
Lₕ
=
L₁₀
×
1,000,000
60 ×
N
2,000
=
216
×
1,000,000
60 × 1,800
Assumptions
•
L₁₀ is the life that 90% of a batch of identical bearings reach or pass.
•
P is the equivalent radial load, steady at a steady speed; C is the catalog rating for one million revolutions.
•
Exponent 3 for ball bearings and 10/3 for roller bearings; C is at least P.
Step-by-step
WE KNOW
B: ball, C = 30 kN, P = 5 kN, N = 1,800 rpm
FIND
life exponent (a), rated life in millions of revolutions (L₁₀), life in hours (Lₕ)
Step 1 · Find life exponent (a)
Use a = 3
Ball bearing: the exponent is 3 (balls touch the races at points).
ball chosen: ball
a
=
3
Step 2 · Find rated life in millions of revolutions (L₁₀)
Use L₁₀ = (C ÷ P)ᵃ
C
is
the
load
90%
of
bearings
carry
for
one
million
revolutions;
life
goes
as
C
P
to
the
power
a
.
L₁₀
=
(
C
P
)
3
L₁₀
=
(
30
5
)
3
L₁₀
=
6
3
L₁₀
=
216
Step 3 · Find life in hours (Lₕ)
Use Lₕ = L₁₀ × 1,000,000 ÷ (60 × N)
The revolutions over the revolutions each hour, 60 times the speed in rpm.
Lₕ
=
216
×
1,000,000
60 × 1,800
Lₕ
=
216
×
1,000,000
108,000
Lₕ
=
216,000,000
108,000
Lₕ = 2,000 h
Check
Put the numbers back in. Both sides should match.
3 = 3 ✓
216
=
(
30
5
)
3
✓
2,000
=
216
×
1,000,000
60 × 1,800
✓
Topic check
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