SME
Machine Design · Problem type
Bearing rating to select
Use this for “A ball bearing must carry 4 kN at 1000 rpm for 20,000 h. What catalog rating C should it have?”
Diagram
Balls
touch
the
races
at
points,
so
a
=
3:
L₁₀
=
(
C
P
)
3
=
10.63
3
=
1,200
million
revolutions.
At
1,000
rpm,
Lₕ
=
L₁₀
×
10
6
60N
=
20,000
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
Lₕ
=
L₁₀
×
1,000,000
60 ×
N
20,000
=
1,200
×
1,000,000
60 × 1,000
L₁₀
=
(
C
P
)
a
1,200
=
(
42.5063
4
)
3
Assumptions
•
The design life is the L₁₀ life: 90% of the bearings chosen reach it.
•
P is the equivalent radial load, steady at a steady speed; pick the next catalog C above the answer.
•
Exponent 3 for ball bearings and 10/3 for roller bearings; C is at least P.
Step-by-step
WE KNOW
B: ball, P = 4 kN, N = 1,000 rpm, Lₕ = 20,000 h
FIND
life exponent (a), design life in millions of revolutions (L₁₀), catalog rating needed (C)
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 design life in millions of revolutions (L₁₀)
Use Lₕ = L₁₀ × 1,000,000 ÷ (60 × N)
The hours times the revolutions each hour, counted in millions.
L₁₀
=
Lₕ
×
60
×
N
1,000,000
L₁₀
=
20,000
×
60
×
1,000
1,000,000
L₁₀
=
1,200,000
×
1,000
1,000,000
L₁₀
=
1,200,000,000
1,000,000
L₁₀
=
1,200
Step 3 · Find catalog rating needed (C)
Use L₁₀ = (C ÷ P)ᵃ
Undo
the
power:
take
the
root
of
the
life,
then
times
the
load
P
.
C
=
P
×
L₁₀
1
3
C
=
4
×
1,
200
1
3
C
=
4
×
10.627
C
=
42.5063
kN
Check
Put the numbers back in. Both sides should match.
3 = 3 ✓
20,000
=
1,200
×
1,000,000
60 × 1,000
✓
1,200
=
(
42.5063
4
)
3
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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