Shafts and bearings: one load, two choices
Introduzido| English | Significado |
|---|---|
| rolling bearings | Bearings in which balls or rollers roll between rings, giving low friction under load. |
| basic dynamic load rating | The load a rolling bearing can carry for one million revolutions with 90% reliability. |
| L10 life | The life that 90% of a group of identical bearings will reach or exceed. |
Strong, stiff and long-lived
- The shaft must be strong and stiff enough, and its bearings must last as long as the pump. Both choices come from the same loads.
- Draw the bending-moment diagram, find the worst section (the gear seat, where moment and notch meet) and step the shaft down where the moment is small.
The DE-Goodman shaft equation
- Solving Goodman for the diameter gives $d = [\tfrac{16n}{\pi}(2K_fM_a/S_e + \sqrt3K_{fs}T_m/S_{ut})]^{1/3}$.
- With the shoulder data and $n = 2$: $d = \mathbf{29.0}$ mm, so choose the standard 30 mm, which gives $n = 2.21$. The peak moment is $1.017 \times 70 = 71.2$ N·m.
Peak moment with R1 = 1.017 kN acting 70 mm from the gear (one decimal, N m)?
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R1 times 0.07 m.
Why bearing life has a cube
- For rolling bearings, the catalogue gives the basic dynamic load rating $C$. The L10 life, the revolutions that 90% survive, is $L_{10} = (C/P)^3 \times 10^6$ for ball bearings.
- Double the load and the life falls to an eighth. A required life fixes $C_{req} = P(L/10^6)^{1/3}$.
Shaft question or bearing question?
Route each design question to the right calculation.
Doubling a ball bearing's load makes its life:
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Life goes as one over P cubed.
Choosing the bearing
- 20 000 h at 180 rpm is $L = \mathbf{2.16 \times 10^8}$ revolutions.
- With $P = 2.94$ kN: $C_{req} = 2.94 \times 216^{1/3} = \mathbf{17.6}$ kN.
- The 6206 ($C = 19.5$ kN, 30 mm bore) gives $L_{10} = \mathbf{27\,000}$ h, a 35% margin, and fits the seat.
Revolutions in 20 000 h at 180 rpm (in units of 10^8, two decimals)?
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180 x 60 x 20 000.
Required C with P = 2.94 kN for that life (one decimal, kN)?
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P times the cube root of 216.
L10 of the 6206, C = 19.5 kN, in hours (nearest thousand)?
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(19.5/2.94)^3 x 10^6 over 10 800.
Lives beyond 90%
- $L_{10}$ is the life 90% of bearings reach. For a pump that must not stop, ask for 99%: the ISO 281 factor $a_1 = 0.25$ scales the life.
- The 6206's 27 000 h becomes $0.25 \times 27\,000 = \mathbf{6750}$ h at 99%, short of 20 000. Holding 99% needs $C \ge 2.94 \times (216/0.25)^{1/3} = \mathbf{28.0}$ kN: a size up.
L10 of 27 000 h with $a_1 = 0.25$ for 99% reliability (h)?
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$0.25 \times 27\,000 = 6750$ h.
Rating needed for 99% at $2.16 \times 10^8$ revolutions and $P = 2.94$ kN (one decimal, kN)?
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$2.94 \times (216/0.25)^{1/3} = 2.94 \times 9.52 = 28.0$ kN.
Worked example: is the shaft stiff enough?
- Treat the 30 mm shaft as simply supported on a 200 mm span with 1565 N at 70 mm: $I = 3.98 \times 10^{-8}$ m⁴.
- The slope at the gear is $Pab(b - a)/3EIL = \mathbf{0.00018}$ rad, inside the 0.0005 limit for good tooth contact. At the near bearing it is 0.00049 rad, within what a ball bearing tolerates.
Slope at the gear: $Pab(b - a)/3EIL$ with $P = 1565$ N, $a = 0.07$, $b = 0.13$, $L = 0.2$ m, $EI = 7952$ N·m² ($\times 10^{-3}$ rad, two decimals)?
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$1565 \times 0.07 \times 0.13 \times 0.06/(3 \times 7952 \times 0.2) = 1.79 \times 10^{-4}$ rad.
Three shaft slips
- Sizing at the peak moment, ignoring the notch beside it. Check both together.
- A bearing chosen by bore alone. Check its rating against the load.
- Shaft slope forgotten. At the gear it should stay below about 0.0005 rad.
A bearing can be chosen from its bore size alone.
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The load rating must meet the required life.
Recap
- DE-Goodman gives 29.0 mm; choose 30 mm ($n = 2.21$).
- $L_{10} = (C/P)^3 \times 10^6$ rev; $C_{req} = 17.6$ kN.
- The 6206 lasts 27 000 h against 20 000 required.