Question 11.12: The bearing lubricant (513 SUS at 100°F) operating point is ......
The bearing lubricant (513 SUS at 100°F) operating point is 135°F. A countershaft is supported by two tapered roller bearings using an indirect mounting. The radial bearing loads are 560 lbf for the left-hand bearing and 1095 for the right-hand bearing. The shaft rotates at 400 rev/min and is to have a desired life of 40 kh. Use an application factor of 1.4 and a combined reliability goal of 0.90. Using an initial K = 1.5, find the required radial rating for each bearing. Select the bearings from Fig. 11–15.
| Manufacturer | Rating Life, revolutions | Weibull Parameters Rating Lives | ||
| {x}_{0} | \theta | b | ||
| 1 | 90(10^{6}) | 0 | 4.48 | 1.5 |
| 2 | 1(10^{6}) | 0.02 | 4.459 | 1.483 |
| Tables 11–2 and 11–3 are based on manufacturer 2. | ||||
Given:
\begin{array}{l}{{F_{r A}=560\,{\mathrm{lbf}}\quad\mathrm{or}\quad2.492\,\mathrm{KN}}}\\ {{F_{r B}=1095\ {\mathrm{lbf}}\quad\mathrm{or}\quad4.873\,\mathrm{KN}}}\end{array}Trial #1: Use K_{A}=K_{B}=1.5 and from Table 11-6 choose an indirect mounting.
{\frac{0.47F_{r A}}{K_{A}}}\lt \ ?\gt {\frac{0.47F_{r B}}{K_{B}}}-(-1)(0){\frac{0.47(2.492)}{1.5}}\lt ?\gt {\frac{0.47(4.873)}{1.5}}
0.781\,\lt \,1.527 Therefore use the upper line of Table 11-6.
F_{a A}=F_{a B}=\frac{0.47F_{r B}}{K_{B}}=1.527\,\mathrm{kN}\begin{array}{l}{{P_{A}=0.4F_{r A}+K_{A}F_{a A}=0.4(2.492)+1.5(1.527)=3.29\,\mathrm{kN}}}\\ {{P_{B}=F_{r B}=4.873\,\mathrm{kN}}}\end{array}
Fig. 11-16: f_{T}=0.8
Fig. 11-17: f_{V}=1.07
Thus, a_{3l}=f_{T}f_{V}=0.8(1.07)=0.856
Individual reliability: R_{i}={\sqrt{0.9}}=0.95
Eq. (11-17):
C_{10}=a_{f}P\left[\frac{L_{D}}{4.48f_{T}f_{v}(1-R_{D})^{2/3}90(10^{6})}\right]^{3/10} ({L}_{D} in revolutions) (11-17)
(C_{10})_{A}=1.4(3.29)\Biggl[{\frac{40\,000(400)(60)}{4.48(0.856)(1-0.95)^{2/3}(90)(10^{6})}}\Biggr]^{0.3}=11.40\,kN(C_{10})_{B}=1.4(4.873)\Biggl[{\frac{40\,000(400)(60)}{4.48(0.856)(1-0.95)^{2/3}(90)(10^{6})}}\Biggr]^{0.3}=16.88\,kN
From Fig. 11-15, choose cone 32 305 and cup 32 305 which provide F_{r}=17.4\,\mathrm{kN} and K = 1.95. With K = 1.95 for both bearings, a second trial validates the choice of cone 32 305 and cup 32 305. Ans.
| Table 11-6 | ||
| Dynamic Equivalent Radial Load Equations for P (Source: Courtesy of The Timken Company.) | ||
| Two-Row Mounting, Fixed or Floating (with No External Thrust, {F_{ae}=0}) Similar Bearing Series
For two-row similar bearing series with no external thrust, {F}_{ae}=0, the dynamic equivalent radial load P equals {{F}}_{rAB} or {{F}}_{rC}. Since {{F}}_{rAB} or {{F}}_{rC} is the radial load on the two-row assembly, the two-row basic dynamic radial load rating, C_{90(2)}, is to be used to calculate bearing life. |
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| OPTIONAL APPROACH FOR DETERMINING DYNAMIC EQUIVALENT RADIAL LOADS
The following is a general approach to determining the dynamic equivalent radial loads and therefore is more suitable for programmable calculators and computer programming. Here a factor m has to be defined as +1 for direct-mounted single row or two-row bearings or −1 for indirect-mounted bearings. Also a sign convention is necessary for the external thrust {{F}}_{a e} as follows: |
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| 1. Single-Row Mounting | ||
| Design | ||
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| Thrust Condition | Thrust Load | Dynamic Equivalent Radial Load |
| {\frac{0.47F_{rA}}{K_{A}}}\leq{\frac{0.47F_{rB}}{K_{B}}}-m F_{ae} | F_{a A}={\frac{0.47F_{r B}}{K_{B}}}-m F_{ae} | P_{A}=0.4F_{r\,A}+K_{A}F_{a\,A} |
| F_{a\,B}=\frac{0.47F_{r\,B}}{K_{B}} | P_{B}=F_{r\,B} | |
| {\frac{0.47F_{rA}}{K_{A}}}\gt{\frac{0.47F_{rB}}{K_{B}}}-m F_{ae} | F_{a\,A}=\frac{0.47F_{r\,A}}{K_{A}} | P_{A}=F_{r\,A} |
| F_{a\,B}={\frac{0.47F_{r\,AB}}{K_{A}}}+m F_{ae} | P_{B}=0.4F_{r\,B}+K_{B}F_{a\,B} | |
| Note: If P_{A}\lt F_{r\,A},{\mathrm{use}}\ P_{A}=F_{rA}\mathrm{or~if}\,P_{B}\lt F_{rB},{\mathrm{~use~}}P_{B}=F_{r{B}}. | ||
| 2. Two-Row Mounting—Fixed Bearing with External Thrust, {F_{ae}} (Similar or Dissimilar Series) | ||
| Design | ||
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| Thrust Condition* | Dynamic Equivalent Radial Load | |
| F_{ae}\leq{\frac{0.6\,F_{r\,AB}}{K}} | P_{A}={\frac{K_{A}}{K_{A}+K_{B}}}(F_{r\,A B}-1.67\,m K_{B}F_{ae}) | |
| P_{B}={\frac{K_{B}}{K_{A}+K_{B}}}(F_{r\,A B}+1.67\,m K_{A}F_{ae}) | ||
| F_{ae}\gt{\frac{0.6\,F_{r\,AB}}{K}} | P_{A}=0.4F_{r\,A B}-m K_{A}F_{a e} | |
| P_{B}=0.4F_{r\,A B}+m K_{B}F_{a e} | ||
| *If m{F}_{ae} is positive, K=K_{B}; If m{F}_{ae} is negative, K=K_{A}. Note: {{F}}_{r\,AB} is the radial load on the two-row assembly. The single-row basic dynamic radial load rating, C_{90}, is to be applied in calculating life by the above equations. |
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