Question 11.20: The expression F^a L = constant can be written using x = L/L......
The expression F^{a}L = constant can be written using x=L/L_{10}, and it can be expressed as F^{a}x=K or F=(1/a)\log K-(1/a)\log x. This is a straight line on a log-log plot, and it is the basis of Fig. 11–5. For the geometric insight provided, produce Fig. 11–5 to scale using Ex. 11–3, and
For point D: find F_{D}=1.2(413)=495.6\ \mathrm{lbf},\,\log\,F_{D},\,x_{D},\,\log x_{D},\,K_{D}
For point B: find x_{B},\log x_{B},\ F_{B},\log F_{B},\ K_{B}
For point A: find F_{A}=F_{B}=C_{10},\,\log\,F_{A},\,K_{10}
and plot to scale. On this plot, also show the line containing C_{10}, the basic load rating, of the selected bearing.
| 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. | ||||
While we made some use of the log F-log x plot in Probs. 11-15 and 11-17, the principal use of Fig. 11-5 is to understand equations (11-6) and (11-7) in the discovery of the catalog basic load rating for a case at hand.
C_{10}=F_{D}\left[\frac{x_{D}}{x_{0}+(\theta-x_{0})(\ln1/R_{D})^{1/b}}\right]^{1/a} (11-6)
C_{10}\doteq F_{D}\left[{\frac{x_{D}}{x_{0}+(\theta-x_{0})(1-R_{D})^{1/b}}}\right]^{1/a}\qquad R\geq0.90 (11-7)
Point D
F_{D}=495.6\mathrm{\,lbf}\log F_{D}=\log495.6=2.70
x_{D}={\frac{30\,000(300)(60)}{10^{6}}}=540
\log x_{D}=\log540=2.73
K_{D}=F_{D}^{3}x_{D}=(495.6)^{3}(540) =65.7(10^{9})\;\mathrm{lbf^{3}}\cdot\operatorname{turns}
\log K_{D}=\log[65.7(10^{9})]=10.82
F_{D} has the following uses: F_{\mathrm{design}},F_{\mathrm{desired}},F_{e} when a thrust load is present. It can include application factor {{a}}_{f}, or not. It depends on context.
Point B
x_{B}=0.02+4.439[\ln(1/0.99)]^{1/1.483}=0.220\,\mathrm{turns}\log x_{B}=\log 0.220=-0.658
F_{B}=F_{D}\left({\frac{x_{D}}{x_{B}}}\right)^{1/3}=495.6\left({\frac{540}{0.220}}\right)^{1/3}=6685\,lbf
Note: Example 11-3 used Eq. (11-7). Whereas, here we basically used Eq. (11-6).
\log F_{B}=\log(6685)=3.825
K_{D}=6685^{3}(0.220)=65.7(10^{9})\;\mathrm{lbf^{3}\cdot\ t u r n s} (as it should)
Point A
F_{A}=F_{B}=C_{10}=6685\,\mathrm{lbf}\log C_{10}=\log(6685)=3.825
x_{A}=1
\log x_{A}=\log(1)=0
K_{10}=F_{A}^{3}x_{A}=C_{10}^{3}(1)=6685^{3}=299(10^{9})\;\mathrm{lbf}^{3}\cdot\mathrm{turns}
Note that K_{D}/K_{10}=65.7(10^{9})/[299(10^{9})]=0.220, which is {{x}}_{B}. This is worth knowing since
K_{10}={\frac{K_{D}}{x_{B}}}\log K_{10}=\log[299(10^{9})]=11.48
Now C_{10}=6685\,lbf=29.748\,\mathrm{kN}, which is required for a reliability goal of 0.99. If we select an angular contact 02-40 mm ball bearing, then C_{10}=31.9\,\mathrm{kN}{=}7169\,\mathrm{lbf}.
