Question 11.16: A 16-tooth pinion drives the double-reduction spur-gear trai......
A 16-tooth pinion drives the double-reduction spur-gear train in the figure. All gears have 25° pressure angles. The pinion rotates ccw at 1200 rev/min and transmits power to the gear train.
The shaft has not yet been designed, but the free bodies have been generated. The shaft speeds are 1200 rev/min, 240 rev/min, and 80 rev/min. A bearing study is commencing with a 10-kh life and a gearbox bearing ensemble reliability of 0.99. An application factor of 1.2 is appropriate. Specify the six bearings.
| 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. | ||||
This problem is rich in useful variations. Here is one.
Decision: Make straight roller bearings identical on a given shaft. Use a reliability goal of (0.99)^{1/6}=0.9983.
Shaft a
\begin{array}{l}{{F_{A}^{r}=(239^{2}+111^{2})^{1/2}=264\,lbf~~~\mathrm{or}~~~1.175\,\mathrm{kN}}}\\ {{F_{B}^{r}=(502^{2}+1075^{2})^{1/2}=1186\,\mathrm{lbf~~~or~~~5.28\mathrm{kN}}}}\end{array}Thus the bearing at B controls
x_{D}=\frac{10\,000(1200)(60)}{10^{6}}=7200.02+4.439[\mathrm{ln}(1/0.9983)]^{1/1.483}=0.080\,26
C_{10}=1.2(5.2)\left(\frac{720}{0.080\,26}\right)^{0.3}=97.2\,\mathrm{kN}
Select either a 02-80 mm with C_{10}=106\,\mathrm{kN} or a 03-55 mm with C_{10}=102\,\mathrm{kN}.
Shaft b
\begin{array}{l}{{F_{C}^{r}=(874^{2}+2274^{2})^{1/2}=2436\ lbf\quad\mathrm{or\quad10.84\mathrm{\,{kN}}}}}\\ {{F_{D}^{r}=(393^{2}+657^{2})^{1/2}=766\ lbf\quad\mathrm{or\quad3.41\mathrm{\,{kN}}}}}\end{array}The bearing at C controls
x_{D}={\frac{10\,000(240)(60)}{10^{6}}}=144C_{10}=1.2(10.84)\left(\frac{144}{0.0826}\right)^{0.3}=122\,\mathrm{kN}
Select either a 02-90 mm with C_{10}=142\,\mathrm{kN} or a 03-60 mm with C_{10}=123\,\mathrm{kN}.
Shaft c
\begin{array}{l}{{F_{E}^{r}=(1113^{2}+2385^{2})^{1/2}=2632\ lbf\quad\mathrm{or\quad11.71\mathrm{\,{kN}}}}}\\ {{F_{F}^{r}=(417^{2}+895^{2})^{1/2}=987\ lbf\quad\mathrm{or\quad4.39\mathrm{\,{kN}}}}}\end{array}The bearing at E controls
x_{D}=10\,000(80)(60/10^{6})=48C_{10}=1.2(11.71)\left(\frac{48}{0.0826}\right)^{0.3}=94.8\,\mathrm{kN}
Select a 02-80 mm with C_{10}=106\,\mathrm{kN} or a 03-60 mm with C_{10}=106\,\mathrm{kN}.