Question 11.14: The worm shaft shown in part a of the figure transmits 1.35 ......
The worm shaft shown in part a of the figure transmits 1.35 hp at 600 rev/min. A static force analysis gave the results shown in part b of the figure. Bearing A is to be an angular-contact ball bearing mounted to take the 555-lbf thrust load. The bearing at B is to take only the radial load, so a straight roller bearing will be employed. Use an application factor of 1.3, a desired life of 25 kh, and a reliability goal, combined, of 0.99. Specify each 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. | ||||
Shafts subjected to thrust can be constrained by bearings, one of which supports the thrust. The shaft floats within the endplay of the second (Roller) bearing. Since the thrust force here is larger than any radial load, the bearing absorbing the thrust is heavily loaded compared to the other bearing. The second bearing is thus oversized and does not contribute measurably to the chance of failure. This is predictable. The reliability goal is not \sqrt{0.99}, but 0.99 for the ball bearing. The reliability of the roller is 1. Beginning here saves effort.
Bearing at A (Ball)
\begin{array}{l}{{F_{r}=(36^{2}+212^{2})^{1/2}=215\ lbf=0.957\,\mathrm{kN}}}\\ {{F_{a}=555\ \mathrm{lbf}=2.47\,\mathrm{kN}}}\end{array}Trial #1:
Tentatively select a 02-85 mm angular-contact with C_{10}=90.4\,\mathrm{kN}\,\mathrm{and}\,C_{0}=63.0\,\mathrm{kN}.
{\frac{F_{a}}{C_{0}}}={\frac{2.47}{63.0}}=0.0392x_{D}=\frac{25\,000(600)(60)}{10^{6}}=900
Table 11-1: X_{2}=0.56,Y_{2}=1.88
F_{e}=0.56(0.957)+1.88(2.47)=5.18\,\mathrm{kN}
F_{D}=f_{A}F_{e}=1.3(5.18)=6.73\;\mathrm{kN}
C_{10}=6.73\left\{\frac{900}{0.02+4.439[\ln(1/0.99)]^{1/1.483}}\right\}^{1/3}=107.7\,{\mathrm{kN}}\gt 90.4\,{\mathrm{kN}}
Trial #2:
Tentatively select a 02-95 mm angular-contact ball with C_{10}=121\,{\mathrm{kN}}\mathop{\mathrm{and}}C_{0}=85\,{\mathrm{kN}}.
{\frac{F_{a}}{C_{0}}}={\frac{2.47}{85}}=0.029Table 11-1: Y_{2}=1.98
F_{e}=0.56(0.957)+1.98(2.47)=5.43\,\mathrm{kN}
F_{D}=1.3(5.43)=7.05\,\mathrm{kN}
C_{10}=7.05\left\{\frac{900}{0.02+4.439[\ln(1/0.99)]^{1/1.483}}\right\}^{1/3}={113{\mathrm{~kN}}}<121{\mathrm{~kN}} O.K.
Select a 02-95 mm angular-contact ball bearing.
Bearing at B (Roller): Any bearing will do since R = 1. Let’s prove it. From Eq. (11-18) when
R=\exp\left(-\{{\frac{x_{D}\left({\frac{a_{f}F_{D}}{C_{10}}}\right)^{a}-x_{0}}{\theta-x_{0}}}\}^{b}\right) (11-18)
\left(\frac{a_{f}F_{D}}{C_{10}}\right)^{3}x_{D}\lt x_{0}\qquad R=1The smallest 02-series roller has a C_{10}=16.8\,\mathrm{kN} for a basic load rating.
\left(\frac{0.427}{16.8}\right)^{3}(900)\lt ?\gt 0.02
0.0148\lt 0.02 ∴ R = 1
Spotting this early avoided rework from {\sqrt{0.99}}=0.995.
Any 02-series roller bearing will do. Same bore or outside diameter is a common choice. (Why?)
| Table 11–1 Equivalent Radial Load Factors for Ball Bearings |
{F}_{a}/{C}_{0} | e | {F}_{a}/({V F}_{r})\leq e | {F}_{a}/({V F}_{r})\gt e | ||
| {X}_{1} | {Y}_{1} | {X}_{2} | {Y}_{2} | |||
| 0.014* | 0.19 | 1.00 | 0 | 0.56 | 2.30 | |
| 0.021 | 0.21 | 1.00 | 0 | 0.56 | 2.15 | |
| 0.028 | 0.22 | 1.00 | 0 | 0.56 | 1.99 | |
| 0.042 | 0.24 | 1.00 | 0 | 0.56 | 1.85 | |
| 0.056 | 0.26 | 1.00 | 0 | 0.56 | 1.71 | |
| 0.070 | 0.27 | 1.00 | 0 | 0.56 | 1.63 | |
| 0.084 | 0.28 | 1.00 | 0 | 0.56 | 1.55 | |
| 0.110 | 0.30 | 1.00 | 0 | 0.56 | 1.45 | |
| 0.17 | 0.34 | 1.00 | 0 | 0.56 | 1.31 | |
| 0.28 | 0.38 | 1.00 | 0 | 0.56 | 1.15 | |
| 0.42 | 0.42 | 1.00 | 0 | 0.56 | 1.04 | |
| 0.56 | 0.44 | 1.00 | 0 | 0.56 | 1.00 | |
| *Use 0.014 if F_{a}/C_{0}\lt 0.014. | ||||||