Holooly Plus Logo
Holooly Plus Logo

Question 11.19: The same 02-30 angular-contact ball bearing as in Prob. 11–1......

The same 02-30 angular-contact ball bearing as in Prob. 11–18 is to be subjected to a two-step loading cycle of 4 min with a loading of 18 kN, and one of 6 min with a loading of 30 kN. This cycle is to be repeated until failure. Estimate the total life in revolutions, hours, and loading cycles.

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.
The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.
Report Answer

Total life in revolutions

Let:

l = total turns

f_{1} = fraction of turns at F_{1}

f_{2} = fraction of turns at F_{2}

From the solution of Prob. 11-18, L_{1}=1.434(10^{6})\;\mathrm{rev\;and}\;L_{2}=0.310(10^{6})\;\mathrm{rev}.

Palmgren-Miner rule:

{\frac{l_{1}}{L_{1}}}+{\frac{l_{2}}{L_{2}}}={\frac{f_{1}l}{L_{1}}}+{\frac{f_{2}l}{L_{2}}}=1

from which

l=\frac{1}{f_{1}/L_{1}+f_{2}/L_{2}}

 

l=\frac{1}{\{0.40/[1.434(10^{6})]\}+\{0.60/[0.310(10^{6})]\}}=451\,585\,rev

Total life in loading cycles

4 min at 2000 rev/min = 8000 rev

{\frac{6\ \mathrm{min}}{10\ \mathrm{min}/\mathrm{cycle}}}\ a t\,2000\ \mathrm{rev/min}={\frac{12\,000\ \mathrm{rev}}{20\,000\ \mathrm{rev/cycle}}}

 

{\frac{451\,585\,\mathrm{rev}}{20\,000\,\mathrm{rev/cycle}}}=22.58\,\mathrm{cycles}

Total life in hours

\left(10{\frac{\mathrm{min}}{\mathrm{cycle}}}\right)\left({\frac{22.58\,\mathrm{cycles}}{60\,\mathrm{min}/\mathrm{h}}}\right)=3.76\,\mathrm{h}

Related Answered Questions

Question: 11.20

The expression F^a L = constant can be written using x = L/L10 , and it can be expressed as F^a x = K or log 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, ...

Verified Answer:

While we made some use of the log F-log x plot in ...
Question: 11.18

Estimate the remaining life in revolutions of an 02-30 mm angular-contact ball bearing already subjected to 200 000 revolutions with a radial load of 18 kN, if it is now to be subjected to a change in load to 30 kN. ...

Verified Answer:

Express Eq. (11-1) as F L^{1/a}={\mathrm{co...
Question: 11.17

Different bearing metallurgy affects bearing life. A manufacturer reports that a particular heat treatment increases bearing life at least threefold. A bearing identical to that of Prob. 11–15 except for the heat treatment, loaded to 18 kN and run at 2000 rev/min, revealed an L10 life of 360 h and ...

Verified Answer:

The horizontal separation of the R = 0.90 loci in ...
Question: 11.16

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 ...

Verified Answer:

This problem is rich in useful variations. Here is...
Question: 11.15

In bearings tested at 2000 rev/min with a steady radial load of 18 kN, a set of bearings showed an L10 life of 115 h and an L80 life of 600 h. The basic load rating of this bearing is 39.6 kN. Estimate the Weibull shape factor b and the characteristic life θ for a two-parameter model. This ...

Verified Answer:

Hoover Ball-bearing Division uses the same 2-param...
Question: 11.14

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 ...

Verified Answer:

Shafts subjected to thrust can be constrained by b...
Question: 11.13

A gear-reduction unit uses the countershaft depicted in the figure. Find the two bearing reactions. The bearings are to be angular-contact ball bearings, having a desired life of 40 kh when used at 200 rev/min. Use 1.2 for the application factor and a reliability goal for the bearing pair of 0.95. ...

Verified Answer:

\begin{array}{l}{{R=\sqrt{0.95}=0.975}}\\ {...
Question: 11.12

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 ...

Verified Answer:

Given: \begin{array}{l}{{F_{r A}=560\,{\mat...
Question: 11.11

The figure is a schematic drawing of a countershaft that supports two V-belt pulleys. The countershaft runs at 1200 rev/min and the bearings are to have a life of 60 kh at a combined reliability of 0.999. The belt tension on the loose side of pulley A is 15 percent of the tension on the tight side. ...

Verified Answer:

The reliability of the individual bearings is [lat...
Question: 11.10

The figure shown is a geared countershaft with an overhanging pinion at C. Select an angularcontact ball bearing from Table 11–2 for mounting at O and a straight roller bearing for mounting at B. The force on gear A is FA = 600 lbf, and the shaft is to run at a speed of 480 rev/min. Solution of the ...

Verified Answer:

For a combined reliability goal of 0.90, use [late...