Question

Estimate the remaining life in revolutions of a 02-30 mm angular-contact ball bearing already subjected to 216000 revolutions with a radial load of 15 kN, if it is now to be subjected to a change in load to 25.05 kN. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.

Answer 1

Radial load of 15 kN, remaining life is 2478519 rev

When load is 25.05, we have 485783 rev

Step-by-step explanation:

For 02-30 mm angular contact bearing, then
`C_(10)=20.3 kN`

Since
`(F_1)^(a)L_1=K` then

`(C_(10))^(a)L_(10)=K`

For a ball bearing, a=3 hence

`(20.3)^(3)* (10^(6))=K`

`K=8.365* 10^(9)`

Life at load 15 kN
`L_1=\frac {K}{F_1^(a)}=\frac {8.365* 10^(9)}{15^(3)}=2478519 rev`

Life load at 25.05 kN
`L_2=\ frac {8.365* 10^(9)}{25.05^(3)}= 532160.6 rev`

In such a scenario, the Palmgren-Miner cycle ratio summation rule is expressed as

`\frac {I_1}{L_1}+\frac {I_2}{L_2}=1`

By substitution

`\frac {216000}{2478519}+ \frac {I_2}{532160.6 }=1`

`I_2=485783.4659 rev`