Question

A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is 40 kh at 520 rev/min. The application factor is 1.4. The radial load is 2600 lbf. The reliability goal is 0.90.

Required:
Determine the C10 value in kN for this application and design factor.

Answer 1

The C10 value for a cylindrical roller bearing can be calculated using the L10 life equation, application factors, and reliability adjustments. Without bearing life adjustment factors and reliability factor tables, exact calculations cannot be made.

Step-by-step explanation:

To determine the C10 value for a cylindrical roller bearing with specific life, speed, load, and reliability requirements, we use bearing life equations and reliability adjustments. The L10 life is the number of revolutions at which 90% of a group of identical bearings will still be operational. The equation for L10 life, in millions of revolutions, is L10 = (C/P)^3, where C is the dynamic load rating and P is the equivalent dynamic bearing load. Since we want to find C and we have the life requirement L10 along with the radial load P, we can rearrange the equation to solve for the dynamic load rating.

The application factor is used to adjust the equivalent dynamic bearing load, and the reliability goal modifies the basic rating life to achieve the desired reliability. This is done by using a reliability factor a1. Unfortunately, without the necessary bearing life adjustment factors and reliability factor tables, we cannot calculate the exact C10 value. Typically, bearing manufacturers provide the required adjustment factors and tables to make these calculations.

Also, we need to consider the RPM of the bearing when calculating the life requirement in hours. The formula for calculating life in hours is L10 (hours) = (L10 (millions of revolutions)) / (speed (RPM) ÷ 60). Since we already have the life requirement in hours and the speed, we can rearrange to find L10 (millions of revolutions) required and then use it to find the C10 value by correcting for the application factor and reliability.

Answer 2

`\mathbf{C_(10) = 137.611 \ kN}`

Step-by-step explanation:

From the information given:

Life requirement = 40 kh = 40
`40 * 10^(3) \ h`

Speed (N) = 520 rev/min

Reliability goal
`(R_D)` = 0.9

Radial load
`(F_D)` = 2600 lbf

To find C10 value by using the formula:

`C_(10)=F_D* \pmatrix \frac{x_D}{x_o +(\theta-x_o) \bigg(In((1)/(R_o)) \bigg)^{(1)/(b)}} \end {pmatrix} ^{^{^{(1)/(a)}`

where;

`x_D = \text{bearing life in million revolution} \\ \\ x_D = (60 * L_h * N)/(10^6) \\ \\ x_D = (60 * 40 * 10^3 * 520)/(10^6)\\ \\ x_D = 1248 \text{ million revolutions}`

`\text{The cyclindrical roller bearing (a)}= (10)/(3)`

The Weibull parameters include:

`x_o = 0.02`

`(\theta - x_o) = 4.439`

`b= 1.483`

∴

Using the above formula:

`C_(10)=1.4* 2600 * \pmatrix \frac{1248}{0.02+(4.439) \bigg(In((1)/(0.9)) \bigg)^{(1)/(1.483)}} \end {pmatrix} ^{^{^{(1)/((10)/(3))}`

`C_(10)=3640 * \pmatrix \frac{1248}{0.02+(4.439) \bigg(In((1)/(0.9)) \bigg)^{(1)/(1.483)}} \end {pmatrix} ^{^{^{(3)/(10)}`

`C_(10) = 3640 * \bigg[(1248)/(0.9933481582)\bigg]^{(3)/(10)}`

`C_(10) = 30962.449 \ lbf`

Recall that:

1 kN = 225 lbf

∴

`C_(10) = (30962.449)/(225)`

`\mathbf{C_(10) = 137.611 \ kN}`