Chapter 11, Problem 1P

Manufacturer	Rating Life, Revolutions	Weibull Parameters Rating Lives
x0	θ	b
1	90(106)	0	4.48	1.5
2	1(106)	0.02	4.459	1.483

Tables 11-2 and 11-3 are based on manufacturer 2.

11-1 A certain application requires a ball bearing with the inner ring rotating, with a design life of 25 kh at a speed of 350 rev/min. The radial load is 2.5 kN and an application factor of 1.2 is appropriate. The reliability goal is 0.90. Find the multiple of rating life required. x_D, and the catalog rating C₁₀ with which to enter a bearing table. Choose a 02-series deep-groove ball bearing from Table 11-2, and estimate the reliability in use.

To determine

The multiple of rating life of the bearing.

The catalog rating of the bearing.

The reliability of the bearing.

Answer to Problem 1P

The multiple of rating life required is $525$.

The catalog rating of the bearing is $25.5\text{kN}$.

The reliability of the bearing is $0.92$.

Explanation of Solution

Write the expression for multiple of rating life for bearing.

$x_D=\frac{L_D}{L_R}$ (I)

Here, multiple of rating life for design is $x_D$, desired life is $L_D$, and rating life is $L_R$.

Write the regression equation for bearing load life.

$\begin{array}{l}{F}_B{x}_B{}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$a$}\right.}=F_D{x}_D{}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$a$}\right.}\\ F_B=F_D{\left(\frac{x_D}{x_B}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$a$}\right.}\end{array}$ (II)

Here, the bearing load for design is $F_D$, the basic bearing load is $F_D$, bearing life in revolutions is $L$, arbitrary constant is $a$ and multiple of rating life for basic is $x_B$.

Write the equation for bearing life.

$L=60\mathcal{l}n$ (III)

Here, the rating life in hour is $\mathcal{l}$ and rating speed in revolutions per minute is $n$.

The following figure shows the relationship between bearing load and dimensionless life in terms of logarithmic values.

Figure-(1)

Write the equation for reliability along a constant load line $AB$.

$\begin{array}{l}{R}_D=\exp \left[-{\left(\frac{x_B-x_0}{\theta -x_0}\right)}^b\right]\\ x_B=x_0+\left(\theta -x_0\right){\left(\ln \frac{1}{R_D}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$b$}\right.}\end{array}$ (IV)

Here, the characteristic parameter is $\theta$, and Weibull parameters are $x_0$ and $b$.

Write the expression for catalog rating in terms of application factor.

$C_{10}=a_f{F}_B$ (V)

Here, catalog rating is $C_{10}$ and application factor is $a_f$.

Conclusion:

Substitute $25\text{kh}$ for $\mathcal{l}$ and $350\text{rpm}$ for $n$ in Equation (III).

$\begin{array}{c}{L}_D=\left(60\text{min}/\text{h}\right)\left(25\text{kh}\right)\left(350\text{rev}/\text{min}\right)\\ =\left(60\text{min}/\text{h}\right)\left(25\text{kh}\right)\left(\frac{1000\text{h}}{1\text{kh}}\right)\left(350\text{rev}/\text{min}\right)\\ =\left(\text{1500000}\text{min}\right)\left(350\text{rev}/\text{min}\right)\\ =525\times {10}^6\text{rev}\end{array}$

Substitute ${10}^6\text{rev}$ for $L_R$ and $525\times {10}^6\text{rev}$ for $L_D$ in Equation (I).

$\begin{array}{l}{x}_D=\frac{525\times {10}^6\text{rev}}{{10}^6\text{rev}}\\ x_D=525\end{array}$

Thus, the multiple of rating life of bearing is $525$.

Substitute $0.90$ for $R_D$, $0.02$ for $x_0$, $4.459$ for $\theta$ and $1.483$ for $b$ in Equation (IV).

$\begin{array}{c}{x}_B=0.02+\left(4.459-0.02\right){\left(\ln \left(\frac{1}{0.90}\right)\right)}^{\frac{1}{1.483}}\\ =0.02+\left(4.457\right){\left(0.1053\right)}^{0.674}\\ =0.02+\left(4.457\right)\left(0.2193\right)\\ =0.997\end{array}$

For ball bearing, the value of constant $a$ is $3$.

Substitute $0.997$ for $x_B$, $525$ for $x_D$, $3$ for $a$ and $2.5\text{kN}$ for $F_D$ in Equation (II).$\begin{array}{c}{F}_B=\left(2.5\text{kN}\right){\left(\frac{525}{0.997}\right)}^{\frac{1}{3}}\\ =\left(2.5\text{kN}\right)\left(\frac{1000\text{N}}{\text{1}\text{kN}}\right){\left(526.57\right)}^{0.33}\\ =\left(2500\text{N}\right)\left(7.908\right)\\ =19770\text{N}\end{array}$

Substitute $1.2$ for $a_f$ and $19770\text{N}$ for $F_B$ in Equation (V).

$\begin{array}{c}{C}_{10}=1.2\times 19770\text{N}\\ =24226.4\text{N}\\ =\left(23724\text{N}\right)\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ =23.7\text{kN}\end{array}$

Refer table 11-2 "Dimensions and Load Rating of Ball Bearing”, to obtain the ball bearing at catalog rating of $25.5\text{kN}$ as $02-35\text{mm}$.

Thus, the catalog rating of the bearing is $25.5\text{kN}$.

Substitute $25.5\text{kN}$ for $C_{10}$, and $1.2$ for $a_f$ in Equation (V).

$\begin{array}{l}25.5\text{kN}=1.2\times F_B\\ F_B=\frac{25.5\text{kN}}{1.2}\\ F_B=21.25\text{kN}\end{array}$

Substitute $21.25\text{kN}$ for $F_B$, $525$ for $x_D$, $3$ for $a$ and $2.5\text{kN}$ for $F_D$ in Equation (II).

$\begin{array}{l}\left(21.25\text{kN}\right)\times x_B{}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\left(2.5\text{kN}\right){\left(525\right)}^{\frac{1}{3}}\\ x_B{}^{{}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\left(\frac{2.5\text{kN}}{21.25\text{kN}}\right){\left(525\right)}^{\frac{1}{3}}\\ x_B{}^{{}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=0.1176\times {\left(525\right)}^{\frac{1}{3}}\\ x_B=0.948\end{array}$

Substitute $0.948$ for $x_B$, $0.02$ for $x_0$, $4.459$ for $\theta$ and $1.483$ for $b$ in Equation (IV).

$\begin{array}{l}R=\exp \left[-{\left(\frac{0.948-0.02}{4.459-0.02}\right)}^{1.483}\right]\\ R=\exp \left[-{\left(\frac{0.928}{4.439}\right)}^{1.483}\right]\\ R=\exp \left[-{\left(0.2\right)}^{1.483}\right]\\ R\approx 0.92\end{array}$

The reliability of the bearing is $0.92$.