Chapter 11, Problem 5P

To determine

The reliability that meets the uses.

Answer to Problem 5P

The reliabilities that meets the uses are $0.953$ for $R_{03-series}$ and $0.94$ for $R_{02-series}$.

Explanation of Solution

Write the expected reliability goal for each bearing.

    $\begin{array}{c}{R}_{02-series}=R_{03-series}\\ =\sqrt{\text{overall}\text{reliability}}\end{array}$ (I)

Here, reliability of 02 series is $R_{02-series}$ and the reliability of 03 series is $R_{03-series}$.

Write the expression for multiple of rating life for bearing.

    $x_D=\frac{L_D}{L_R}$ (II)

Here, multiple of rating life for design is $x_D$, desired life is $L_D$, and the 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}$ (III)

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.

    $L=60\mathcal{l}n$ (IV)

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

Write the equation for reliability along a constant load line.

    $\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}$ (V)

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.

    $C_{10}=a_f{F}_B$ (VI)

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

Write the expression for reliability for 02 series angular ball bearing.

    $R_{02-series}=\exp \left[-\left(\frac{x_D{\left(\frac{a_f{F}_D}{C_{10}}\right)}^a-x_o}{\theta -x_o}\right)\right]$ (VII)

Write the expression for reliability for 03 series angular ball bearing.

    $R_{03-series}=\exp \left[-\left(\frac{x_D{\left(\frac{a_f{F}_D}{C_{10}}\right)}^a-x_o}{\theta -x_o}\right)\right]$ (VIII)

Write the expression for overall reliability.

    $R=R_{02-series}\times R_{03-series}$ (IX)

Here, overall reliability is $R$.

Conclusion:

Substitute $0.90$ for $\text{overall}\text{reliability}$ in Equation (I).

    $\begin{array}{c}{R}_{02-series}=R_{03-series}\\ =\sqrt{0.90}\\ =0.95\end{array}$

Substitute $40\text{kh}$ for $\mathcal{l}$ and $520\text{rpm}$ for $n$ in Equation (II).

    $\begin{array}{c}{L}_D=\left(60\text{min}/\text{h}\right)\left(40\text{kh}\right)\left(520\text{rev}/\text{min}\right)\\ =\left(60\text{min}/\text{h}\right)\left(40\text{kh}\right)\left(\frac{1000\text{h}}{1\text{kh}}\right)\left(520\text{rev}/\text{min}\right)\\ =\left(\text{2400000}\text{min}\right)\left(520\text{rev}/\text{min}\right)\\ =1248\times {10}^6\text{rev}\end{array}$

Substitute ${10}^6\text{rev}$ for $L_R$ and $1248\times {10}^6\text{rev}$ for $L_D$ in Equation (II).

    $\begin{array}{l}{x}_D=\frac{1248\times {10}^6\text{rev}}{{10}^6\text{rev}}\\ x_D=1248\end{array}$

Thus, the multiple of rating life of bearing is $1248$.

Substitute $0.95$ for $R_D$, $0.02$ for $x_0$, $4.459$ for $\theta$ and $1.483$ for $b$ in Equation (V).

    $\begin{array}{c}{x}_B=0.02+\left(4.459-0.02\right){\left(\ln \frac{1}{0.95}\right)}^{\frac{1}{1.483}}\\ =0.02+\left(4.457\right){\left(0.051293\right)}^{0.674}\\ =0.02+\left(4.457\right)\left(0.135076\right)\\ =0.622\end{array}$

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

Substitute $0.622$ for $x_B$, $1248$ for $x_D$, $3$ for $a$ and $725\text{lbf}$ for $F_D$ in Equation (III).

    $\begin{array}{c}{F}_B=\left(725\text{lbf}\right){\left(\frac{1248}{0.622}\right)}^{\frac{1}{3}}\\ =\left(725\text{lbf}\right){\left(2006.4308\right)}^{\frac{1}{3}}\\ =\left(725\text{lbf}\right)\left(12.612\right)\\ =9144.2075\text{lbf}\end{array}$

Substitute $1.4$ for $a_f$ and $9144.2075\text{lbf}$ for $F_B$ in Equation (VI).

    $\begin{array}{c}{C}_{10}=1.4\times 9144.2075\text{lbf}\\ =\left(12801.8905\text{lbf}\right)\left(\frac{4.45\text{N}}{1\text{lbf}}\right)\\ =\left(56968.41273\text{N}\right)\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ \approx 57\text{kN}\end{array}$

Refer table 11-2 “Dimensions and Load Rating of Ball Bearing” to obtain the ball bearing at catalog rating of $57\text{kN}$ as $6\text{5}\text{mm}$ with catalog life $63.7\text{kN}$.

Substitute $0.02$ for $x_o$, $1248$ for $x_D$, $3$ for $a$, $1.4$ for $a_f$ $4.439$ for $\theta -x_o$ and $725\text{lbf}$ for $F_D$ in Equation (II).

    $\begin{array}{l}{R}_{02-series}=\exp \left[-{\left(\frac{1248{\left(\frac{1.4\left(725\text{lbf}\right)}{63.7\text{kN}}\right)}^a-0.02}{4.439}\right)}^{1.483}\right]\\ =\exp \left[-{\left(\frac{1248{\left(\frac{1.4\left(\left(\frac{4.44822\text{N}}{1\text{lbf}}\right)\times 725\text{lbf}\right)}{\left(63.7\text{kN}\right)\left(\frac{1000\text{N}}{1\text{kN}}\right)}\right)}^a-0.02}{4.439}\right)}^{1.483}\right]\\ =\exp \left(-{\left(\frac{0.4244}{4.439}\right)}^{1.483}\right)\\ =0.97\end{array}$

For 03 series cylindrical roller bearing:

Substitute $0.90$ for $\text{overall}\text{reliability}$ in Equation (I).

    $\begin{array}{c}{R}_{02-series}=R_{03-series}\\ =\sqrt{0.90}\\ =0.95\end{array}$

Substitute $40\text{kh}$ for $\mathcal{l}$ and $520\text{rpm}$ for $n$ in Equation (II).

    $\begin{array}{c}{L}_D=\left(60\text{min}/\text{h}\right)\left(40\text{kh}\right)\left(520\text{rev}/\text{min}\right)\\ =\left(60\text{min}/\text{h}\right)\left(40\text{kh}\right)\left(\frac{1000\text{h}}{1\text{kh}}\right)\left(520\text{rev}/\text{min}\right)\\ =\left(\text{2400000}\text{min}\right)\left(520\text{rev}/\text{min}\right)\\ =1248\times {10}^6\text{rev}\end{array}$

Substitute ${10}^6\text{rev}$ for $L_R$ and $1248\times {10}^6\text{rev}$ for $L_D$ in Equation (II).

    $\begin{array}{l}{x}_D=\frac{1248\times {10}^6\text{rev}}{{10}^6\text{rev}}\\ x_D=1248\end{array}$

Substitute $0.95$ for $R_D$, $0.02$ for $x_0$, $4.439$ for $\theta$ and $1.483$ for $b$ in Equation (IV).

    $\begin{array}{c}{x}_B=0.02+\left(4.439-0.02\right){\left(\ln \frac{1}{0.95}\right)}^{\frac{1}{1.483}}\\ =0.02+\left(4.419\right){\left(0.05129\right)}^{0.674}\\ =0.02+\left(4.419\right)\left(0.135248\right)\\ =0.617\end{array}$

For roller bearing, the value of constant $a$ is $10/3$.

Substitute $0.617$ for $x_B$, $1248$ for $x_D$, $10/3$ for $a$ and $2235\text{lbf}$ for $F_D$ in Equation (II).

    $\begin{array}{c}{F}_B=\left(2235\text{lbf}\right){\left(\frac{1248}{0.617}\right)}^{\frac{3}{10}}\\ =\left(2235\text{lbf}\right){\left(2022.6914\right)}^{0.3}\\ =\left(2235\text{lbf}\right)\left(9.81248\right)\\ =21930.8928\text{lbf}\end{array}$

Substitute $1.4$ for $a_f$ and $21930.8928\text{lbf}$ for $F_B$ in Equation (V).

    $\begin{array}{l}{C}_{10}=1.4\times 21930.8928\text{lbf}\\ C_{10}=\left(30703.24992\text{lbf}\right)\left(\frac{4.45\text{N}}{1\text{lbf}}\right)\\ C_{10}=\left(136629.4621\text{N}\right)\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ C_{10}=136.6\text{kN}\end{array}$

Refer to table 11-3 “Dimensions and basic Load Rating for cylindrical roller bearing:03 series” to obtain the cylindrical bearing at catalog rating of $138\text{kN}$ as $65\text{mm}$.

Substitute $0.02$ for $x_o$, $1248$ for $x_D$, $3$ for $a$, $1.4$ for $a_f$ $4.439$ for $\theta -x_o$, $136.6\text{kN}$ for $C_{10}$ and $2235\text{lbf}$ for $F_D$ in Equation (II).

    $\begin{array}{l}{R}_{02-series}=\exp \left[-{\left(\frac{1248{\left(\frac{1.4\left(2235\text{lbf}\right)}{136.6\text{kN}}\right)}^a-0.02}{4.439}\right)}^{1.483}\right]\\ =\exp \left[-{\left(\frac{1248{\left(\frac{1.4\left(\left(\frac{4.44822\text{N}}{1\text{lbf}}\right)\times 2235\text{lbf}\right)}{\left(136.6\text{kN}\right)\left(\frac{1000\text{N}}{1\text{kN}}\right)}\right)}^a-0.02}{4.439}\right)}^{1.483}\right]\\ =0.953\end{array}$

Substitute $0.953$ for $R_{03-series}$ and $0.97$ for $R_{02-series}$ in Equation (VIII).

    $\begin{array}{c}R=0.97\times 0.953\\ =0.924\end{array}$

Here, $0.924$>, so the overall reliability exceeds the reliability goal. Therefore, this combination has no use.

Take $0.94$ for $R_{02-series}$ and $0.917$ for $R_{03-series}$.

Substitute $0.917$ for $R_{03-series}$ and $0.94$ for $R_{02-series}$ in Equation (VIII).

    $\begin{array}{c}R=0.94\times 0.917\\ =0.861\end{array}$

Here, $0.861<0.90$, so the overall reliability is less than reliability goal . Therefore, this combination can be used.

Take $0.94$ for $R_{02-series}$ and $0.953$ for $R_{03-series}$.

Substitute $0.953$ for $R_{03-series}$ and $0.94$ for $R_{02-series}$ in Equation (VIII).

    $\begin{array}{c}R=0.94\times 0.953\\ =0.89582\end{array}$

Here, $0.89582<0.90$, so the overall reliability is less than the reliability goal. Therefore, this combination can be used.

Thus, the reliabilities that meets the uses are $0.953$ for $R_{03-series}$ and $0.94$ for $R_{02-series}$.