Chapter 11, Problem 4P

Problems 11–2 and 11–3 raise the question of the reliability of the bearing pair on the shaft. Since the combined reliabilities R is R₁R₂, what is the reliability of the two bearings (probability that either or both will not fail) as a result of your decisions in Probs. 11–2 and 11–3? What does this mean in setting reliability goals for each of the bearings of the pair on the shaft?

To determine

The reliability of two bearing.

Answer to Problem 4P

The reliability of two bearing is $0.94$.

Setting the reliability goals for each bearing depends on the combined reliability of the bearings.

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 below 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 $a_f$.

Write the expression for combined reliabilities.

    $R=R_1{R}_2$ (VI)

Here, combined reliability is $R$, reliability of first bearing is $R_1$ and the reliability of second bearing is $R_2$.

Conclusion:

For first bearing.

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

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

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

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

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

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

    $\begin{array}{c}{x}_B=0.02+\left(4.459-0.02\right){\left(\ln \frac{1}{0.90}\right)}^{\frac{1}{1.483}}\\ =0.02+\left(4.457\right){\left(0.1053\right)}^{0.674}\\ =0.02+\left(4.457\right)\left(0.2192\right)\\ =0.9969\end{array}$

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

Substitute $0.9969$ for $x_B$, $840$ for $x_D$, $3$ for $a$ and $725\text{lbf}$ for $F_D$ in Equation (II).

    $\begin{array}{c}{F}_B=\left(725\text{lbf}\right){\left(\frac{1248}{0.9969}\right)}^{\frac{1}{3}}\\ =\left(725\text{lbf}\right){\left(1251.88083\right)}^{\frac{1}{3}}\\ =\left(725\text{lbf}\right)\left(10.777\right)\\ =7813.74\text{lbf}\end{array}$

Substitute $1.4$ for $a_f$ and $7813.74\text{lbf}$ for $F_B$ in Equation (V).

    $\begin{array}{c}{C}_{10}=1.4\times 7813.74\text{lbf}\\ =\left(10939.23\text{lbf}\right)\left(\frac{4.45\text{N}}{1\text{lbf}}\right)\\ =\left(48679.57\text{N}\right)\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ =48.67\text{kN}\end{array}$

Refer table 11-2 “Dimensions and Load Rating of Ball Bearing” to obtain the ball bearing at catalog rating of $55.9\text{kN}$ as $02-60\text{mm}$.

Thus, the catalog rating of bearing is $55.9\text{kN}$.

Substitute $55.9\text{kN}$ for $C_{10}$ and $1.2$ for $a_f$ in Equation (V).

    $\begin{array}{l}55.9\text{kN}=1.4\times F_B\\ F_B=\frac{55.9\text{kN}}{1.4}\\ F_B=39.928\text{kN}\end{array}$

Substitute $39.928\text{kN}$ for $F_B$, $1248$ for $x_D$, $3$ for $a$ and $725\text{lbf}$ for $F_D$ in Equation (II).

    $\begin{array}{l}\left(39.928\text{kN}\right)\times x_B{}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\left(725\text{lbf}\right){\left(1248\right)}^{\frac{1}{3}}\\ \left(39.928\text{kN}\right)\times \left(\frac{1000\text{N}}{1\text{kN}}\right)\times x_B{}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\left(725\text{lbf}\right)\left(\frac{4.45\text{N}}{1\text{lbf}}\right){\left(1248\right)}^{\frac{1}{3}}\\ \left(39928\text{N}\right)\times x_B{}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\left(3226.25\text{N}\right){\left(1248\right)}^{\frac{1}{3}}\\ x_B{}^{{}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\left(\frac{3226.25\text{N}}{39928\text{N}}\right){\left(1248\right)}^{\frac{1}{3}}\end{array}$

    $\begin{array}{l}{x}_B{}^{{}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=0.08\times {\left(1248\right)}^{\frac{1}{3}}\\ x_B={\left(0.08\times 10.766\right)}^3\\ x_B={\left(0.86128\right)}^3\\ x_B=0.6389\end{array}$

Substitute $0.6389$ for $x_B$, $0.02$ for $x_0$, $4.459$ for $\theta$ and $1.483$ for $b$ in Equation (IV).

    $\begin{array}{c}{R}_1=\exp \left[-{\left(\frac{0.6389-0.02}{4.459-0.02}\right)}^{1.483}\right]\\ =\exp \left[-{\left(\frac{0.6189}{4.439}\right)}^{1.483}\right]\\ =\exp \left[-{\left(0.1394\right)}^{1.483}\right]\\ \approx 0.94\end{array}$

Thus, the reliability of the bearing is $0.94$.

For second bearing:

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

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

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

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

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

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

    $\begin{array}{l}{x}_B=0.02+\left(4.459-0.02\right){\left(\ln \frac{1}{0.90}\right)}^{\frac{1}{1.483}}\\ x_B=0.02+\left(4.457\right){\left(0.1053\right)}^{0.674}\\ x_B=0.02+\left(4.457\right)\left(0.2192\right)\\ x_B=0.997\end{array}$

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

Substitute $0.997$ for $x_B$, $1248$ for $x_D$, $10/3$ for $a$ and $2235\text{lbf}$ for $F_D$ in Equation (II).

    $\begin{array}{c}{F}_B=\left(2235\text{lbf}\right){\left(\frac{1248}{0.997}\right)}^{\frac{3}{10}}\\ =\left(2235\text{lbf}\right){\left(1251.75\right)}^{0.3}\\ =\left(2235\text{lbf}\right)\left(8.496\right)\\ =18988.6\text{lbf}\end{array}$

Substitute $1.4$ for $a_f$ and $18988.6\text{lbf}$ for $F_B$ in Equation (V).

    $\begin{array}{l}{C}_{10}=1.4\times 18988.6\text{lbf}\\ C_{10}=\left(26584.04\text{lbf}\right)\left(\frac{4.45\text{N}}{1\text{lbf}}\right)\\ C_{10}=\left(118298.98\text{N}\right)\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ C_{10}=118.29\text{kN}\end{array}$

Refer to table 11-2 “Dimensions and Load Rating of Ball Bearing” to obtain the ball bearing at catalog rating of $123\text{kN}$ as $03-60\text{mm}$.

Thus, the catalog rating of the bearing is $123\text{kN}$.

Substitute $123\text{kN}$ for $C_{10}$, and $1.4$ for $a_f$ in Equation (VI).

    $\begin{array}{l}123\text{kN}=1.4\times F_B\\ F_B=\frac{123\text{kN}}{1.4}\\ F_B=87.857\text{kN}\end{array}$

Substitute $87.857\text{kN}$ for $F_B$, $1248$ for $x_D$, $10/3$ for $a$ and $2235\text{lbf}$ for $F_D$ in Equation (III).

    $\begin{array}{l}\left(87.857\text{kN}\right)\times x_B{}^{\frac{3}{10}}=\left(2235\text{lbf}\right){\left(1248\right)}^{\frac{3}{10}}\\ \left(87.857\text{kN}\right)\times \left(\frac{1000\text{N}}{1\text{kN}}\right)\times x_B{}^{\frac{3}{10}}=\left(2235\text{lbf}\right)\left(\frac{4.45\text{N}}{1\text{lbf}}\right){\left(1248\right)}^{\frac{3}{10}}\\ \left(87857\text{N}\right)\times x_B{}^{\frac{3}{10}}=\left(9945.75\text{N}\right){\left(1248\right)}^{\frac{3}{10}}\\ x_B{}^{\frac{3}{10}}=\left(\frac{9945.75\text{N}}{87857\text{N}}\right){\left(1248\right)}^{\frac{3}{10}}\end{array}$

    $\begin{array}{l}{x}_B{}^{\frac{3}{10}}=0.1131\times {\left(1248\right)}^{\frac{3}{10}}\\ x_B={\left(0.1131\times 8.489\right)}^{\frac{10}{3}}\\ x_B={\left(0.9601\right)}^{\frac{10}{3}}\\ x_B=0.873\end{array}$

Substitute $0.873$ for $x_B$, $0.02$ for $x_0$, $4.459$ for $\theta$ and $1.483$ for $b$ in Equation (III)

    $\begin{array}{c}{R}_2=\exp \left[-{\left(\frac{0.873-0.02}{4.459-0.02}\right)}^{1.483}\right]\\ =\exp \left[-{\left(\frac{0.853}{4.439}\right)}^{1.483}\right]\\ =\exp \left[-{\left(0.1922\right)}^{1.483}\right]\\ \approx 0.917\end{array}$

Thus, the reliability of the bearing is $0.917$.

Substitute $0.94$ for $R_1$ and $0.917$ for $Rl2$ in Equation (VI).

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

Reliability goal can be achieved by setting the reliability of one bearing and according to it the reliability of other bearing is achieved.

Substitute $0.861$ for $R$ in Equation (VI).

    $\begin{array}{l}{R}_1{R}_2=0.861\\ R_1=\frac{0.861}{R_2}\end{array}$

Thus, the setting the reliability goals for each bearing depends on the combined reliability.