Chapter 11, Problem 23P

11–22 to

11–26 An 02-series single-row deep-groove ball bearing is to be selected from Table 11–2 for the application conditions specified in the table. Assume Table 11–1 is applicable if needed. Specify the smallest bore size from Table 11–2 that can satisfy these conditions.

To determine

The bore size of single-row deep-grove ball bearing.

Answer to Problem 23P

The bore size of single-row deep-grove ball bearing is $02\text{-}90\text{mm}$.

Explanation of Solution

Write the expression for Design life.

    $\left(x_D\right)=\frac{60L_D{N}_D}{L_{10}}$                                                                      (I)

Here, desired life is $L_D$, desires speed is $N_D$ and rating life $\left({10}^6\right)$ is $L_{10}$.

Write the expression for catalog rating life.

    $\left(C_{10}\right)=a_f{F}_r{\left[\frac{x_D}{x_0+\left(\theta -x_0\right){\left(\ln \left(\frac{1}{R}\right)\right)}^{\frac{1}{b}}}\right]}^{\frac{1}{a}}$                                (II)

Here, application factor is $a_f$, reliability is $R$, radial load is $F_r$, design life is $x_D$.and $a$,$b$ and $x_0$ are constant.

Write the expression for ration of axial load and $C_o$.

    $k_1=\frac{F_a}{C_o}$                                                                                 (III)

Here, axial force is $F_a$ and $C_o$ is a constant.

Write the expression for load factor.

    $k_2=\frac{F_a}{\left(V{F}_r\right)}$                                                                            (IV)

Here, axial force is $F_a$, radial load is $F_r$ and rotation factor is $V$.

Write the expression for equivalent radial load.

    $F_e=X_{2}V{F}_r+Y_2{F}_a$                                                                   (V)

Here, axial force is $F_a$, radial load is $F_r$,rotation factor is $V$ and $X_2$, $Y_2$ are constants.

Conclusion:

Substitute $10\times {10}^3\text{h}$ for $L_D$, $400\text{rev}/\min$ for $N_D$ and ${10}^6$ for $L_{10}$ in Equation (I).

    $\begin{array}{c}\left(x_D\right)=\frac{60\times 10\times {10}^3\text{h}\times 400\text{rev}/\min }{{10}^6}\\ =\frac{240\times {10}^6}{{10}^6}\\ =240\end{array}$

Refer to table of “Weibull parameters Rating lives for manufacturer 2” to obtain $x_0=0.02$, $\left(\theta -x_0\right)=4.439$ and $b=1.483$.

Substitute $1$ for $a_f$, $8\text{kN}$ for $F_e$, $240$ for $x_D$, $0.02$ for $x_0$, $4.439$ for $\left(\theta -x_0\right)$, $1.483$ for $b$, $0.99$ for $R$ and $3$ for $a$ in Equation (II).

    $\begin{array}{c}\left(C_{10}\right)=\left(1\right)\left(8\text{kN}\right){\left[\frac{240}{0.02+\left(4.439\right){\left(\ln \left(\frac{1}{0.99}\right)\right)}^{\frac{1}{1.483}}}\right]}^{\frac{1}{3}}\\ =1\times 8\text{kN}{\left[\frac{240}{0.2196}\right]}^{\frac{1}{3}}\\ =82.4\text{kN}\end{array}$

Refer to Table $11-2$ “Dimensions and Load Ratings for Single-Row $02\text{-}$ Series Deep-Groove and Angular-Contact Ball Bearings” for $02\text{-}80\text{mm}$ deep groove to obtain $C_{10}=70.2\text{kN}$ and $C_o=45.0\text{kN}$.

Substitute $2\text{kN}$ for $F_a$ and $45\text{kN}$ for $C_o$ in Equation (III).

    $\begin{array}{c}\frac{F_a}{C_o}=\frac{2\text{kN}}{45\text{kN}}\\ =0.044\end{array}$

Refer to Table $11-1$ “Equivalent Radial Load Factor for Ball Bearings”, to obtain $e=0.243$ at at $\frac{F_a}{C_o}=0.088$.

Substitute $2\text{kN}$ for $F_a$, $1$ for $V$ and $8\text{kN}$ for $F_r$ in Equation (IV).

    $\begin{array}{c}\frac{F_a}{\left(V{F}_r\right)}=\frac{2\text{kN}}{1\left(8\text{kN}\right)}\\ =0.25\end{array}$

Refer to Table $11-1$ “Equivalent Radial Load Factor for Ball Bearings”, at $\frac{F_a}{C_o}=0.044$ and $\frac{F_a}{C_o}>e$ to obtain $X_2=0.56$ and $Y_2=1.85$.

Substitute $2\text{kN}$ for $F_a$, $1$ for $V$, $8\text{kN}$ for $F_r$, $0.56$ for $X_2$ and $1.85$ for $Y_2$ in Equation (V).

    $\begin{array}{c}{F}_e=\left(0.56\times 1\times 8\text{kN}\right)+1.85\left(2\text{kN}\right)\\ =4.48+3.7\\ =8.18\text{kN}\end{array}$

Substitute $1$ for $a_f$, $8.18\text{kN}$ for $F_e$, $240$ for $x_D$, $0.02$ for $x_0$, $4.439$ for $\left(\theta -x_0\right)$, $1.483$ for $b$, $0.99$ for $R$ and $3$ for $a$ in Equation (II).

    $\begin{array}{c}\left(C_{10}\right)=\left(1\right)\left(8.18\text{kN}\right){\left[\frac{240}{0.02+\left(4.439\right){\left(\ln \left(\frac{1}{0.99}\right)\right)}^{\frac{1}{1.483}}}\right]}^{\frac{1}{3}}\\ =\left(1\right)\left(8.18\text{kN}\right){\left[\frac{240}{0.2196}\right]}^{\frac{1}{3}}\\ =84.26\text{kN}\end{array}$

Since $C_{10}=84.26\text{kN}$, that is greater than $70.2\text{kN}$, so this bearing cannot be expected to carry load.

Refer to Table $11-2$ “Dimensions and Load Ratings for Single-Row $02-$ Series Deep-Groove and Angular-Contact Ball Bearings” for $02\text{-}85\text{mm}$ deep groove to obtain $C_{10}=83.2\text{kN}$ and.$C_o=53.0\text{kN}$.

Substitute $2\text{kN}$ for $F_a$ and $53\text{kN}$ for $C_o$ in Equation (III).

    $\begin{array}{c}\frac{F_a}{C_o}=\frac{2\text{kN}}{53\text{kN}}\\ =0.038\end{array}$

Refer to Table $11-1$ “Equivalent Radial Load Factor for Ball Bearings”, to obtain $e=0.234$ at $\frac{F_a}{C_o}=0.088$.

Substitute $2\text{kN}$ for $F_a$, $1$ for $V$ and $8\text{kN}$ for $F_r$ in Equation (IV).

    $\begin{array}{c}\frac{F_a}{\left(V{F}_r\right)}=\frac{2\text{kN}}{1\left(8\text{kN}\right)}\\ =0.25\end{array}$

Refer to Table $11-1$ “Equivalent Radial Load Factor for Ball Bearings”, at $\frac{F_a}{C_o}=0.038$ and $\frac{F_a}{C_o}>e$ to obtain $X_2=0.56$ and $Y_2=1.89$.

Substitute $2\text{kN}$ for $F_a$, $1$ for $V$, $8\text{kN}$ for $F_r$, $0.56$ for $X_2$ and $1.89$ for $Y_2$ in Equation (V).

    $\begin{array}{c}{F}_e=\left(0.56\times 1\times 8\text{kN}\right)+1.89\left(2\text{kN}\right)\\ =4.48+3.78\\ =8.26\text{kN}\end{array}$

Substitute $1$ for $a_f$, $8.26\text{kN}$ for $F_e$, $240$ for $x_D$, $0.02$ for $x_0$, $4.439$ for $\left(\theta -x_0\right)$, $1.483$ for $b$, $0.99$ for $R$ and $3$ for $a$ in Equation (II).

    $\begin{array}{c}\left(C_{10}\right)=\left(1\right)\left(8.26\text{kN}\right){\left[\frac{240}{0.02+\left(4.439\right){\left(\ln \left(\frac{1}{0.99}\right)\right)}^{\frac{1}{1.483}}}\right]}^{\frac{1}{3}}\\ =\left(1\right)\left(8.26\text{kN}\right){\left[\frac{240}{0.2196}\right]}^{\frac{1}{3}}\\ =85.08\text{kN}\end{array}$

Since $C_{10}=85.08\text{kN}$, that is greater than $83.2\text{kN}$, so this bearing cannot be expected to carry load.

Refer to Table $11-2$ “Dimensions and Load Ratings for Single-Row $02-$ Series Deep-Groove and Angular-Contact Ball Bearings” for $02-90\text{mm}$ deep groove to obtain $C_{10}=95.26\text{kN}$ and.$C_o=62\text{kN}$.

Substitute $2\text{kN}$ for $F_a$ and $62\text{kN}$ for $C_o$ in Equation (III).

    $\begin{array}{c}\frac{F_a}{C_o}=\frac{2\text{kN}}{62\text{kN}}\\ =0.032\end{array}$

Refer to Table $11-1$ “Equivalent Radial Load Factor for Ball Bearings”, to obtain $e=0.226$ at $\frac{F_a}{C_o}=0.032$.

Substitute $2\text{kN}$ for $F_a$, $1$ for $V$ and $8\text{kN}$ for $F_r$ in Equation (IV).

    $\begin{array}{c}\frac{F_a}{\left(V{F}_r\right)}=\frac{2\text{kN}}{1\left(8\text{kN}\right)}\\ =0.25\end{array}$

Refer to Table $11-1$ “Equivalent Radial Load Factor for Ball Bearings”, at $\frac{F_a}{C_o}=0.032$ and $\frac{F_a}{C_o}>e$ to obtain $X_2=0.56$ and $Y_2=1.95$.

Substitute $2\text{kN}$ for $F_a$, $1$ for $V$, $8\text{kN}$ for $F_r$, $0.56$ for $X_2$ and $1.95$ for $Y_2$ in Equation (V).

    $\begin{array}{c}{F}_e=\left(0.56\times 1\times 8\text{kN}\right)+1.95\left(2\text{kN}\right)\\ =4.48\text{kN}+3.9\text{kN}\\ =8.38\text{kN}\end{array}$

Substitute $1$ for $a_f$, $8.38\text{kN}$ for $F_e$, $240$ for $x_D$, $0.02$ for $x_0$, $4.439$ for $\left(\theta -x_0\right)$, $1.483$ for $b$, $0.99$ for $R$ and $3$ for $a$ in Equation (II).

    $\begin{array}{c}\left(C_{10}\right)=\left(1\right)\left(8.38\text{kN}\right){\left[\frac{240}{0.02+\left(4.439\right){\left(\ln \left(\frac{1}{0.99}\right)\right)}^{\frac{1}{1.483}}}\right]}^{\frac{1}{3}}\\ =\left(1\right)\left(8.38\text{kN}\right){\left[\frac{240}{0.2196}\right]}^{\frac{1}{3}}\\ =86.314\text{kN}\end{array}$

Since $C_{10}=86.3\text{kN<95}\text{.6}\text{kN}$, so this bearing can expected to carry load.

Thus, the bore size of single-row deep-grove ball bearing is $02\text{-}90\text{mm}$.