Chapter 11, Problem 34P

A gear-reduction unit uses the countershaft depicted in the figure. Find the two bearing reactions. The beatings are to be angular-contact ball bearings, having a desired life of 50 kh when used at 300 rev/min. Use 1.2 for the application factor and a reliability goal for the bearing pair of 0.96, assuming distribution data from manufacturer 2 in Table 11-6. Select the bearings from Table 11-2.

Problem 11-34

Dimensions in inches.

To determine

The reaction load at point O

The reaction load at point C

The bore diameter of angular contact ball bearing at point O.

The bore diameter of angular contact ball bearing at point C.

Answer to Problem 34P

The reaction load at point O is $112\text{lbf}$.

The reaction load at point C is $297\text{lbf}$.

The bore diameter of angular contact ball bearing at point O is 17 mm.

The bore diameter of angular contact ball bearing at point C is 35 mm.

Explanation of Solution

Write the expression for net force in y direction at point A.

$F_A^y=F_1\sin \theta$ (I)

Here, net force in y direction at point A is $F_A^y$, applied load on point A is $F_1$ and the angle of applied force with horizontal direction is $\theta$.

Write the expression for net force in z direction at point A.

$F_A^z=F_1\cos \theta$ (II)

Write the expression for net force in y direction at point B.

$F_B^y=F\sin {\theta }_B$ (III)

Here, net force in y direction at point B is $F_B^y$, applied load on point B is $F$ and the angle of applied force with horizontal direction is ${\theta }_B$.

Write the expression for net force in z direction at point B.

$F_B^z=F\cos {\theta }_B$ (IV)

Here, net force in y direction at point B is $F_B^y$.

Write the expression for net torque on shaft.

$\begin{array}{l}{\displaystyle \sum T}=0\\ \left\{F_A^z\cdot r_1-F_B^z\cdot r_2\right\}=0\end{array}$ (V)

Here, radius of gear at point A is $r_1$ and radius of pulley at point B is $r_2$.

Write the expression for moment equilibrium in $z$ direction at $O$.

$\begin{array}{l}{\displaystyle \sum M_O^z}=0\\ -\left(F_A^y\cdot r_A\right)-\left(F_B^y\cdot r_B\right)+\left(R_C^y\cdot r_C\right)=0\end{array}$ (VI)

Here, the moment is $M$, the reaction force at C in y direction is $R_C^y$, the distance between A and O is $r_A$, the distance between B and O is $r_B$ and the distance between C and O is $r_C$.

Write the expression for moment equilibrium in $y$ direction at $O$.

$\begin{array}{l}{\displaystyle \sum M_O^y}=0\\ \left(F_A^z\cdot r_A\right)-\left(F_B^z\cdot r_B\right)+\left(R_C^z\cdot r_C\right)=0\end{array}$ (VII)

Here, the reaction force at C in z direction is $R_C^z$.

Write the expression for static force equilibrium in $y$ direction.

$\begin{array}{l}{\displaystyle \sum F^y}=0\\ R_O^y-F_A^y-F_B^y+R_C^y=0\end{array}$ (VIII)

Here, the force is $F$ and the reaction force at O in y direction is $R_O^y$.

Write the expression for static force equilibrium in $z$ direction.

$\begin{array}{l}{\displaystyle \sum F^z}=0\\ R_O^z-F_A^z+F_B^z+R_C^z=0\end{array}$ (IX)

Here, the force is $F$ and the reaction force at O in z direction is $R_O^z$.

Write the expression for resultant force at point O.

$R_O=\sqrt{{\left(R_O^y\right)}^2+{\left(R_O^z\right)}^2}$ (X)

Here, the resultant force at point O is $R_O$.

Write the expression for resultant force at point C.

$R_C=\sqrt{{\left(R_C^y\right)}^2+{\left(R_C^z\right)}^2}$ (XI)

Here, the resultant force at point C is $R_C$.

Write the expression for desired reliability.

$R_D=\sqrt{R_1\times R_2}$ (XII)

Here, the desired reliability is $R_D$, the reliability of component 1 is $R_1$ and the reliability of component 2 is $R_2$.

Write the expression for multiple of dating life.

$x_D=\frac{L_D\times n_D\times 60}{L_{10}}$ (XIII)

Here, the multiple of rating life is $x_D$, the desired life is $L_D$, the desired speed is $n_D$ and life hours of 90% component is $L_{10}$.

Write the expression for catalogue load rating at point O.

$C_{10}^O=a_f{R}_O{\left[\frac{x_D}{x_0+\left(\theta -x_0\right){\left\{\ln \left(\frac{1}{R_D}\right)\right\}}^{\frac{1}{b}}}\right]}^{\frac{1}{a}}$ (XIV)

Here the catalogue load rating is $C_{10}^O$, the application factor is $a_f$, the reaction load is $R_O$, minimum value of Weibull life measure is $x_0$, the characteristic parameter is $\theta$, the shape parameter for skewness is $b$ and the factor of power is $a$.

Write the expression for catalogue load rating at point C.

$C_{10}^C=a_f{R}_C{\left[\frac{x_D}{x_0+\left(\theta -x_0\right){\left\{\ln \left(\frac{1}{R_D}\right)\right\}}^{\frac{1}{b}}}\right]}^{\frac{1}{a}}$ (XV)

Here the catalogue load rating is $C_{10}^C$.

Conclusion:

Substitute $240\text{lbf}$ for $F_1$ and $20°$ for $\theta$ in Equation (I).

$\begin{array}{c}{F}_A^y=240\text{lbf}\sin 20°\\ =\left(240\times 0.342\right)\text{lbf}\\ \simeq 82.1\text{lbf}\end{array}$

Substitute $240\text{lbf}$ for $F_1$ and $20°$ for $\theta$ in Equation (II).

$\begin{array}{c}{F}_A^z=240\text{lbf}\cos 20°\\ =\left(240\times 0.9396\right)\\ \simeq 226\text{lbf}\end{array}$

Substitute $25°$ for ${\theta }_B$ in Equation (III).

$F_B^y=F\sin 25°$

Substitute $25°$ for ${\theta }_B$ in Equation (IV).

$F_B^z=F\cos 25°$

Substitute $225.52\text{lbf}$ for $F_A^z$, $12\text{in}$ for $r_1$, $6\text{in}$ for $r_2$ and $F\cos 25°$ for $F_B^z$ in Equation (V).

$\begin{array}{l}\left\{225.52\text{lbf}\cdot 12\text{in}-F\cos 25°\cdot 6\text{in}\right\}=0\\ F=\frac{225.52\text{lbf}\times 12\text{in}}{\cos 25°\times 6\text{in}}\\ F=\frac{2706.24}{5.4378}\\ F\simeq 498\text{lbf}\end{array}$

Substitute $25°$ for ${\theta }_B$ and $498\text{lbf}$ for $F$ in Equation (III).

$\begin{array}{c}{F}_B^y=498\text{lbf}\sin 25°\\ =\left(498\times 0.42262\right)\text{lbf}\\ \simeq 210\text{lbf}\end{array}$

Substitute $25°$ for ${\theta }_B$ and $498\text{lbf}$ for $F$ in Equation (IV).

$\begin{array}{c}{F}_B^z=498\text{lbf}\cos 25°\\ =\left(498\times 0.90631\right)\text{lbf}\\ \simeq 451\text{lbf}\end{array}$

Substitute $82.1\text{lbf}$ for $F_A^y$, $210\text{lbf}$ for $F_B^y$, 16 in for $r_A$, 30 in for $r_B$ and 42 in for $r_C$ in Equation (VI).

$\begin{array}{l}-\left(82.1\text{lbf}\cdot 16\text{in}\right)-\left(210\text{lbf}\cdot 30\text{in}\right)+\left(R_C^y\cdot 42\text{in}\right)=0\\ R_C^y=\frac{\left(1313.6\text{lbf}\cdot \text{in}\right)+\left(6300\text{lbf}\cdot \text{in}\right)}{42\text{in}}\\ R_C^y\simeq 181\text{lbf}\end{array}$

Substitute $82.1\text{lbf}$ for $F_A^y$, $210\text{lbf}$ for $F_B^y$ and $181\text{lbf}$ for $R_C^y$ in Equation (VIII)

$\begin{array}{l}{R}_O^y-F_A^y-F_B^y+R_C^y=0\\ R_O^y=82.11\text{lbf}+210\text{lbf}-181\text{lbf}\\ R_O^y\simeq 111.1\text{lbf}\end{array}$

Substitute $226\text{lbf}$ for $F_A^z$, $451\text{lbf}$ for $F_B^z$, 16 in for $r_A$, 30 in for $r_B$ and 42 in for $r_C$ in Equation (VII).

$\begin{array}{l}\left(226\text{lbf}\cdot 16\text{in}\right)-\left(451\text{lbf}\cdot 30\text{in}\right)-\left(R_C^z\cdot 42\text{in}\right)=0\\ R_C^z=\frac{\left(3616\text{lbf}\cdot \text{in}\right)-\left(13530\text{lbf}\cdot \text{in}\right)}{42\text{in}}\\ R_C^z\simeq -236\text{lbf}\end{array}$

Substitute $226\text{lbf}$ for $F_A^z$, $451\text{lbf}$ for $F_B^z$ and $236\text{lbf}$ for $R_C^z$ in Equation (IX)

$\begin{array}{l}{R}_O^z-F_A^z+F_B^z+R_C^z=0\\ R_O^z=226\text{lbf-451}\text{lbf+}236\text{lbf}\\ R_O^y\simeq 11\text{lbf}\end{array}$

Substitute $111.1\text{lbf}$ for $R_O^y$ and $11\text{lbf}$ for $R_O^z$ in Equation (X).

$\begin{array}{c}{R}_O=\sqrt{{\left(111.1\text{lbf}\right)}^2+{\left(11\text{lbf}\right)}^2}\\ =\sqrt{12464.21}\\ \simeq 112\text{lbf}\end{array}$

Thus the reaction load at point O is $112\text{lbf}$

Substitute $181\text{lbf}$ for $R_C^y$ and $236\text{lbf}$ for $R_C^z$ in Equation (XI).

$\begin{array}{c}{R}_C=\sqrt{{\left(181\text{lbf}\right)}^2+{\left(236\text{lbf}\right)}^2}\\ =\sqrt{88457}\\ \simeq 297\text{lbf}\end{array}$

Thus the reaction load at point C is $297\text{lbf}$

Substitute 0.96 for $\left(R_1\times R_2\right)$ in Equation (XII).

$\begin{array}{c}{R}_D=\sqrt{0.96}\\ =0.98\end{array}$

Substitute $50000\text{h}$ for $L_D$, $300\text{rev/min}$ for $n_D$ and ${10}^6\text{h}$ for $L_{10}$ in Equation (XIII).

$\begin{array}{c}{x}_D=\frac{50000\text{h}\times 300\text{rev}/\text{min}\times 60}{{10}^6\text{h}}\\ =\frac{9\times {10}^8}{{10}^6}\\ =900\end{array}$

Refer to Table 11-6 “Typical Weibull Parameters for Two Manufacturers” to obtain values of $x_O$ as 0.02, $\theta$ as 4.459 and $b$ as 1.1483.

Substitute $1.2$ for $a_f$, $112\text{lbf}$ for $R_O$, 900 for $x_D$, 0.02 for $x_0$, 4.459 for $\theta$, 1.483 for $b$, 0.98 for $R_D$ and 3 for $a$ in Equation (XIV).

$\begin{array}{c}{C}_{10}^O=1.2\times 112\text{lbf}{\left[\frac{900}{0.02+\left(4.459-0.02\right){\left\{\ln \left(\frac{1}{0.98}\right)\right\}}^{\frac{1}{1.483}}}\right]}^{\frac{1}{3}}\\ \simeq 1860\text{lbf}\left(\frac{1\text{N}}{0.2248\text{lbf}}\right)\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ \simeq 8.28\text{kN}\end{array}$

Refer to Table 11-2 “Dimensions and Load Ratings for Single-Row 02-Series Deep-Groove and Angular-Contact Ball Bearing” to obtain 17 mm bore size angular contact ball bearing corresponding to $8.28\text{kN}$ of catalogue load rating.

Thus the Bore diameter of angular contact ball bearing at point O is 17 mm.

Substitute $1.2$ for $a_f$, $297\text{lbf}$ for $R_C$, 900 for $x_D$, 0.02 for $x_0$, 4.459 for $\theta$, 1.483 for $b$, 0.98 for $R_D$ and 3 for $a$ in Equation (XV).

$\begin{array}{c}{C}_{10}^C=1.2\times 297\text{lbf}{\left[\frac{900}{0.02+\left(4.459-0.02\right){\left\{\ln \left(\frac{1}{0.98}\right)\right\}}^{\frac{1}{1.483}}}\right]}^{\frac{1}{3}}\\ \simeq 4932\text{lbf}\left(\frac{1\text{N}}{0.2248\text{lbf}}\right)\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ \simeq 21.9\text{kN}\end{array}$

Refer to Table 11-2 “Dimensions and Load Ratings for Single-Row 02-Series Deep-Groove and Angular-Contact Ball Bearing” to obtain 35 mm bore size angular contact ball bearing corresponding to $21.9\text{kN}$ of catalogue load rating.

Thus Bore diameter of angular contact ball bearing at point C is 35 mm.