Chapter 13, Problem 44P

The figure shows a 10 diametral pitch 18-tooth 20° straight bevel pinion driving a 30-tooth gear. The transmitted load is 25 lbf. Find the bearing reactions at C and D on the output shaft if D is to take both radial and thrust loads.

Problem 13–44

Dimensions in inches.

To determine

The bearing reaction at $A$.

The bearing reaction at $B$.

Answer to Problem 44P

The bearing reaction at $A$ is ${\overrightarrow{F}}_A=\left(94.4\text{lbf}\right)\widehat{i}-\left(64.2\text{lbf}\right)\widehat{j}-\left(1209.9\text{lbf}\right)\widehat{k}$.

The bearing reaction at $B$ is ${\overrightarrow{F}}_B=-\left(222.8\text{lbf}\right)\widehat{i}+\left(815.6\text{lbf}\right)\widehat{k}$.

Explanation of Solution

The figure below shows the forces acting at the bevel gear and pinion assembly.

Figure-(1)

Write the expression for the diameter of gear 2.

    $d_2=\frac{N_2}{p_d}$                                                                 (I)

Here, the diametrical pitch is $p_d$ and the number of teeth on gear 2 is $N_2$.

Write the expression for the diameter of gear 3.

    $d_3=\frac{N_3}{p_d}$                                                                 (II)

Here, the number of teeth on gear 3 is $N_3$.

Write the expression for the pitch angle for gear 2.

    $\gamma ={\tan }^{-1}\left(\frac{\left(\frac{d_2}{2}\right)}{\frac{d_3}{2}}\right)$                                                          (III)

Write the expression for the pitch angle for gear 3.

    $\Gamma =90°-\gamma$                                                                  (IV)

Write the expression fort the pitch radius at the mid-point of the bevel gear.

    $r_{av}=\frac{d_3}{2}-\frac{l_{s2}\sin \gamma }{2}$                                                   (V)

Here, the lateral side of the gear 2 is $l_{s2}$.

Write the expression for $DE$.

    $DE=DK+\frac{l_{s2}\sin \gamma }{2}$                                               (VI)

Here, the distance between the points $D$ and $K$ is $DK$.

Write the expression for the radial load.

    $W^r=W^t\tan \varphi \cos \Gamma$                                                       (VII)

Here, the pressure angle is $\varphi$.

Write the expression for the axial load.

    $W^a=W^t\tan \varphi \cdot \sin \Gamma$                                                        (VIII)

Write the expression of the load in vector form.

    $\overrightarrow{W}=-{\overrightarrow{W}}^r\widehat{i}-{\overrightarrow{W}}^a\widehat{j}+W^t\widehat{k}$                                                       (IX)

Write the expression for the position vector of $DG$.

    ${\overrightarrow{R}}_{DG}=r_{av}\left(\widehat{i}\right)+\left(DE\right)\widehat{j}$

Write the expression for the position vector of $DC$.

    ${\overrightarrow{R}}_{DC}=-DC\widehat{j}$

Write the expression for the force on the bearing $C$.

    ${\overrightarrow{F}}_C={\overrightarrow{F}}_C^x\widehat{i}+{\overrightarrow{F}}_C^z\widehat{k}$                                                               (X)

Here, the force on bearing $C$ in x direction is $F_C^x$ and the force on bearing $C$ in y direction is $F_C^y$.

Write the expression for the moment about $D$ in vector form.

    ${\overrightarrow{R}}_{DG}\times \overrightarrow{W}+{\overrightarrow{R}}_{DC}\times {\overrightarrow{F}}_C+T\widehat{j}=0$                                               (XI)

Here, the torque is $T$.

Write the expression for the force equilibrium for the set of bearings.

    ${\overrightarrow{F}}_C+{\overrightarrow{F}}_D+\overrightarrow{W}=0$                                                                  (XII)

Substitute ${\overrightarrow{F}}_C^x\widehat{i}+{\overrightarrow{F}}_C^z\widehat{k}$ for ${\overrightarrow{F}}_C$, $-DC\widehat{j}$ for ${\overrightarrow{R}}_{DC}$, $r_{av}\left(\widehat{i}\right)+\left(DE\right)\widehat{j}$ for ${\overrightarrow{R}}_{DG}$, $-{\overrightarrow{W}}^r\widehat{i}-{\overrightarrow{W}}^a\widehat{j}+W^t\widehat{k}$ for $\overrightarrow{W}$ in Equation (XII).

    $\begin{array}{l}\left\{\begin{array}{l}\left(r_{av}\left(\widehat{i}\right)+\left(DE\right)\widehat{j}\right)\times \left(-{\overrightarrow{W}}^r\widehat{i}-{\overrightarrow{W}}^a\widehat{j}+W^t\widehat{k}\right)\\ +\left(-DC\widehat{j}\right)\times \left({\overrightarrow{F}}_C^x\widehat{i}+{\overrightarrow{F}}_C^z\widehat{k}\right)+T\widehat{j}\end{array}\right\}=0\\ \left\{\begin{array}{l}\left[-r_{av}\cdot W^a\right]\widehat{k}+\left[r_{av}\cdot W^t\left(-\widehat{j}\right)\right]+DE\cdot \left(-W^r\left(-\widehat{k}\right)\right)+DE\cdot W^t\left(\widehat{i}\right)\\ +\left[\left(-DC\right)F_C^x\left(-\widehat{k}\right)+\left(-DC\right)F_C^z\widehat{i}\right]+T\widehat{j}\end{array}\right\}=0\end{array}$     (XIII)

Conclusion:

Substitute $18$ for $N_2$ and $10{\text{in}}^{-1}$ for $p_d$ in Equation (I).

    $\begin{array}{c}{d}_2=\frac{18}{10{\text{in}}^{-1}}\\ =1.8\text{in}\end{array}$

Substitute $30$ for $N_3$ and $10{\text{in}}^{-1}$ for $p_d$ in Equation (II).

    $\begin{array}{c}{d}_3=\frac{30}{10{\text{in}}^{-1}}\\ =3\text{in}\end{array}$

Substitute $3\text{in}$ for $d_3$ and $1.8\text{in}$ for $d_2$ in Equation (III).

    $\begin{array}{c}\gamma ={\tan }^{-1}\left(\frac{\left(\frac{1.8\text{in}}{2}\right)}{\frac{3\text{in}}{2}}\right)\\ ={\tan }^{-1}\left(\frac{0.9\text{in}}{1.5\text{in}}\right)\\ ={\tan }^{-1}\left(0.6\right)\\ =30.96°\end{array}$

Substitute $30.96°$ for $\gamma$ in Equation (IV).

    $\begin{array}{c}\Gamma =90°-30.96°\\ =59.04°\end{array}$

Substitute $3\text{in}$ for $d_3$, $0.5\text{in}$ for $l_{s2}$ and $59.04°$ for $\Gamma$ in Equation (V).

    $\begin{array}{c}{r}_{av}=\frac{3\text{in}}{2}-\frac{\left(0.5\text{in}\right)\sin 59.04°}{2}\\ =1.5\text{in}-0.214\text{in}\\ =1.286\text{in}\end{array}$

Substitute $\frac{9}{16}\text{in}$ for $DK$, $0.5\text{in}$ for $l_{s2}$ and $59.04°$ for $\Gamma$ in Equation (VI).

    $\begin{array}{c}DE=\frac{9}{16}\text{in}+\frac{\left(0.5\text{in}\right)\sin \left(59.04°\right)}{2}\\ =\frac{9}{16}\text{in}+0.214\text{in}\\ =0.6911\text{in}\end{array}$

Substitute $25\text{lbf}$ for $W^t$, $20°$ for $\varphi$ and $59.04°$ for $\Gamma$ in Equation (VII).

    $\begin{array}{c}{W}^r=25\text{lbf}\tan 20°\cos 59.04°\\ =9.099\cos 59.04°\text{lbf}\\ =4.681\text{lbf}\end{array}$

Substitute $25\text{lbf}$ for $W^t$, $20°$ for $\varphi$ and $59.04°$ for $\Gamma$ in Equation (VIII).

    $\begin{array}{c}{W}^a=\left(25\text{lbf}\right)\tan 20°\cdot \sin 59.04°\\ =9.099\sin 59.04°\text{lbf}\\ =7.803\text{lbf}\end{array}$

Substitute $25\text{lbf}$ for $W^t$, $7.803\text{lbf}$ for $W^a$ and $4.681\text{lbf}$ for $W^r$ in Equation (IX).

    $\overrightarrow{W}=-\left(4.681\text{lbf}\right)\widehat{i}-\left(7.803\text{lbf}\right)\widehat{j}+\left(25\text{lbf}\right)\widehat{k}$

Substitute $25\text{lbf}$ for $W^t$, $7.803\text{lbf}$ for $W^a$, $1.286\text{in}$ for $r_{av}$, $0.6911\text{in}$ for $DE$, $4.681\text{lbf}$ for $W^r$  and $\frac{5}{8}\text{in}$ for $DC$ in Equation (XIII).

    $\begin{array}{l}\left\{\begin{array}{l}\left[-\left(1.286\text{in}\right)\cdot \left(7.803\text{lbf}\right)\right]\widehat{k}+\left[\left(1.286\text{in}\right)\cdot \left(25\text{lbf}\right)\left(-\widehat{j}\right)\right]\\ +\left[\left(-\frac{5}{8}\text{in}\right)F_C^x\left(-\widehat{k}\right)+\left(-\frac{5}{8}\text{in}\right)F_C^z\widehat{i}\right]+\\ T\widehat{j}+0.6911\text{in}\cdot \left(-\left(4.681\text{lbf}\right)\left(-\widehat{k}\right)\right)\\ +0.6911\text{in}\cdot \left(25\text{lbf}\right)\left(\widehat{i}\right)\end{array}\right\}=0\\ \left\{\begin{array}{l}\left[-10.034\text{lbf}\cdot \text{in}\right]\widehat{k}+\left[32.15\text{lbf}\cdot \text{in}\left(-\widehat{j}\right)\right]+\left(3.235\text{lbf}\cdot \text{in}\right)\widehat{k}\\ +\left(17.2775\text{lbf}\cdot \text{in}\right)\widehat{i}+\left[\begin{array}{l}\left(-0.625\text{in}\right)F_C^x\left(-\widehat{k}\right)\\ +\left(-0.625\text{in}\right)F_C^z\widehat{i}\end{array}\right]+T\widehat{j}\end{array}\right\}=0\\ \left\{\begin{array}{l}\left(17.2775\text{lbf}\cdot \text{in}-0.625\text{in}\cdot F_C^z\right)\widehat{i}+\left(T-32.15\text{lbf}\cdot \text{in}\right)\widehat{j}\\ +\left(3.235\text{lbf}\cdot \text{in}-10.034\text{lbf}\cdot \text{in}+\left(0.625\text{in}\right)F_C^x\right)\widehat{k}\end{array}\right\}=0\end{array}$       (XIV)

Compare the $\widehat{i}$ coordinates of Equation (XIV).

    $\begin{array}{l}17.2775\text{lbf}\cdot \text{in}-0.625\text{in}\cdot F_C^z=0\\ F_C^z=27.65\text{lbf}\end{array}$

Compare the $\widehat{j}$ coordinates of Equation (XIV).

    $\begin{array}{l}T-32.15\text{lbf}\cdot \text{in}=0\\ T=32.15\text{lbf}\cdot \text{in}\end{array}$

Compare the $\widehat{k}$ coordinates of Equation (XIV).

    $\begin{array}{l}3.235\text{lbf}\cdot \text{in}-10.034\text{lbf}\cdot \text{in}+\left(0.625\text{in}\right)F_C^x=0\\ F_C^x=10.88\text{lbf}\end{array}$

Substitute $-222.8\text{lbf}$ for ${\overrightarrow{F}}_B^x$ and $815.6\text{lbf}$ for ${\overrightarrow{F}}_B^z$ in Equation (X).

    $\begin{array}{c}{\overrightarrow{F}}_B=\left(-222.8\text{lbf}\right)\widehat{i}+\left(815.6\text{lbf}\right)\widehat{k}\\ =-\left(222.8\text{lbf}\right)\widehat{i}+\left(815.6\text{lbf}\right)\widehat{k}\end{array}$

Thus, the bearing reaction at $B$ is ${\overrightarrow{F}}_B=-\left(222.8\text{lbf}\right)\widehat{i}+\left(815.6\text{lbf}\right)\widehat{k}$.

Substitute $-\left(222.8\text{lbf}\right)\widehat{i}+\left(815.6\text{lbf}\right)\widehat{k}$ for ${\overrightarrow{F}}_B$ and $\left(128.4\text{lbf}\right)\widehat{i}-\left(64.2\text{lbf}\right)\widehat{j}+\left(394.3\text{lbf}\right)\widehat{k}$ for $\overrightarrow{W}$ in Equation (XII).

    $\begin{array}{l}\left\{\begin{array}{l}{\overrightarrow{F}}_A+\left[-\left(222.8\text{lbf}\right)\widehat{i}+\left(815.6\text{lbf}\right)\widehat{k}\right]\\ +\left[\left(128.4\text{lbf}\right)\widehat{i}-\left(64.2\text{lbf}\right)\widehat{j}+\left(394.3\text{lbf}\right)\widehat{k}\right]\end{array}\right\}=0\\ {\overrightarrow{F}}_A=\left(222.8\text{lbf}-128.4\text{lbf}\right)\widehat{i}+\left(64.2\text{lbf}\right)\widehat{j}-\left(815.6\text{lbf}+394.3\text{lbf}\right)\widehat{k}\\ {\overrightarrow{F}}_A=\left(94.4\text{lbf}\right)\widehat{i}-\left(64.2\text{lbf}\right)\widehat{j}-\left(1209.9\text{lbf}\right)\widehat{k}\end{array}$

Thus, the bearing reaction at $A$ is ${\overrightarrow{F}}_A=\left(94.4\text{lbf}\right)\widehat{i}-\left(64.2\text{lbf}\right)\widehat{j}-\left(1209.9\text{lbf}\right)\widehat{k}$.