Chapter 13, Problem 43P

The figure shows a 16T 20° straight bevel pinion driving a 32T gear, and the location of the bearing centerlines. Pinion shaft a receives 2.5 hp at 240 rev/min. Determine the bearing reactions at A and B if A is to take both radial and thrust loads.

Problem 13–43

Dimensions in inches.

To determine

The bearing reaction at $A$.

The bearing reaction at $B$.

Answer to Problem 43P

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 input torque.

    $T_{in}=\frac{63025H}{n}$                                                          (I)

Here, the power is $H$ and the speed of pinion is $n$.

Write the expression for the pitch angle for gear 2.

    $\gamma ={\tan }^{-1}\left(\frac{l_{3y}}{l_{2x}}\right)$                                                          (II)

Here, the distance between the gear 3 and y axis is $l_{3y}$ and distance between the gear 2 and x axis is $l_{2x}$.

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

    $r_{av}=BK-\frac{l_{s3}\sin \gamma }{2}$                                                   (III)

Here, the lateral side of the gear 3 is $l_{s3}$, the pitch angle for gear 2 is $\gamma$ and the distance between the points $B$ and $K$ is $BK$.

Write the expression for the tangential load.

    $W^t=\frac{T_{in}}{r_{av}}$                                                                  (IV)

Write the expression for the pitch angle for gear 3.

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

Write the expression for the radial load.

    $W^r=W^t\tan \varphi \cos \gamma$                                                      (VI)

Here, the pressure angle is $\varphi$.

Write the expression for the axial load.

    $W^a=W^t\tan \varphi \cdot \sin \gamma$                                                         (VII)

Write the expression of the load in vector form.

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

Calculate the value of $a$.

    $a=BK+\frac{l_{s3}\cos \gamma }{2}$                                                                (IX)

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

    ${\overrightarrow{R}}_{AG}=r_{av}\left(-\widehat{i}\right)+\left(a+AB\right)\widehat{j}$

Here, the distance between the points $A$ and $B$  is $AB$.

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

    ${\overrightarrow{R}}_{AB}=AB\widehat{j}$

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

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

Here, the force on bearing $B$ in x direction is $F_B^x$ and the force on bearing $B$ in y direction is $F_B^y$.

Write the expression for the moment about gear 4 in vector form.

    ${\overrightarrow{R}}_{AG}\times \overrightarrow{W}+{\overrightarrow{R}}_{AB}\times {\overrightarrow{F}}_B+T\widehat{j}=0$                                               (XI)

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

    ${\overrightarrow{F}}_A+{\overrightarrow{F}}_B+\overrightarrow{W}=0$                                                                  (XII)

Substitute ${\overrightarrow{F}}_B^x\widehat{i}+{\overrightarrow{F}}_B^z\widehat{k}$ for ${\overrightarrow{F}}_B$, $AB\widehat{j}$ for ${\overrightarrow{R}}_{AB}$, $r_{av}\left(-\widehat{i}\right)+\left(a+AB\right)\widehat{j}$ for ${\overrightarrow{R}}_{AG}$, ${\overrightarrow{W}}^r\widehat{i}-{\overrightarrow{W}}^a\widehat{j}+W^t\widehat{k}$ for $\overrightarrow{W}$ in Equation (XI).

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

Conclusion:

Substitute $2.5\text{hp}$ for $H$ and $240\text{rev}/\min$ for $n$ in Equation (I).

    $\begin{array}{c}{T}_{in}=\frac{63025\left(2.5\text{hp}\right)}{240\text{rev}/\min }\\ =\frac{157562.5}{240}\\ =656.5\text{lbf}\cdot \text{in}\end{array}$

Substitute $4\text{in}$ for $l_{3y}$ and $2\text{in}$ for $l_{2x}$ in Equation (II).

    $\begin{array}{c}\gamma ={\tan }^{-1}\left(\frac{4\text{in}}{2\text{in}}\right)\\ ={\tan }^{-1}\left(2\right)\\ =26.565°\end{array}$

Substitute $2\text{in}$ for $BK$, $1.5\text{in}$ for $l_{s3}$ and $26.565°$ for $\gamma$ in Equation (III).

    $\begin{array}{c}{r}_{av}=2\text{in}-\frac{\left(1.5\text{in}\right)\sin \left(26.565°\right)}{2}\\ =2\text{in}-\frac{0.6708\text{in}}{2}\\ =1.665\text{in}\end{array}$

Substitute $656.5\text{lbf}\cdot \text{in}$ for $T_{in}$ and $1.665\text{in}$ for $r_{av}$ in Equation (IV).

    $\begin{array}{c}{W}^t=\frac{656.5\text{lbf}\cdot \text{in}}{1.665\text{in}}\\ =394.29\text{lbf}\\ \approx 394.3\text{lbf}\end{array}$

Substitute $26.565°$ for $\gamma$ in Equation (V).

    $\begin{array}{c}\Gamma =90°-26.565°\\ =63.435°\end{array}$

Substitute $394.3\text{lbf}$ for $W^t$, $20°$ for $\varphi$ and $26.565°$ for $\gamma$ in Equation (VI).

    $\begin{array}{c}{W}^r=\left(394.3\text{lbf}\right)\tan \left(20°\right)\cos \left(26.565°\right)\\ =\left(394.3\text{lbf}\right)\times 0.32555\\ \approx 128.4\text{lbf}\end{array}$

Substitute $394.3\text{lbf}$ for $W^t$, $20°$ for $\varphi$ and $26.565°$ for $\gamma$ in Equation (VII).

    $\begin{array}{c}{W}^a=\left(394.3\text{lbf}\right)\tan \left(20°\right)\cdot \sin \left(26.565°\right)\\ =\left(394.3\text{lbf}\right)\times 0.16277\\ \approx 64.2\text{lbf}\end{array}$

Substitute $64.2\text{lbf}$ for $W^a$, $128.4\text{lbf}$ for $W^r$ and $394.3\text{lbf}$ for $W^t$ in Equation (VII).

    $\overrightarrow{W}=\left(128.4\text{lbf}\right)\widehat{i}-\left(64.2\text{lbf}\right)\widehat{j}+\left(394.3\text{lbf}\right)\widehat{k}$

Substitute $2\text{in}$ for $BK$, $1.5\text{in}$ for $l_{s3}$ and $26.565°$ for $\gamma$ in Equation (IX).

    $\begin{array}{c}a=2\text{in}+\frac{\left(1.5\text{in}\right)\cos \left(26.565°\right)}{2}\\ =2\text{in}+\frac{0.6708\text{in}}{2}\\ \approx 2.671\text{in}\end{array}$

Substitute $2.671\text{in}$ for $a$, $2.5\text{in}$ for $AB$, $1.665\text{in}$ for $r_{av}$, $394.3\text{lbf}$ for $W^t$, $64.2\text{lbf}$ for $W^a$ and $128.4\text{lbf}$ for $W^r$ in Equation (XIII).

    $\begin{array}{l}\left\{\begin{array}{l}\left[\left(2.671\text{in}+2.5\text{in}\right)\left(394.3\text{lbf}\right)+\left(2.5\text{in}\right)\cdot {\overrightarrow{F}}_B^z\right]\widehat{i}\\ +\left[\left(1.665\text{in}\right)\cdot 394.3\text{lbf}+T\right]\widehat{j}\\ +\left[\begin{array}{l}\left(1.665\text{in}\right)\cdot \left(64.2\text{lbf}\right)-\left(2.5\text{in}\right)\cdot {\overrightarrow{F}}_B^x\\ -\left(2.671\text{in}+2.5\text{in}\right)\left(128.4\text{lbf}\right)\end{array}\right]\widehat{k}\end{array}\right\}=0\\ \left\{\begin{array}{l}\left[2038.9253\text{lbf}\cdot \text{in}+\left(2.5\text{in}\right)\cdot {\overrightarrow{F}}_B^z\right]\widehat{i}\\ +\left[656.5095\text{lbf}\cdot \text{in}+T\right]\widehat{j}\\ +\left[\begin{array}{l}106.893\text{lbf}\cdot \text{in}-663.95564\text{lbf}\cdot \text{in}\\ -\left(2.5\text{in}\right)\cdot {\overrightarrow{F}}_B^x\end{array}\right]\widehat{k}\end{array}\right\}=0\\ \left\{\begin{array}{l}\left[2038.9253\text{lbf}\cdot \text{in}+\left(2.5\text{in}\right)\cdot {\overrightarrow{F}}_B^z\right]\widehat{i}\\ +\left[6656.5095\text{lbf}\cdot \text{in}+T\right]\widehat{j}\\ +\left[\begin{array}{l}-557.06264\text{lbf}\cdot \text{in}\\ -\left(2.5\text{in}\right)\cdot {\overrightarrow{F}}_B^x\end{array}\right]\widehat{k}\end{array}\right\}=0\end{array}$            (XIV)

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

    $\begin{array}{l}2038.9253\text{lbf}\cdot \text{in}+\left(2.5\text{in}\right)\cdot {\overrightarrow{F}}_B^z=0\\ {\overrightarrow{F}}_B^z=\frac{-2038.9253\text{lbf}\cdot \text{in}}{2.5\text{in}}\\ {\overrightarrow{F}}_B^z\approx 815.6\text{lbf}\end{array}$

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

    $\begin{array}{l}656.5095\text{lbf}\cdot \text{in}+T=0\\ T=-656.5095\text{lbf}\cdot \text{in}\\ T\approx -656.5\text{lbf}\cdot \text{in}\end{array}$

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

    $\begin{array}{l}-557.06264\text{lbf}\cdot \text{in}-\left(2.5\text{in}\right)\cdot {\overrightarrow{F}}_B^x=0\\ {\overrightarrow{F}}_B^x=\frac{-557.06264\text{lbf}\cdot \text{in}}{2.5\text{in}}\\ {\overrightarrow{F}}_B^x\approx -222.8\text{lbf}\end{array}$

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 ${\overrightarrow{F}}_A=\left(94.4\text{lbf}\right)\widehat{i}-\left(64.2\text{lbf}\right)\widehat{j}-\left(1209.9\text{lbf}\right)\widehat{k}$.