Chapter 18, Problem 1P

To determine

The design of the input shaft including complete specification of gear, bearings, key, retaining rings and shaft.

Answer to Problem 1P

The torque on the shaft $2$ is $60\text{lb}\cdot \text{ft}$.

The torque on the shaft $3$ is $270\text{lb}\cdot \text{ft}$.

The torque on the shaft $5$ is $1215\text{lb}\cdot \text{ft}$.

The number of tooth on gear $2$ is $16$.

The number of tooth on gear $3$ is $72$.

Explanation of Solution

Assume the following values for the suitable design.

    Power to be delivered =$20\text{hp}$.

    Input speed = $1750\text{rpm}$.

    Output speed = $82-85\text{rev}/\min$.

    Extension of the shaft =$4\text{in}$.

    Gear box size = $14\text{in}\times 14\text{in}\times 22\text{in}$.

    Gear bearing life $>12000\text{hrs}$.

    Pressure angle = $20°$.

    Modulus = $4\text{mm}$.

Here, take the value of output speed as $85\text{rev}/\text{min}$.

Write the expression for the ratio of input and output speeds.

    $e=\frac{{\omega }_5}{{\omega }_2}$                                                           (I)

Here, the ratio is $e$, the input speed of the shaft is ${\omega }_2$, the output speed of the shaft is ${\omega }_5$.

Write the expression for the identical inline condition.

    $e=\frac{N_2}{N_3}$                                                         (II)

Here, the number of teeth of second gear is $N_2$ and the number of teeth of third gear is $N_3$.

Write the expression for the ratio of speed for a reverted compound gear train.

    $\frac{N_2}{N_3}=\frac{N_4}{N_5}$                                                        (III)

Here, the number of teeth on the fourth gear is $N_4$ and the number of teeth on the fifth gear is $N_5$.

Write the expression for the ratio of speed of fourth and fifth gear.

    $\frac{N_4}{N_5}=e^{\frac{1}{2}}$                                                             (IV)

Substitute $e^{\frac{1}{2}}$ for $\frac{N_4}{N_5}$ in Equation (III).

    $\frac{N_2}{N_3}=e^{\frac{1}{2}}$                                                                 (V)

Write the expression for the number of teeth on second gear.

    ${\text{N}}_2\text{ =}\frac{\text{2k}}{{\text{(1 +2m) sin}}^2\varphi }\left(\sqrt{\left({\text{m}}^2\text{+}\left(1+2m\right){\text{sin}}^2\varphi \right)}\right)$      (VI)

Here, the module is $m$ and the pressure angle is $\varphi$.

Write the expression for checking the limit of the output speed.

    ${\omega }_5=\left(\frac{N_2}{N_3}\right)\times {\omega }_2$                                                     (VII)

Write the expression for speed of third shaft.

  ${\omega }_3=\left(\frac{N_2}{N_3}\times {\omega }_2\right)$                                                   (VIII)

Write the expression for speed of third shaft.

    ${\omega }_4=\left(\frac{N_2}{N_3}\times {\omega }_2\right)$                                                     (IX)

Write the expression for the torque at gear $2$.

    $T_2=\frac{H}{{\omega }_2}$                                                                   (X)

Here, the power to be delivered is $H$.

Write the expression for the torque at gear $3$.

    $T_3=T_2\frac{{\omega }_2}{{\omega }_3}$                                                      (XI)

Here, the torque on the gear three is $T_3$.

Write the expression for the torque at gear $5$.

    $T_5=T_2\frac{{\omega }_2}{{\omega }_5}$                                                   (XII)

Here, the torque on the gear five is $T_5$.

Conclusion:

Substitute $85\text{rev}/\min$ for ${\omega }_5$ and $1750\text{rev}/\min$ for ${\omega }_2$ in Equation (I).

    $\begin{array}{c}e=\left(\frac{85\text{rev}/\min }{1750\text{rev}/\min }\right)\\ =\frac{1}{20.59}\end{array}$

Substitute $\frac{1}{20.59}$ for $e$ in Equation (V).

    $\begin{array}{l}\sqrt{\frac{1}{20.59}}=\frac{N_2}{N_3}\\ \frac{N_2}{N_3}=\frac{1}{4.54}\end{array}$                                    (XIII)

Substitute $20°$ for $\varphi$ and $4\text{mm}$ for $m$ in Equation (VI).

    $\begin{array}{c}{\text{N}}_2\text{ =}\frac{\text{2}\left(1\right)}{\text{(1 +2}\left(\text{4}\text{mm}\right){\text{) sin}}^{2}20°}\left(\sqrt{\left({\left(4\text{mm}\right)}^2\text{+}\left(1+2\left(4\text{mm}\right)\right){\text{sin}}^{2}20°\right)}\right)\\ =15.4\\ =16\end{array}$

Substitute $16$ for $N_2$ in Equation (XIII).

    $\begin{array}{l}\frac{16}{N_3}=\frac{1}{4.54}\\ N_3=16\times 4.54\\ N_3=72.64\\ N_3=73\end{array}$

Substitute $16$ for $N_2$ and $1750\text{rev}/\min$ for ${\omega }_2$, and $72$ for $N_3$ in Equation (VII).

    $\begin{array}{c}{\omega }_5={\left(\frac{16}{72}\right)}^2\times \left(1750\text{rev}/\min \right)\\ =0.049\times \left(1750\text{rev}/\min \right)\\ =86.42\text{rev}/\min \end{array}$

Substitute $16$ for $N_2$, $1750\text{rev}/\min$ for ${\omega }_2$, and $72$ for $N_3$ in Equation (VIII).

    $\begin{array}{c}{\omega }_3=\left(\left(\frac{16}{72}\right)\times \left(1750\text{rev}/\min \right)\right)\\ =0.222\times \left(1750\text{rev}/\min \right)\\ =388.9\text{rev}/\min \end{array}$

Substitute $20\text{hp}$ for $H$ and $1750\text{rev}/\min$ for ${\omega }_2$ in equation (X).

    $\begin{array}{c}{T}_2=\frac{\left(20\text{hp}\right)}{\left(1750\text{rev}/\min \right)}\\ =\frac{\left(20\text{hp}\right)\left(\frac{550\text{lb}\cdot \text{ft}/\text{s}}{1\text{hp}}\right)}{\left(1750\text{rev}/\min \right)\left(\frac{2\pi \text{rad}}{1\text{rev}}\right)\left(\frac{1}{\left(60\text{sec}\right)}\right)}\\ =60\text{lb}\cdot \text{ft}\end{array}$

Substitute $60\text{lb}\cdot \text{ft}$ for $T_2$, $1750\text{rev}/\min$ for ${\omega }_2$, and $388.9\text{rev}/\min$ for ${\omega }_3$ in Equation (XI).

    $\begin{array}{c}{T}_3=\left(60\text{lb}\cdot \text{ft}\right)\times \left(\frac{1750\text{rev}/\min }{388.9\text{rev}/\min }\right)\\ =\left(60\text{lb}\cdot \text{ft}\right)\times 4.49\\ =269.4\text{lb}\cdot \text{ft}\end{array}$

Substitute $60\text{lb}\cdot \text{ft}$ for $T_2$, $1750\text{rev}/\min$ for ${\omega }_2$, and $388.9\text{rev}/\min$ for ${\omega }_3$ in Equation (XII).

    $\begin{array}{c}{T}_5=\left(60\text{lb}\cdot \text{ft}\right)\times \left(\frac{1750\text{rev}/\min }{86.42\text{rev}/\min }\right)\\ =\left(60\text{lb}\cdot \text{ft}\right)\times 20.24\\ =1214.4\text{lb}\cdot \text{ft}\end{array}$