Chapter 13, Problem 36P

(a)

To determine

The pitch diameter for gear 2.

The pitch diameter for gear 3.

The pitch diameter for gear 4.

The pitch diameter for gear 5.

Answer to Problem 36P

The pitch diameter for gear 2 is $2.5\text{in}$.

The pitch diameter for gear 3 is $7.33\text{in}$.

The pitch diameter for gear 4 is $2.5\text{in}$.

The pitch diameter for gear 5 is $7.33\text{in}$.

Explanation of Solution

Write the expression for the compound reverted gear train.

$N_2+N_3=N_4+N_5$ (I)

Here, the numbers of teeth on gear 3 is $N_3$, the numbers of teeth on gear 2 is $N_2$, the numbers of teeth on gear 5 is $N_5$ and the numbers of teeth on gear 4 is $N_4$.

Write the expression for the ratio of output speed to input speed.

$A=\frac{{\omega }_o}{{\omega }_i}$ (II)

Here, the ratio of output speed to input speed is $A$, the input speed of gear is ${\omega }_i$ and output speed of gear is ${\omega }_o$.

Write the expression for the ratio of output speed to input speed in terms of number of teeth.

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

Write the expression for the smallest number of teeth on pinion without interference.

$N_P=\frac{2k}{\left(1+2m\right){\sin }^2\varphi }\left(m+\sqrt{m^2+\left(1+2m\right){\sin }^2\varphi }\right)$ (IV)

Here, the number of teeth on pinion is $N_P$, the pressure angle is $\varphi$, the gear ratio is $m$ and the depth of teeth is $k$.

Write the expression for the number of teeth on gear 3.

$\frac{N_2}{N_3}=\frac{1}{m}$ (V)

Write the expression for the number of teeth on gear 5.

$\frac{N_4}{N_5}=\frac{1}{m}$ (VI)

Substitute $\frac{{\omega }_o}{{\omega }_i}$ for $A$ in Equation (III).

$\frac{{\omega }_o}{{\omega }_i}=\frac{N_2}{N_3}\left(\frac{N_4}{N_5}\right)$ (VII)

Write the pitch diameter for gear 2.

$d_2=\frac{N_2}{P}$ (VIII)

Here, the pitch is $P$.

Write the pitch diameter for gear 4.

$d_4=\frac{N_4}{P}$ (IX)

Write the pitch diameter for gear 3.

$d_3=\frac{N_3}{P}$ (X)

Write the pitch diameter for gear 5.

$d_5=\frac{N_5}{P}$ (XI)

Conclusion:

Assume the output speed of the gear train is $300\text{rev}/\text{min}$.

Substitute $300\text{rev}/\text{min}$ for ${\omega }_o$ and $2500\text{rev}/\text{min}$ for ${\omega }_i$ in Equation (II).

$\begin{array}{c}A=\frac{300\text{rev}/\text{min}}{2500\text{rev}/\text{min}}\\ =\frac{1}{8.333}\end{array}$

Substitute $N_2$ for $N_4$, $N_3$ for $N_5$ and $\left(\frac{1}{8.333}\right)$ for $A$ in Equation (III).

$\begin{array}{l}\frac{1}{8.333}=\frac{N_2}{N_3}\left(\frac{N_2}{N_3}\right)\\ {\left(\frac{N_2}{N_3}\right)}^2=\frac{1}{8.333}\\ \left(\frac{N_2}{N_3}\right)=\frac{1}{\sqrt{8.333}}\\ \left(\frac{N_2}{N_3}\right)=\frac{1}{2.887}\end{array}$

Substitute $\left(\frac{1}{2.887}\right)$ for $\frac{N_2}{N_3}$ in Equation (V).

$\begin{array}{l}\frac{1}{m}=\frac{1}{2.887}\\ m=2.887\end{array}$

Substitute $1$ for $k$, $2.887$ for $m$ and $20°$ for $\varphi$ in Equation (IV).

$\begin{array}{c}{N}_P=\frac{2\left(1\right)}{\left(1+2\left(2.887\right)\right){\left(\sin 20°\right)}^2}\left(2.887+\sqrt{\begin{array}{l}{\left(2.887\right)}^2\\ +\left(1+2\left(2.887\right)\right){\left(\sin 20°\right)}^2\end{array}}\right)\\ =2.52\left(6.05\right)\\ =15.24\\ \approx 15\end{array}$

Substitute $15$ for $N_2$ and $2.887$ for $m$ in Equation (V).

$\begin{array}{l}\frac{15}{N_3}=\frac{1}{2.887}\\ N_3=15\times 2.887\\ =43.30\\ \approx 43\end{array}$

Substitute $15$ for $N_4$ and $2.887$ for $m$ in Equation (VI).

$\begin{array}{l}\frac{15}{N_5}=\frac{1}{2.887}\\ N_5=15\times 2.887\\ =43.30\\ \approx 43\end{array}$

Substitute $15$ for $N_2$, $15$ for $N_4$, $43$ for $N_3$, $43$ for $N_5$ and $2500\text{rev}/\text{min}$ for ${\omega }_i$ in Equation (VII).

$\begin{array}{l}\frac{{\omega }_o}{2500\text{rev}/\text{min}}=\frac{15}{43}\left(\frac{15}{43}\right)\\ \frac{{\omega }_o}{2500\text{rev}/\text{min}}=0.121\\ {\omega }_o=\left(2500\text{rev}/\text{min}\right)\left(0.121\right)\\ =302.5\text{rev}/\text{min}\end{array}$

Therefore, the speed is greater than $300\text{rev}/\text{min}$.

Assume number of teeth on gear 3 and gear 5 is equal to 44 teeth.

Substitute $15$ for $N_2$, $15$ for $N_4$, $44$ for $N_3$, $44$ for $N_5$ and $2500\text{rev}/\text{min}$ for ${\omega }_i$ in Equation (VII).

$\begin{array}{l}\frac{{\omega }_o}{2500\text{rev}/\text{min}}=\frac{15}{44}\left(\frac{15}{44}\right)\\ \frac{{\omega }_o}{2500\text{rev}/\text{min}}=0.116\\ {\omega }_o=\left(2500\text{rev}/\text{min}\right)\left(0.116\right)\\ {\omega }_o=290\text{rev}/\text{min}\end{array}$

Therefore, the assumption is correct.

Substitute $15$ for $N_2$ and $6{\text{in}}^{-1}$ for $P$ in Equation (VIII).

$\begin{array}{c}{d}_2=\frac{15}{6{\text{in}}^{-1}}\\ =2.5\text{in}\end{array}$

Thus, the pitch diameter for gear 2 is $2.5\text{in}$.

Substitute $15$ for $N_2$ and $6{\text{in}}^{-1}$ for $P$ in Equation (IX).

$\begin{array}{c}{d}_4=\frac{15}{6{\text{in}}^{-1}}\\ =2.5\text{in}\end{array}$

Thus, the pitch diameter for gear 4 is $2.5\text{in}$.

Substitute $44$ for $N_3$ and $6{\text{in}}^{-1}$ for $P$ in Equation (X).

$\begin{array}{c}{d}_3=\frac{44}{6{\text{in}}^{-1}}\\ =7.33\text{in}\end{array}$

Thus, the pitch diameter for gear 3 is $7.33\text{in}$.

Substitute $44$ for $N_5$ and $6{\text{in}}^{-1}$ for $P$ in Equation (XI).

$\begin{array}{c}{d}_5=\frac{44}{6{\text{in}}^{-1}}\\ =7.33\text{in}\end{array}$

Thus, the pitch diameter for gear 5 is $7.33\text{in}$.

(b)

To determine

The pitch line velocity for gear 2.

The pitch line velocity for gear 3.

The pitch line velocity for gear 4.

The pitch line velocity for gear 5.

Answer to Problem 36P

The pitch line velocity for gear 2 is $1636\text{ft}/\text{min}$.

The pitch line velocity for gear 3 is $1636\text{ft}/\text{min}$.

The pitch line velocity for gear 4 is $558\text{ft}/\text{min}$.

The pitch line velocity for gear 5 is $558\text{ft}/\text{min}$.

Explanation of Solution

Write the pitch line velocity for gear 2.

$V_2=\pi d_2{n}_2$ (XII)

Here, the rotational speed of gear 2 is $n_2$.

Write the pitch line velocity for gear 3.

$V_3=\pi d_3{n}_3$ (XIII)

Here, the rotational speed of gear 3 is $n_3$.

Write the pitch line velocity for gear 4.

$V_4=\pi d_4{n}_4$ (XIV)

Here, the rotational speed of gear 4 is $n_4$.

Write the pitch line velocity for gear 5.

$V_5=\pi d_5{n}_5$ (XV)

Here, the rotational speed of gear 5 is $n_5$.

Write the rotational speed of gear 4.

$n_4=\frac{N_2}{N_3}{n}_2$ (XVI)

Write the rotational speed of gear 5.

$n_5=\frac{N_4}{N_5}{n}_4$ (XVII)

Conclusion:

Substitute $2.5\text{in}$ for $d_2$ and $2500\text{rev}/\min$ for $n_2$ in Equation (XII).

$\begin{array}{c}{V}_2=\pi \left(2.5\text{in}\right)\left(2500\text{rev}/\min \right)\\ =\pi \left(2.5\text{in}\right)\left(2500\text{rev}/\min \right)\left(\frac{1\text{ft}}{12\text{in}}\right)\\ \approx 1636\text{ft}/\text{min}\end{array}$

Thus, the pitch line velocity for gear 2 is $1636\text{ft}/\text{min}$.

Substitute $2.5\text{in}$ for $d_3$ and $2500\text{rev}/\min$ for $n_3$ in Equation (XIII).

$\begin{array}{c}{V}_3=\pi \left(2.5\text{in}\right)\left(2500\text{rev}/\min \right)\\ =\pi \left(2.5\text{in}\right)\left(2500\text{rev}/\min \right)\left(\frac{1\text{ft}}{12\text{in}}\right)\\ \approx 1636\text{ft}/\text{min}\end{array}$

Thus, the pitch line velocity for gear 3 is $1636\text{ft}/\text{min}$.

Substitute $15$ for $N_2$, $44$ for $N_3$ and $2500\text{rev}/\text{min}$ for $n_2$ in Equation (XVI).

$\begin{array}{c}{n}_4=\frac{15}{44}\times 2500\text{rev}/\min \\ =852.27\text{rev}/\min \end{array}$

Substitute $15$ for $N_4$, $44$ for $N_5$ and $2500\text{rev}/\text{min}$ for $n_3$ in Equation (XVII).

$\begin{array}{c}{n}_5=\frac{15}{44}\times 2500\text{rev}/\min \\ =852.27\text{rev}/\min \end{array}$

Substitute $2.5\text{in}$ for $d_4$ and $852.27\text{rev}/\min$ for $n_4$ in Equation (XIV).

$\begin{array}{c}{V}_4=\pi \left(2.5\text{in}\right)\left(852.27\text{rev}/\min \right)\\ =\pi \left(2.5\text{in}\right)\left(852.27\text{rev}/\min \right)\left(\frac{1\text{ft}}{12\text{in}}\right)\\ =557.809\text{ft}/\text{min}\\ =558\text{ft}/\text{min}\end{array}$

Thus, the pitch line velocity for gear 4 is $558\text{ft}/\text{min}$.

Substitute $2.5\text{in}$ for $d_5$ and $852.27\text{rev}/\min$ for $n_5$ in Equation (XV).

$\begin{array}{c}{V}_5=\pi \left(2.5\text{in}\right)\left(852.27\text{rev}/\min \right)\\ =\pi \left(2.5\text{in}\right)\left(852.27\text{rev}/\min \right)\left(\frac{1\text{ft}}{12\text{in}}\right)\\ =557.809\text{ft}/\text{min}\\ =558\text{ft}/\text{min}\end{array}$

Thus, the pitch line velocity for gear 5 is $558\text{ft}/\text{min}$.

(c)

To determine

The tangential force on gear 2.

The radial force on gear 2.

The total force on gear 2.

The tangential force on gear 3.

The radial force on gear 3.

The total force on gear 3.

The tangential force on gear 4.

The radial force on gear 4.

The total force on gear 4.

The tangential force on gear 5.

The radial force on gear 5.

The total force on gear 5.

Answer to Problem 36P

The tangential force on gear 2 is $504.3\text{lbf}$.

The radial force on gear 2 is $184\text{lbf}$.

The total force on gear 2 is $537\text{lbf}$.

The tangential force on gear 3 is $504.3\text{lbf}$.

The radial force on gear 3 is $184\text{lbf}$.

The total force on gear 3 is $537\text{lbf}$.

The tangential force on gear 4 is $1478\text{lbf}$.

The radial force on gear 4 is $538\text{lbf}$.

The total force on gear 4 is $1573\text{lbf}$.

The tangential force on gear 5 is $1478\text{lbf}$.

The radial force on gear 5 is $538\text{lbf}$.

The total force on gear 5 is $1573\text{lbf}$.

Explanation of Solution

Write the tangential load on the gear 2.

$W_2^t=33000\frac{H}{V_2}$ (XVIII)

Here, the power in $\text{hp}$ is $H$.

Write the tangential load on the gear 3.

$W_3^t=33000\frac{H}{V_3}$ (XIX)

Write the tangential load on the gear 4.

$W_4^t=33000\frac{H}{V_4}$ (XX)

Write the tangential load on the gear 5.

$W_5^t=33000\frac{H}{V_5}$ (XXI)

Write the radial load on gear 2.

$W_2^r=W_2^t\tan \varphi$ . (XXII)

Write the total load on gear 2.

$W_2=\frac{W_2^t}{\cos \varphi }$ (XXIII)

Write the radial load on gear 3.

$W_3^r=W_3^t\tan \varphi$ . (XXIV)

Write the total load on gear 3.

$W_3=\frac{W_3^t}{\cos \varphi }$ (XXV)

Write the radial load on gear 4.

$W_4^r=W_4^t\tan \varphi$ . (XXVI)

Write the total load on gear 4.

$W_4=\frac{W_4^t}{\cos \varphi }$ (XXVII)

Write the radial load on gear 5.

$W_5^r=W_5^t\tan \varphi$ . (XXVIII)

Write the total load on gear 5.

$W_5=\frac{W_5^t}{\cos \varphi }$ (XXIX)

Conclusion:

Substitute $25\text{hp}$ for $H$ and $1636\text{ft}/\text{min}$ for $V_2$ in Equation (XVIII).

$\begin{array}{c}{W}_2^t=33000\frac{25\text{hp}}{1636\text{ft}/\text{min}}\\ =\frac{825000}{1636}\text{lbf}\\ =504.3\text{lbf}\end{array}$

Thus, the tangential force on gear 2 is $504.3\text{lbf}$.

Substitute $25\text{hp}$ for $H$ and $1636\text{ft}/\text{min}$ for $V_2$ in Equation (XIX).

$\begin{array}{c}{W}_3^t=33000\frac{25\text{hp}}{1636\text{ft}/\text{min}}\\ =\frac{825000}{1636}\text{lbf}\\ =504.3\text{lbf}\end{array}$

Thus, the tangential force on gear 3 is $504.3\text{lbf}$.

Substitute $25\text{hp}$ for $H$ and $558\text{ft}/\text{min}$ for $V_2$ in Equation (XX).

$\begin{array}{c}{W}_4^t=33000\frac{25\text{hp}}{558\text{ft}/\text{min}}\\ =\frac{825000}{558}\text{lbf}\\ \approx 1478\text{lbf}\end{array}$

Thus, the tangential force on gear 4 is $1478\text{lbf}$.

Substitute $25\text{hp}$ for $H$ and $558\text{ft}/\text{min}$ for $V_2$ in Equation (XXI).

$\begin{array}{c}{W}_5^t=33000\frac{25\text{hp}}{558\text{ft}/\text{min}}\\ =\frac{825000}{558}\text{lbf}\\ \approx 1478\text{lbf}\end{array}$

Thus, the tangential force on gear 5 is $1478\text{lbf}$.

Substitute $504.3\text{lbf}$ for $W_2^t$ and $20°$ for $\varphi$ in Equation (XXII).

$\begin{array}{c}{W}_2^r=\left(504.3\text{lbf}\right)\tan 20°\\ =504.3\text{lbf}\times \text{0}\text{.3639}\\ \text{=183}\text{.55}\text{lbf}\\ \approx 184\text{lbf}\end{array}$

Thus, the radial force on gear 2 is $184\text{lbf}$.

Substitute $504.3\text{lbf}$ for $W_2^t$ and $20°$ for $\varphi$ in Equation (XXIII).

$\begin{array}{c}{W}_2=\frac{504.3\text{lbf}}{\cos 20°}\\ =\frac{504.3\text{lbf}}{0.94}\\ =536.67\text{lbf}\\ \approx 537\text{lbf}\end{array}$

Thus, the total force on gear 2 is $537\text{lbf}$.

Substitute $504.3\text{lbf}$ for $W_3^t$ and $20°$ for $\varphi$ in Equation (XXIV).

$\begin{array}{c}{W}_3^r=\left(504.3\text{lbf}\right)\tan 20°\\ =504.3\text{lbf}\times \text{0}\text{.3639}\\ \text{=183}\text{.55}\text{lbf}\\ \approx 184\text{lbf}\end{array}$

Thus, the radial force on gear 3 is $184\text{lbf}$.

Substitute $504.3\text{lbf}$ for $W_3^t$ and $20°$ for $\varphi$ in Equation (XXV).

$\begin{array}{c}{W}_3=\frac{504.3\text{lbf}}{\cos 20°}\\ =\frac{504.3\text{lbf}}{0.94}\\ =536.67\text{lbf}\\ \approx 537\text{lbf}\end{array}$

Thus, the total force on gear 3 is $537\text{lbf}$.

Substitute $1478\text{lbf}$ for $W_4^t$ and $20°$ for $\varphi$ in Equation (XXVI).

$\begin{array}{c}{W}_4^r=\left(1478\text{lbf}\right)\tan 20°\\ =1478\text{lbf}\times \text{0}\text{.3639}\\ \text{=537}\text{.84}\text{lbf}\\ \approx 538\text{lbf}\end{array}$

Thus, the radial force on gear 4 is $538\text{lbf}$.

Substitute $1478\text{lbf}$ for $W_4^t$ and $20°$ for $\varphi$ in Equation (XXVII).

$\begin{array}{c}{W}_4=\frac{1478\text{lbf}}{\cos 20°}\\ =\frac{1478\text{lbf}}{0.94}\\ =1572.34\text{lbf}\\ \approx 1573\text{lbf}\end{array}$

Thus, the total force on gear 4 is $1573\text{lbf}$.

Substitute $1478\text{lbf}$ for $W_5^t$ and $20°$ for $\varphi$ in Equation (XXVIII).

$\begin{array}{c}{W}_5^r=\left(1478\text{lbf}\right)\tan 20°\\ =1478\text{lbf}\times \text{0}\text{.3639}\\ \text{=537}\text{.84}\text{lbf}\\ \approx 538\text{lbf}\end{array}$

Thus, the radial force on gear 5 is $538\text{lbf}$.

Substitute $1478\text{lbf}$ for $W_5^t$ and $20°$ for $\varphi$ in Equation (XXIX).

$\begin{array}{c}{W}_5=\frac{1478\text{lbf}}{\cos 20°}\\ =\frac{1478\text{lbf}}{0.94}\\ =1572.34\text{lbf}\\ \approx 1573\text{lbf}\end{array}$

Thus, the total force on gear 5 is $1573\text{lbf}$.

(d)

To determine

The input torque.

Answer to Problem 36P

The input torque is $630.375\text{lbf}\cdot \text{in}$.

Explanation of Solution

Write the input torque.

$T_i=W_2^t\left(\frac{d_2}{2}\right)$ (XXX)

Conclusion:

Substitute $504.3\text{lbf}$ for $W_2^t$ and $2.5\text{in}$ for $d_2$ in Equation (XXX).

$\begin{array}{c}{T}_i=\left(504.3\text{lbf}\right)\left(\frac{2.5\text{in}}{2}\right)\\ =\left(504.3\text{lbf}\right)\left(1.25\text{in}\right)\\ =630.375\text{lbf}\cdot \text{in}\end{array}$

Thus, the input torque is $630.375\text{lbf}\cdot \text{in}$.

(e)

To determine

The output torque.

Answer to Problem 36P

The output torque is $5422.8\text{lbf}\cdot \text{in}$.

Explanation of Solution

Write the output torque.

$T_O=T_i\left(\frac{N_5}{N_2}\right)$ (XXXI)

Conclusion:

Substitute $630.375\text{lbf}\cdot \text{in}$ for $T_i$, $44$ for $N_5$ and $15$ for $N_2$ in Equation (XXXI).

$\begin{array}{c}{T}_O=\left(630.375\text{lbf}\cdot \text{in}\right){\left(\frac{44}{15}\right)}^2\\ =\left(630.375\text{lbf}\cdot \text{in}\right){\left(2.933\right)}^2\\ =\left(630.375\text{lbf}\cdot \text{in}\right)\left(8.602\right)\\ =5422.79\text{lbf}\cdot \text{in}\\ =5422.8\text{lbf}\cdot \text{in}\end{array}$

Thus, the output torque is $5422.8\text{lbf}\cdot \text{in}$.