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Q. 9-2

Figure 9–5 represents the power flow through a gear drive to multiple output shafts. Gear A is mounted on an electric motor that has a rotational speed of 1500 rpm (CW). Gear A drives a simple gear train consisting of gears B, C, D, and E. Gear A transfers the input power through each gear to the shaft on which it is mounted. The power levels delivered from shafts B, C, D, and E are 3 hp, 8 hp, 3 hp, and 3 hp, respectively. All gears have a diametral pitch of 10 and the following numbers of teeth:

$N_A = 60$    $N_B = 30$      $N_C = 90$          $N_D = 30$          $N_E = 30$      $P_d = 8$

Determine the following data for the drive system:
a. The pitch diameter of each gear
b. The center distance of each gear mesh
c. The rotational speed of each output shaft
d. Neglecting efficiency losses, find the power requirement of the electric motor
e. The torque through each output shaft and the tangential force on the gear teeth

Verified Solution

a. Pitch Diameter for each gear:

$D_A=\frac{N_A}{P_d}=\frac{60}{10}$ = 6.000 in        $D_B=\frac{N_B}{P_d}=\frac{30}{10}$ = 3.000 in        $D_C=\frac{N_C}{P_d}=\frac{90}{10}$ = 9.000 in

$D_D=\frac{N_D}{P_d}=\frac{30}{10}$ = 3.000 in        $D_E=\frac{N_E}{P_d}=\frac{30}{10}$ = 3.000 in

b. The center distance for each gear mesh:

$CD_{A-B}=\frac{D_A+D_B}{2}=\frac{6.0 in + 3.0 in}{2}=$ 4.500 in                        $CD_{B-C}=\frac{D_B+D_C}{2}=\frac{3.0 in + 9.0 in}{2}=$ 6.000 in

$CD_{C-D}=\frac{D_C+D_D}{2}=\frac{9.0 in + 3.0 in}{2}=$ 6.000 in                        $CD_{C-E}=\frac{D_C+D_E}{2}=\frac{9.0 in + 3.0 in}{2}=$ 6.000 in

c. The next item we want to solve for is the rotational speed of each gear of the gear train. Let’s step through one gear mesh at a time, starting with gear $A$ that drives gear $B$ :

$VR_{A-B}=\frac{n_A}{n_B}=\frac{N_B}{N_A}= =\frac{30}{60}=\frac{1}{2}$

The equation $VR_{A-B}=\frac{n_A}{n_B}$ is used to solve for the rotational speed of gear $B$. The velocity ratio gives a speed increase from gear $A$ to gear $B$:

$n_B=\frac{n_A}{VR_{A-B}}=\frac{1500 rpm}{0.5}=3000$ rpm

Likewise, considering gear $B$ driving gear $C$, the velocity ratio is:

$VR_{B-C}=\frac{n_B}{n_C}=\frac{N_C}{N_B}= =\frac{90}{30}=3$

The rotation speed of gear $C$: $n_C=\frac{n_B}{VR_{B-C}}=\frac{3000 rpm}{3}=1000$ rpm

It is important to note, however, that using the velocity ratio equation from gear $A$ to gear $C$ would give us the same result:

$VR_{A-C}=\frac{n_A}{n_C}= \frac{N_B}{N_A}.\frac{N_C}{N_B}= \frac{30}{60}.\frac{90}{30}= 1.5$

The rotation speed of gear $C$: $n_C=\frac{n_A}{VR_{A-C}}=\frac{1500 rpm}{1.5}=1000$ rpm

The rotational speed of gear $D$ and gear $E$ will be the same since both gears have the same number of teeth and are driven by gear $C$.

$VR_{C-D}=\frac{n_C}{n_D}=\frac{N_D}{N_C}=\frac{30}{90}=\frac{1}{3}$

$n_{E}=n_{D}=\frac{n_C}{VR_{C-D}}=\frac{1000rpm}{0.333}=3000$ rpm

Again, we could have used the velocity ratio equation from gear $A$ to gear $D$ or $E$. That would have given us the same results:

$V R_{A-D}=\frac{n_{A}}{n_{D}}=\frac{N_{B}}{N_{A}} \cdot \frac{N_{C}}{N_{B}} \cdot \frac{N_{D}}{N_{C}}=\frac{30}{60} \cdot \frac{90}{30} \cdot \frac{30}{90}=\frac{1}{2}$
The rotational speed of gear $\mathrm{D}$ and gear $\mathrm{E}$  $n_{D}=n_{E}=\frac{n_{A}}{V R_{A-D}}=\frac{1500 \mathrm{rpm}}{0.5}=1000 \mathrm{rpm}$
d. The power required by the motor, assuming no losses in the system, is equal to the sum of the power required by each drive shaft. This can be written as:
$\begin{array}{c}\text { Power }_{\text {int }}=\text { Power }_{\text {out }} \\\\P_{\text {in }}=P_{B}+P_{C}+P_{D}+P_{E} \\\\P_{\text {in }}=3 \mathrm{hp}+8 \mathrm{hp}+3 \mathrm{hp}+3 \mathrm{hp}=17 \mathrm{hp}\end{array}$
e. The next part of the problem is asking us to find the torque transmitted through each output shaft and the tangential force on the gear teeth. Start with gear $E$ and work through to the drive motor gear $A$. The torque transmitted through shaft $E$ can be found using the required output power and the rotational speed of shaft.
$\begin{array}{c}P_{\text {out } E}=T_{E^{.} n_{E}} \\\\T_{E}=\frac{P_{\text {out } E}}{n_{E}}=\frac{3.0 \mathrm{hp}}{3000 \mathrm{rev} / \mathrm{min}} \cdot \frac{33000 \cdot\frac{\mathrm{lb} \cdot \mathrm{ft}}{\mathrm{min}}}{1 \mathrm{hp}} \cdot \frac{1 \mathrm{rev}}{2 \pi \mathrm{rad}} \cdot \frac{12 \mathrm{in}}{1 \mathrm{ft}}=63.0 \mathrm{lb} \cdot \mathrm{in}\end{array}$
The tangential gear force, $W_{t C/E}$, is the force gear $\mathrm{C}$ applies to gear $\mathrm{E}$ to obtain the torque required to transmit the required power through shaft $E$. The tangential force, $W_{t E/ C}$ is the equal to but opposite the reaction force gear $\mathrm{E}$ applies to gear $\mathrm{C}$. The tangential gear force can be determined:
$W_{I D/ C}=W_{t E /C}=\frac{T_{E}}{\left(\frac{D_{E}}{2}\right)}=\frac{63.0 \mathrm{lb} \cdot \mathrm{in}}{\left(\frac{3.0 \mathrm{in}}{2}\right)}=42.0 \mathrm{lb}$
Since gear $E$ has the same rotational speed, power requirement, and pitch diameter as gear $D$, the torque through shaft $E$ and the tangential gear tooth force is the same. This is illustrated in Figure 9-6.
$\begin{array}{c}T_{D}=T_{E}=63.0 \mathrm{lb} \cdot \mathrm{in} \\W_{t C/ D}=W_{t D /C}=42.0 \mathrm{lb}\end{array}$
$8 \mathrm{hp}$ is transmitted through shaft $\mathrm{C}$ at a rotational speed of $1000 \mathrm{rpm}$. The torque through shaft $\mathrm{C}$ can then be determined by rearranging this equation:
$P_{\text {out } C}=T_{C} \cdot n_{C}$

$T_{C}=\frac{P_{\text {out } C}}{n_{C}}=\frac{8.0 \mathrm{hp}}{1000 \mathrm{rev} / \mathrm{min}} \cdot \frac{33000 \frac{\mathrm{lb} \cdot \mathrm{ft}}{\mathrm{min}}}{1 \mathrm{hp}} \cdot \frac{1 \mathrm{rev}}{2 \pi \mathrm{rad}} \cdot \frac{12 \mathrm{in}}{1 \mathrm{ft}}=504.2 \mathrm{lb} \cdot \mathrm{in}$
The total torque required to drive gears$\mathrm{C}, \mathrm{D}$, and $\mathrm{E}$ :
$\begin{array}{c}T_{C \text { total }}=T_{C}+W_{t E/ C}\cdot\left(\frac{D_{C}}{2}\right)+W_{t D / C}\cdot\left(\frac{D_{C}}{2}\right) \\\\T_{C \text { total }}=504.2\mathrm{lb} \cdot \text { in }+42.0 \mathrm{lb}\cdot\left(\frac{9.0 \mathrm{in}}{2}\right)+42.0\mathrm{lb} \cdot\left(\frac{9.0 \mathrm{in}}{2}\right)=882.3 \mathrm{lb} \cdot \mathrm{in}\end{array}$
We can verify this torque, using the sum of the power required by shafts $C, D$, and $E$.
$P_{C \text { total }}=P_{C}+P_{D}+P_{E}=8 \mathrm{hp}+3 \mathrm{hp}+3 \mathrm{hp}=14 \mathrm{hp}$
The total torque required to drive gears $\mathrm{C}, \mathrm{D}$, and $\mathrm{E}$ :
$T_{C \text { total }}=\frac{P_{C \text { total }}}{n_{C}}=\frac{14 \mathrm{hp}}{1000 \mathrm{rev} / \mathrm{min}} \cdot \frac{33000 \frac{\mathrm{lb} \cdot \mathrm{ft}}{\mathrm{min}}}{1 \mathrm{hp}} \cdot \frac{1 \mathrm{rev}}{2 \pi \mathrm{rad}} \cdot \frac{12 \mathrm{in}}{1 \mathrm{ft}}=882.3 \mathrm{Ib} \cdot \mathrm{in}$
We can see this gives us the same value as was shown above. This torque is used to find the tangential force gear $\mathrm{C}$ applies to gear $\mathrm{B}$ :
$W_{t C / B}=\frac{T_{C \text { total }}}{\left(\frac{D_{C}}{2}\right)}=\frac{882.3 \mathrm{lb} \cdot \mathrm{in}}{\left(\frac{9.0 \mathrm{in}}{2}\right)}=196.0 \mathrm{lb}$
The tangential force gear $\mathrm{C}$ applies to gear $\mathrm{B}$ is equal to the reaction force gear $\mathrm{B}$ applies to gear $\mathrm{C}$.

$W_{t C/B} = W_{t B/C} = 196.0$ lb

This is shown in Figure 9–7.

Shaft $B$ transmits 3 hp at a rotational speed of $3000 \mathrm{rpm}$. The torque applied to shaft $B$ to drive the output:
$T_{B}=\frac{P_{B}}{n_{B}}=\frac{3.0 \mathrm{hp}}{3000 \mathrm{rev} / \mathrm{min}} \cdot \frac{33000 \frac{\mathrm{lb} \cdot \mathrm{ft}}{\mathrm{min}}}{1 \mathrm{hp}} \cdot \frac{1 \mathrm{rev}}{2 \pi \mathrm{rad}} \cdot \frac{12 \mathrm{in}}{1 \mathrm{ft}}=63.0 \mathrm{lb} \cdot \mathrm{in}$
The total torque required to drive gears $B, C, D$, and $E$

$T_{B \text { total }}=T_{B}+W_{t} C /B \cdot\left(\frac{D_{B}}{2}\right)=63.0 \mathrm{lb} \cdot \mathrm{in}+196.0 \mathrm{lb} \cdot\left(\frac{3.0 \mathrm{in}}{2}\right)=357.1 \mathrm{lb} \cdot \mathrm{in}$
We can verify this torque using the sum of the power required by shafts $B, C, D$, and $E$.
$P_{B \text { total }}=P_{B}+P_{C}+P_{D}+P_{E}=3 \mathrm{hp}+8 \mathrm{hp}+3 \mathrm{hp}+3 \mathrm{hp}=17 \mathrm{hp}$
The total torque required to drive gears $B, C, D$, and $E$ :
$T_{B \text { total }}=\frac{P_{B \text { total }}}{n_{C}}=\frac{17 \mathrm{hp}}{1000 \mathrm{rev} / \mathrm{min}} \cdot \frac{33000 \frac{\mathrm{Ib} \cdot \mathrm{ft}}{\mathrm{min}}}{1 \mathrm{hp}} \cdot \frac{1 \mathrm{rev}}{2 \pi \mathrm{rad}} \cdot \frac{12 \mathrm{in}}{1 \mathrm{ft}}=357.1 \mathrm{lb} \cdot \mathrm{in}$
This gives us the same result as was shown above. This torque is used to find the tangential force gear $B$ applies to gear $A$ :
$W_{t B/ A}=\frac{T_{B \text { total }}}{\left(\frac{D_{B}}{2}\right)}=\frac{357.1 \mathrm{lb} \cdot \mathrm{in}}{\left(\frac{3.0 \mathrm{in}}{2}\right)}=238.1 \mathrm{lb}$
The tangential force gear $B$ applies to gear $A$ is equal to the reaction force gear $A$ applies to gear $B$ and is shown in Figure 9-8.
$W_{t B / A}=W_{t A/ B}=238.1 \mathrm{lb}$
The required torque of the motor is:
$T_{\text {Motor }}=T_{A}=W_{t A/ B} \cdot\left(\frac{D_{A}}{2}\right)=238.1 \mathrm{lb} \cdot \mathrm{in} \cdot\left(\frac{6.0 \mathrm{in}}{2}\right)=714.3 \mathrm{lb} \cdot \mathrm{in}$
The motor torque and rotational speed can be used to verify the input power requirement.
$P_{\text {Motor }}=T_{\text {Motor }} \cdot n_{\text {Motor }}=714.3 \mathrm{lb} \cdot \mathrm{in} \cdot 1500 \mathrm{rpm} \cdot \frac{1 \mathrm{hp}}{33000 \frac{\mathrm{lb} \cdot \mathrm{ft}}{\mathrm{min}}} \cdot \frac{2 \pi \mathrm{rad}}{1 \mathrm{rev}} \cdot \frac{1 \mathrm{ft}}{12 \mathrm{in}}=17 \mathrm{hp}$