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 :

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:

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:

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.

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}