The problem describes a gear drive system with multiple output shafts (B, C, D, and E) driven by gear A, which is connected to an electric motor. We are given the rotational speed of gear A (1500 rpm), the power delivered by each output shaft (3 hp, 8 hp, 3 hp, 3 hp), the number of teeth on each gear ($N_A = 60, N_B = 30, N_C = 90, N_D = 30, N_E = 30$), and the diametral pitch ($P_d = 10$). We need to determine: a. The pitch diameter of each gear. b. The center distance of each gear mesh. c. The rotational speed of each output shaft. d. The power requirement of the electric motor (neglecting efficiency losses). e. The torque through output shafts E and D and the tangential force on the gear teeth of gear C.

1. Problem Description

The problem describes a gear drive system with multiple output shafts (B, C, D, and E) driven by gear A, which is connected to an electric motor. We are given the rotational speed of gear A (1500 rpm), the power delivered by each output shaft (3 hp, 8 hp, 3 hp, 3 hp), the number of teeth on each gear ($N_A = 60, N_B = 30, N_C = 90, N_D = 30, N_E = 30$), and the diametral pitch ($P_d = 10$). We need to determine:

a. The pitch diameter of each gear.

b. The center distance of each gear mesh.

c. The rotational speed of each output shaft.

d. The power requirement of the electric motor (neglecting efficiency losses).

e. The torque through output shafts E and D and the tangential force on the gear teeth of gear C.

2. Solution Steps

a. Pitch diameter of each gear:

The pitch diameter ($D$) is related to the number of teeth ($N$) and the diametral pitch ($P_d$) by the formula:

$D = \frac{N}{P_d}$

$D_A = \frac{N_A}{P_d} = \frac{60}{10} = 6$ inches

$D_B = \frac{N_B}{P_d} = \frac{30}{10} = 3$ inches

$D_C = \frac{N_C}{P_d} = \frac{90}{10} = 9$ inches

$D_D = \frac{N_D}{P_d} = \frac{30}{10} = 3$ inches

$D_E = \frac{N_E}{P_d} = \frac{30}{10} = 3$ inches

b. Center distance of each gear mesh:

The center distance ($C$) between two gears is half the sum of their pitch diameters.

$C = \frac{D_1 + D_2}{2}$

Center distance between A and B: $C_{AB} = \frac{D_A + D_B}{2} = \frac{6+3}{2} = 4.5$ inches

Center distance between A and C: $C_{AC} = \frac{D_A + D_C}{2} = \frac{6+9}{2} = 7.5$ inches

Center distance between A and D: $C_{AD} = \frac{D_A + D_D}{2} = \frac{6+3}{2} = 4.5$ inches

Center distance between A and E: $C_{AE} = \frac{D_A + D_E}{2} = \frac{6+3}{2} = 4.5$ inches

c. Rotational speed of each output shaft:

The rotational speed ($n$) of a gear is inversely proportional to the number of teeth (or pitch diameter).

$\frac{n_A}{n_B} = \frac{N_B}{N_A}$

$n_B = n_A \cdot \frac{N_A}{N_B} = 1500 \cdot \frac{60}{30} = 3000$ rpm

$n_C = n_A \cdot \frac{N_A}{N_C} = 1500 \cdot \frac{60}{90} = 1000$ rpm

$n_D = n_A \cdot \frac{N_A}{N_D} = 1500 \cdot \frac{60}{30} = 3000$ rpm

$n_E = n_A \cdot \frac{N_A}{N_E} = 1500 \cdot \frac{60}{30} = 3000$ rpm

d. Power requirement of the electric motor:

The total power delivered to the output shafts is the sum of the individual powers. Since we are neglecting losses, the power required from the motor is equal to the total output power.

$P_{motor} = P_B + P_C + P_D + P_E = 3 + 8 + 3 + 3 = 17$ hp

e. Torque through output shafts E and D and the tangential force on the gear teeth of gear C:

Torque is related to power and speed by the formula:

$P = T \cdot \omega$, where $P$ is power, $T$ is torque, and $\omega$ is angular speed.

Also, $\omega = 2\pi n$, where $n$ is the rotational speed in revolutions per second. However, we usually use the formula directly with rpm if the power is in horsepower.

$T = \frac{63025 P}{n}$, where T is in inch-pounds, P is in horsepower, and n is in rpm.

$T_E = \frac{63025 \cdot P_E}{n_E} = \frac{63025 \cdot 3}{3000} = 63.025$ in-lb

$T_D = \frac{63025 \cdot P_D}{n_D} = \frac{63025 \cdot 3}{3000} = 63.025$ in-lb

Tangential force on the gear teeth of gear C:

The torque is also related to the tangential force ($F_t$) and the radius ($r$) of the gear by the formula:

$T = F_t \cdot r$, where $r = D/2$. Therefore, $F_t = \frac{T}{r} = \frac{2T}{D}$

However, it is important to use the torque ON gear C. That torque is equal to the torque APPLIED by gear A. The torque ON gear A is equal to the power supplied to all 4 output gears.

$T_A = \frac{63025 \cdot P_{motor}}{n_A} = \frac{63025 \cdot 17}{1500} = 714.28$ in-lb

$F_t = \frac{T_A}{r_A} = \frac{T_A}{(D_A / 2)} = \frac{2 \cdot 714.28}{6} = 238.09$ lb

3. Final Answer

a. Pitch diameter of each gear: $D_A = 6$ inches, $D_B = 3$ inches, $D_C = 9$ inches, $D_D = 3$ inches, $D_E = 3$ inches

b. Center distance of each gear mesh: $C_{AB} = 4.5$ inches, $C_{AC} = 7.5$ inches, $C_{AD} = 4.5$ inches, $C_{AE} = 4.5$ inches

c. Rotational speed of each output shaft: $n_B = 3000$ rpm, $n_C = 1000$ rpm, $n_D = 3000$ rpm, $n_E = 3000$ rpm

d. Power requirement of the electric motor: $17$ hp

e. Torque through output shafts E and D: $T_E = 63.025$ in-lb, $T_D = 63.025$ in-lb, Tangential force on the gear teeth of gear C: $238.09$ lb