A bevel pinion rotates at 600 rev/min and transmits 5 hp to a gear. The problem requires determining the bearing forces on the gearshaft, given the mounting distances, bearing locations, average pitch radii of the pinion and gear, and the fact that bearings A and C take the thrust loads. The number of teeth for pinion is 15 and for the gear is 45 with $P=5 \text{ teeth/in}$.

1. Problem Description

A bevel pinion rotates at 600 rev/min and transmits 5 hp to a gear. The problem requires determining the bearing forces on the gearshaft, given the mounting distances, bearing locations, average pitch radii of the pinion and gear, and the fact that bearings A and C take the thrust loads. The number of teeth for pinion is 15 and for the gear is 45 with $P=5 \text{ teeth/in}$.

2. Solution Steps

First, we calculate the pitch angles $\gamma$ and $\Gamma$.

$$\gamma = \arctan \frac{N_p}{N_g}$$

$$\Gamma = 90^{\circ} - \gamma$$

where $N_p$ is the number of teeth on the pinion and $N_g$ is the number of teeth on the gear.

$$\gamma = \arctan \frac{15}{45} = \arctan \frac{1}{3} = 18.435^{\circ}$$

$$\Gamma = 90^{\circ} - 18.435^{\circ} = 71.565^{\circ}$$

Next, we calculate the pitch radii of the pinion and gear.

$$r_p = \frac{N_p}{2P} = \frac{15}{2(5)} = 1.5 \text{ in}$$

$$r_g = \frac{N_g}{2P} = \frac{45}{2(5)} = 4.5 \text{ in}$$

The transmitted load $W_t$ can be found using the power and speed:

$$W_t = \frac{33000 H}{V} = \frac{33000 H}{\pi d n / 12} = \frac{63025 H}{n r}$$

where $H$ is the power in hp, $n$ is the speed in rev/min, $r$ is the pitch radius in inches, and $d$ is the pitch diameter in inches. Note that $d=2r$. We can use either pinion or gear to solve for $W_t$.

$$W_t = \frac{63025(5)}{600(1.5)} = 350.14 \text{ lb}$$

This $W_t$ acts tangent to the pitch circle and is the same for both pinion and gear.

Next, we can calculate the radial and axial forces on the gear:

$$W_r = W_t \tan \phi \cos \Gamma$$

$$W_a = W_t \tan \phi \sin \Gamma$$

where $\phi$ is the pressure angle. The pressure angle is not given, but we can assume that it is $20^{\circ}$.

$$W_r = 350.14 \tan 20^{\circ} \cos 71.565^{\circ} = 350.14 (0.364) (0.316) = 40.28 \text{ lb}$$

$$W_a = 350.14 \tan 20^{\circ} \sin 71.565^{\circ} = 350.14 (0.364) (0.949) = 120.97 \text{ lb}$$

$W_t$ acts on the gear in the upward direction.

$W_r$ acts on the gear in the left direction.

$W_a$ acts on the gear in the right direction (thrust).

Summing forces in the $x$, $y$, and $z$ direction.

$\sum F_x = 0$: $A_x + C_x - W_a = 0$.

$\sum F_y = 0$: $D_y - A_y - W_t = 0$.

$\sum F_z = 0$: $A_z - D_z - W_r = 0$.

Summing moments.

About $D_y$: $6.625A_y - 3.875W_t = 0$

About $D_z$: $6.625A_z - 3.875W_r = 0$

Therefore, $A_y = \frac{3.875}{6.625} W_t = \frac{3.875}{6.625} 350.14 = 205.00 \text{ lb}$

$A_z = \frac{3.875}{6.625} W_r = \frac{3.875}{6.625} 40.28 = 23.46 \text{ lb}$

$D_y = A_y + W_t = 205.00 + 350.14 = 555.14 \text{ lb}$

$D_z = A_z - W_r = 23.46 - 40.28 = -16.82 \text{ lb}$

Sum moments about the $x$ axis. $A_x + C_x = 120.97 \text{ lb}$. We know bearing A and C should take the thrust loads, so we can simply assume that the thrust load at A is close to 0 and at C close to $120.97 \text{ lb}$.

3. Final Answer

$A_x = 0 \text{ lb}$

$A_y = 205.00 \text{ lb}$

$A_z = 23.46 \text{ lb}$

$C_x = 120.97 \text{ lb}$

$D_y = 555.14 \text{ lb}$

$D_z = -16.82 \text{ lb}$