Question

A tractor PTO shaft, through a spur-gear reduction, drives a small grain auger, the PTO running at the standard 540 rpm. The driving gear on the PTO shaft has a 0.18 m pitch circle diameter.
The auger requires 3.5 kW, transmitted through this gear. The normal force along the line of action between the meshing teeth is 760 N. Assume involute spur gears, single tooth contact, and neglect friction.
The pressure angle (in degrees) is ________. (Rounded off to two decimal places)
(Take \(\pi = 3.14\))

Hint:

Find the tangential force on the gear from the power and pitch-line speed, then use \(\cos(\phi) = F_t/F_n\).

Answer 1

Correct Answer: 25.13

Step 1: Find the angular speed of the driving gear.
The PTO shaft (and the gear on it) turns at 540 rpm, so
\[ \omega = \frac{2\pi N}{60} = \frac{2 \times 3.14 \times 540}{60} = 56.52\ rad/s \]

Step 2: Find the torque delivered through the gear.
\[ T = \frac{P}{\omega} = \frac{3500}{56.52} = 61.92\ N.m \]

Step 3: Find the tangential (driving) force at the pitch circle.
Pitch radius \(r = 0.18/2 = 0.09\) m.
\[ F_t = \frac{T}{r} = \frac{61.92}{0.09} = 688.06\ N \]

Step 4: Relate the tangential force to the normal force through the pressure angle.
For an involute gear tooth, the normal (line-of-action) force and the tangential force are related by \(F_t = F_n \cos\phi\), where \(\phi\) is the pressure angle.
\[ \cos\phi = \frac{F_t}{F_n} = \frac{688.06}{760} = 0.9053 \] \[ \phi = \cos^{-1}(0.9053) = 25.13^{\circ} \]

Final Answer:
The pressure angle of the spur gear is \[ \boxed{25.13^{\circ}} \]