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 pitch-line (tangential) velocity of the gear.
At 540 rpm on a 0.18 m pitch circle:
$v = \dfrac{\pi D N}{60} = \dfrac{3.14 \times 0.18 \times 540}{60} = 5.087$ m/s.

Step 2: Get the tangential force straight from power and speed.
Power is force times velocity along the direction of motion, so
$F_t = \dfrac{P}{v} = \dfrac{3500}{5.087} = 688.0$ N.

Step 3: Bring in the normal force along the line of action.
For involute teeth in single contact with friction ignored, the normal force $F_n$ makes an angle $\phi$ (the pressure angle) with the direction of motion, so its component along the motion direction is the tangential force: $F_t = F_n\cos\phi$.

Step 4: Solve for the pressure angle.
$\cos\phi = F_t/F_n = 688.0/760 = 0.9053$.
$\phi = \cos^{-1}(0.9053)$.
Since $\cos 25^{\circ} = 0.9063$ and $\cos 26^{\circ} = 0.8988$, interpolating for $0.9053$ gives $\phi \approx 25.13^{\circ}$.

Final Answer:
The gear pair operates at a pressure angle of \[ \boxed{25.13^{\circ}} \]