Question:

A tractor PTO runs at a constant speed of 540 rpm to drive a rotary tiller through a flange coupling having a shear pin parallel to the shaft. The shear pin axis is located 60 mm from the PTO shaft axis. The allowable shear stress of the pin material is 200 MPa. For overload safety, the pin must fail at 150% of the rated torque to protect the gearbox. The tiller requires 40 kW under normal load at the rated PTO speed. Neglecting bending, stress concentration effects, and other losses, the pin diameter, in mm, is nearest to (take \(\pi = 3.14\))

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Work out the overload torque first, then convert it to a shear force on the pin.
Updated On: Aug 6, 2026
  • 7.5
  • 10.6
  • 6.1
  • 8.6
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The Correct Option is B

Solution and Explanation

Step 1: Find the rated torque delivered by the PTO shaft.
Angular speed \(\omega = \dfrac{2\pi N}{60} = \dfrac{2 \times 3.14 \times 540}{60} = 56.52\ \text{rad/s}\).
Rated power \(P = 40\ \text{kW} = 40000\ \text{W}\).
Rated torque \(T_{rated} = \dfrac{P}{\omega} = \dfrac{40000}{56.52} = 707.7\ \text{N.m}\).

Step 2: Apply the overload factor to get the failure torque.
The pin must shear at 150% of rated torque, so \(T_{fail} = 1.5 \times 707.7 = 1061.6\ \text{N.m}\).

Step 3: Convert the failure torque into a shear force on the pin.
The pin acts at a radius \(r = 60\ \text{mm} = 0.06\ \text{m}\) from the shaft axis.
\(F = \dfrac{T_{fail}}{r} = \dfrac{1061.6}{0.06} = 17693\ \text{N}\).

Step 4: Size the pin from the allowable shear stress.
Shear area needed \(A = \dfrac{F}{\tau} = \dfrac{17693}{200 \times 10^6} = 8.847\times10^{-5}\ \text{m}^2\).
\(A = \dfrac{\pi d^2}{4}\), so \(d = \sqrt{\dfrac{4A}{\pi}} = \sqrt{\dfrac{4 \times 8.847\times10^{-5}}{3.14}} = \sqrt{1.127\times10^{-4}} = 0.01062\ \text{m}\).

Step 5: Convert to millimetres and check the wrong options.
\(d \approx 10.6\ \text{mm}\).
Option A (7.5 mm) or C (6.1 mm) would come from using the rated torque without the 150% overload factor.

Final Answer:
The shear pin diameter should be nearest to 10.6 mm. \[ \boxed{d \approx 10.6\ \text{mm}} \]
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