Question Bank • QB_FM_ADV_013 NAT • 2 Marks

Section 2: Farm Machinery

Machine Design

Subtopic: Measurement of force, stress, torque, speed, displacement and acceleration on machine elements – shafts, couplings, keys, bearings and knuckle joints

Question Statement:

A tractor drawbar knuckle joint connects a trailer hitch to the tractor drawbar. The joint transmits a maximum tensile draft pull of $P = 36\text{ kN}$. The knuckle pin is subjected to double shear. If the allowable shear stress for the pin material is $\tau_{\text{allow}} = 60\text{ MPa}$, the minimum required diameter of the knuckle pin in $\text{mm}$ is ________ (round off to 2 decimal places).

Answer Key:

19.54

Step-by-Step Solution & Explanation

``` Knuckle Joint Double Shear Schematic: [Fork End] ===================== | Pin | <--- Shear Plane 1 --------------------- | Eye End | --------------------- | Pin | <--- Shear Plane 2 ===================== [Fork End] ``` **Step-by-step Calculation:** 1. In a knuckle joint, the pin fails across two parallel shear planes (double shear): $F_{\text{shear per plane}} = \frac{P}{2} = \frac{36000\text{ N}}{2} = 18000\text{ N}$ 2. Shear stress relationship: $\tau = \frac{P}{2 \times \frac{\pi}{4} d^2} = \frac{2 P}{\pi d^2} \le \tau_{\text{allow}}$ 3. Solving for pin diameter $d$: $d^2 = \frac{2 P}{\pi \tau_{\text{allow}}} = \frac{2 \times 36000}{\pi \times (60 \times 10^6\text{ N/m}^2)} = \frac{72000}{1.88496 \times 10^8} = 3.8197 \times 10^{-4}\text{ m}^2$ $d = \sqrt{3.8197 \times 10^{-4}} = 0.019544\text{ m} = 19.54\text{ mm}$
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