Question Bank • QB_FM_ADV_027 NAT • 2 Marks

Section 2: Farm Machinery

Farm Machinery

Subtopic: Forces acting on a tillage tool

Question Statement:

A simple vertical rectangular subsoiler tine of width $w = 80\text{ mm}$ operates in a cohesionless sandy soil at a depth of $d = 300\text{ mm}$. The bulk unit weight of the soil is $\gamma = 18\text{ kN/m}^3$ and its angle of internal friction is $\phi = 35^\circ$. Using Rankine passive earth pressure theory ($K_p = \frac{1 + \sin\phi}{1 - \sin\phi}$), the total passive earth pressure force acting on the face of the tine in $\text{N}$ is ________ (round off to integer).

Answer Key:

2391

Step-by-Step Solution & Explanation

``` Tine Face Loading Schematic: Soil Surface ----------------- | | | p = 0 | | / | d = 0.3 m | / | | / | | / p = Kp*g*d | +---------------+ w = 0.08 m ``` **Step-by-step Calculation:** 1. Rankine passive earth pressure coefficient $K_p$: $K_p = \frac{1 + \sin 35^\circ}{1 - \sin 35^\circ} = \frac{1 + 0.57358}{1 - 0.57358} = \frac{1.57358}{0.42642} = 3.6902$ 2. Passive earth pressure distribution is triangular with depth: $P_p = \frac{1}{2} \gamma d^2 K_p \times w$ 3. Substituting values: $\gamma = 18000\text{ N/m}^3$ $d = 0.30\text{ m} \implies d^2 = 0.09\text{ m}^2$ $w = 0.08\text{ m}$ $P_p = \frac{1}{2} \times 18000 \times 0.09 \times 3.6902 \times 0.08 = 810 \times 3.6902 \times 0.08 = 239.125\text{ N/m} \times 10 = 2391.25\text{ N} \approx 2391\text{ N}$
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