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A 30 cm square bearing plate settles by 8 mm in the plate load test on cohesionless soil when the intensity of loading is 180 kN/m2. The settlement of a shallow foundation of 1.5 m square under the same intensity of loading will be nearly
1. 30 mm
2. 26 mm
3. 22 mm
4. 18 mm

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Correct Answer - Option 3 : 22 mm

Concept:

For sandy soil or cohesionless soil, we know that

\(\frac{{{S_f}}}{{{S_p}}} = \;{\left\{ {\frac{{{B_f}\left( {{B_p}\; + \;0.3} \right)}}{{{B_p}\left( {{B_f}\; + \;0.3} \right)}}} \right\}^2}\)

Where,  Bp = Width of plate, Sp = Settlement of plate, Bf = Width of foundation, and Sf = Settlement of foundation

Calculation:

Given, Bp = 30 cm, Sp = 8 mm, Bf = 1.5 m

\(\frac{{{S_f}}}{8} = {\left\{ {\frac{{1.5\left( {0.3\; + \;0.3} \right)}}{{0.3\left( {1.5\; + \;0.3} \right)}}} \right\}^2}\)

Sf = 22.22 mm ≈ 22 mm

Note:

For clay of cohesive soils, the relation between Settlement of foundation and plate is:

\(\frac{{{S_f}}}{{{S_p}}} = \frac{{{B_f}}}{{{B_p}}}\)

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