Question:
A plate load test was conducted on clay on a \(300\) mm diameter plate. If the plate settlement was \(18\) mm at a pressure of \(100\) kPa, then the settlement (in mm) of a \(2\) m wide rectangular footing at the same pressure will be
A plate load test was conducted on clay on a \(300\) mm diameter plate. If the plate settlement was \(18\) mm at a pressure of \(100\) kPa, then the settlement (in mm) of a \(2\) m wide rectangular footing at the same pressure will be
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For plate load tests:
\[
\boxed{
\text{Clay: }
S_f=S_p\left(\frac{B_f}{B_p}\right)
}
\]
For sandy soils, settlement relationships are different.
Updated On: Jul 23, 2026
- \(15\)
- \(30\)
- \(60\)
- \(120\)
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Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Concept:
For cohesive soils (clay), the settlement of a footing is directly proportional to its width.
The relationship between footing settlement and plate settlement is
\[
\boxed{
S_f=S_p\left(\frac{B_f}{B_p}\right)
}
\]
where
\[
S_f=\text{Settlement of footing},
\]
\[
S_p=\text{Settlement of plate},
\]
\[
B_f=\text{Width of footing},
\]
\[
B_p=\text{Width (diameter) of plate}.
\]
Step 1: Write the given data. \[ S_p=18\text{ mm} \] \[ B_p=300\text{ mm}=0.3\text{ m} \] \[ B_f=2\text{ m} \]
Step 2: Apply the settlement relationship. \[ S_f = 18\left(\frac{2}{0.3}\right) = 18\times6.667 = 120\text{ mm} \] Hence, \[ \boxed{S_f=120\text{ mm}} \] Therefore, the correct option is \[ \boxed{(D)\;} \]
Step 1: Write the given data. \[ S_p=18\text{ mm} \] \[ B_p=300\text{ mm}=0.3\text{ m} \] \[ B_f=2\text{ m} \]
Step 2: Apply the settlement relationship. \[ S_f = 18\left(\frac{2}{0.3}\right) = 18\times6.667 = 120\text{ mm} \] Hence, \[ \boxed{S_f=120\text{ mm}} \] Therefore, the correct option is \[ \boxed{(D)\;} \]
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