A plate load test is used to determine:
Try it first: pick an option, then read the explanation.
Correct answer: C. Both (a) and (b)
Asked in SSC JE 2009; SSC JE 2012
Detailed Explanation
Because it measures both load and settlement directly at foundation level, the plate load test gives information on both design criteria for a foundation.
Why “Both (a) and (b)” Is Correct
From the load–settlement curve we read the ultimate bearing capacity (where the curve breaks or settlement becomes very large) and the pressure for a given settlement, which is scaled up to the actual footing size. So the test is used for both — option (c).
| Item (IS 1888) | Value |
|---|---|
| Plate | Square or circular, 300 mm to 750 mm, mild steel, at least 25 mm thick |
| Pit width | At least 5 × plate width |
| Seating load | About 7 kN/m² (0.07 kg/cm²), released before the test |
| Loading | Steps of about one-fifth of the expected safe load |
| Readings | Settlement of the plate on dial gauges, until it nearly stops for each step |
| Result | Load–settlement curve → ultimate bearing capacity and settlement |
Why the Other Options Are Wrong
- A. Bearing capacity of foundations: Correct but incomplete; settlement is also found.
- B. Settlement of foundations: Correct but incomplete; bearing capacity is also found.
- D. None of the above: The test clearly gives both.
Key Point
Plate load test → bearing capacity and settlement.
Key Concepts for Students
- For sand, settlement of footing S_f = S_p [B_f(B_p + 0.3)/B_p(B_f + 0.3)]².
- For clay, settlement is roughly proportional to width.
- The test does not reveal weak layers deeper than about 2B_p.
Practice more: this is question 40 in Shallow and Deep Foundation MCQs – Part 3. Try the whole set and check every answer instantly.