The ultimate consolidation settlement of a structure resting on a soil:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 2015
Detailed Explanation
The final (ultimate) consolidation settlement of a normally consolidated clay layer is given by
\[ \Delta H = \frac{C_c\,H}{1+e_0}\,\log_{10}\!\left(\frac{\sigma'_0 + \Delta\sigma}{\sigma'_0}\right) \]
Why “decreases with the increase in the initial void ratio” Is Correct
Look at where e₀ appears: in the denominator, as (1 + e₀). If the compression index, layer thickness and stresses are kept the same, a larger initial void ratio makes (1 + e₀) larger and the settlement smaller. That is option (a), the answer expected from the formula.
In real soils a high e₀ usually comes with a high Cc, so very loose clays do settle a lot. But this question is testing the formula with the other variables held constant.
Why the Other Options Are Wrong
- B. decreases with the decrease in the plastic limit: The plastic limit does not appear in the settlement formula.
- C. increases with the increase in the initial void ratio: In the formula e₀ is in the denominator, so increasing it decreases the settlement (other terms fixed).
- D. increases with the decrease in the porosity of the soil: Lower porosity means a lower e₀, and in the formula alone that would raise 1/(1 + e₀). But a denser soil with low porosity also has a much smaller Cc, so in reality its settlement is smaller, not larger. The accepted answer is (a), the direct effect of e₀ in the formula.
Key Point
ΔH ∝ Cc·H/(1 + e₀) × log(stress ratio). With Cc fixed, higher e₀ → smaller ΔH.
Key Concepts for Students
- Settlement increases with Cc, layer thickness H and the stress increment Δσ.
- For overconsolidated clays, use Cr up to σ′c and Cc beyond it.
- Thick clay layers are divided into sub-layers, each with its own σ′₀ and e₀.
Practice more: this is question 21 in Soil Consolidation MCQs – Part 2. Try the whole set and check every answer instantly.