A clay deposit of thickness 10 cm and void ratio 0.5 undergoes settlement and its final void ratio is now 0.2. The thickness (cm) of the settlement layer is ______.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 29/01/2018 (Morning)
Detailed Explanation
During one-dimensional compression the volume of solids stays the same; only the voids shrink. If we take the volume of solids as 1, the total volume is 1 + e. So the ratio of the change in thickness to the original thickness equals the change in void ratio divided by (1 + e₀).
\[ \frac{\Delta H}{H} = \frac{\Delta e}{1+e_0} \]
\[ \Delta H = 10 \times \frac{0.5 - 0.2}{1 + 0.5} = 10 \times \frac{0.3}{1.5} = 2\ \text{cm} \]
Why “2” Is Correct
The void ratio falls by 0.3 out of a total of 1.5, which is one fifth of the volume. One fifth of 10 cm is 2 cm of settlement, leaving a layer 8 cm thick.
Why the Other Options Are Wrong
- A. 1: 1 cm would need Δe = 0.15, half of the actual change in void ratio.
- B. 1.5: 1.5 cm does not follow from ΔH = H·Δe/(1 + e₀) with these values; it would need Δe = 0.225.
- D. 2.5: 2.5 cm would need Δe/(1 + e₀) = 0.25, i.e. Δe = 0.375.
Key Point
ΔH/H = Δe/(1 + e₀). Use the initial void ratio in the denominator.
Key Concepts for Students
- This relation holds because the soil is confined laterally (one-dimensional compression).
- Combined with Cc it gives ΔH = Cc·H·log(σ′f/σ′₀)/(1 + e₀).
- Combined with av it gives ΔH = av·Δσ′·H/(1 + e₀) = mv·Δσ′·H.
Practice more: this is question 14 in Soil Consolidation MCQs – Part 1. Try the whole set and check every answer instantly.