Clay 4 m, 90% in 4 hours, time in years

Principle: the same soil has the same $c_v$ and, for the same $U$, the same $T_v$. So

$$\frac{t_{field}}{t_{lab}} = \left(\frac{H_{dr,field}}{H_{dr,lab}}\right)^2$$

Drainage paths

• Field: sand on top, impervious rock at the bottom, so single drainage: $H_{dr} = 4$ m $= 4000$ mm.
• Laboratory: the 20 mm specimen in an oedometer drains at top and bottom, so $H_{dr} = 10$ mm.

Time

$$t_{field} = 4\ \text{h} \times \left(\frac{4000}{10}\right)^2 = 4 \times 160{,}000 = 640{,}000\ \text{h}$$ $$t_{field} = \frac{640000}{24 \times 365} = 73.1\ \text{years}$$

Answer: about 73 years for the building to reach 90% of its final settlement.

If the specimen were drained at one face only, the answer would be about 18 years.