Clay 10 m, 2.54 cm sample, half settlement

Note: the second half of the problem is unclear in print, and the laboratory time for half the final settlement is not given. The usual form is: "the sample took $t_{lab}$ to reach half its final settlement; find the time for the building to reach half its final settlement". The scale relation is solved below, and the field time follows from the lab time.

Principle: same clay, same $c_v$, same $U = 50\%$ (so same $T_v = 0.196$). Then $t \propto H_{dr}^2$.

Drainage paths (both drained top and bottom)

• Field: $H_{dr} = 10/2 = 5$ m $= 500$ cm.
• Laboratory: $H_{dr} = 2.54/2 = 1.27$ cm.

Time ratio

$$\frac{t_{field}}{t_{lab}} = \left(\frac{500}{1.27}\right)^2 = 1.55\times10^5$$

So $t_{field} = 1.55\times10^{5}\, t_{lab}$.

Coefficient of consolidation from the lab test

$$c_v = \frac{0.196\,(1.27)^2}{t_{lab}} = \frac{0.316}{t_{lab}}\ \text{cm}^2/\text{unit time}$$

Field time

$$t_{field} = \frac{0.196\,(500)^2}{c_v}$$

Illustration: if the sample reached half its final settlement in 10 minutes, then $t_{field} = 1.55\times10^{6}$ min $\approx 2.95$ years.

Answer: the building reaches half of its final settlement in $1.55\times10^5$ times the lab time (about 2.95 years for a 10-minute lab test).