Time-settlement curve, 3 m clay

Given: single-drained layer, $H_{dr} = 3$ m $= 300$ cm. Settlement 5 cm at $t = 60$ days corresponds to $U = 20\%$.

Step 1: Final settlement

$$S_{final} = \frac{5}{0.20} = 25\ \text{cm}$$

Step 2: Coefficient of consolidation. For $U=20\%$, $T_v = \frac{\pi}{4}(0.20)^2 = 0.0314$.

$$c_v = \frac{T_v H_{dr}^2}{t} = \frac{0.0314 \times 300^2}{60} = 47.1\ \text{cm}^2/\text{day}$$

Step 3: Time for various $U$. $t = T_v H_{dr}^2 / c_v = T_v \times 1910.0$ days. Use $T_v = \frac{\pi}{4}U^2$ for $U<60\%$ and $T_v = 1.781 - 0.933\log_{10}(100-U\%)$ for $U>60\%$. Settlement $= U \times 25$ cm.

Step 4: Settlement at 3 years ($t = 1095$ days):

$$T_v = \frac{c_v t}{H_{dr}^2} = \frac{47.1 \times 1095}{300^2} = 0.573 \;\Rightarrow\; U \approx 80\%, \quad S \approx 20\ \text{cm}$$

Plot: draw time (days, 0 to 1095) on the horizontal axis and settlement (cm) downward on the vertical axis; join the points above with a smooth curve. The curve is steep at first and flattens towards 25 cm.

 t (days)  0   200  400  600  800  1000 1095
 S (cm)    0
           |  *
        5  |    *
       10  |        *
       15  |             *
       20  |                    *    *  <- 3 yr
       25  - - - - - - - - - - - - - - (final)

Answer: the structure settles about 20 cm in 3 years; the curve approaches the final value of 25 cm.