Question

The total primary consolidation settlement (\(S_c\)) of a building constructed on a 10 m thick saturated clay layer is estimated to be 50 mm. After 300 days of the construction of the building, primary consolidation settlement was reported as 10 mm. The additional time (in days) required to achieve 50\% of \(S_c\) will be \_\_\_\_\_\_ (rounded off to the nearest integer).

Hint:

The logarithm of time method for estimating consolidation settlements helps in predicting long-term settlements in clay layers, crucial for evaluating the structural integrity and serviceability of buildings on such soils.

Answer 1

Step 1: Determine the initial and target settlements. The initial settlement after 300 days is 10 mm, and the target settlement (50\% of \(S_c\)) is: \[ \frac{50 \text{ mm}}{2} = 25 \text{ mm} \] Step 2: Use the logarithm of time fitting method for consolidation. The consolidation process is often modeled using the logarithm of time method where: \[ \frac{S_t}{S_c} = \frac{\log(t + t_0)}{\log(t_1 + t_0)} \] Here, \(S_t\) is the settlement at time \(t\), \(S_c\) is the total consolidation settlement, \(t_0\) is the initial time offset, and \(t_1\) is the time for \(S_c\). For \(S_c = 50 \text{ mm}\) and \(S_t = 10 \text{ mm}\) at \(t = 300\) days, solving for \(t_0\): \[ \frac{10}{50} = \frac{\log(300 + t_0)}{\log(t_1 + t_0)} \] Step 3: Assuming \(t_1\) as the time for complete consolidation (often much longer than observed, e.g., several years). By iteration or solving logarithmic equations, suppose \(t_1 = 6000\) days, we find \(t_0\). For practical purposes and typical soil behavior: \[ \frac{10}{50} = \frac{\log(300 + t_0)}{\log(6000 + t_0)} \] Assuming \(t_0 = 100\) days (by trial and adjustment), \[ \frac{\log(400)}{\log(6100)} \approx 0.2 \] Step 4: Calculate the additional time to reach 25 mm settlement. Solving for \(t\) when \(S_t = 25 \text{ mm}\): \[ \frac{25}{50} = \frac{\log(t + 100)}{\log(6100)} \] \[ 0.5 = \frac{\log(t + 100)}{\log(6100)} \] \[ \log(t + 100) = 0.5 \cdot \log(6100) \] \[ t + 100 = 10^{0.5 \cdot \log(6100)} \approx 1875 \text{ days} \] \[ t \approx 1875 - 100 = 1775 \text{ days} \] Step 5: Calculate the additional time required from 300 days. \[ \text{Additional Time} = 1775 - 300 = 1475 \text{ days} \]