Question

For a clayey soil stratum, the time required for degree of consolidation from 25 % to 50 % is 30 days.

The total time (in days) required for 90 % degree of consolidation of the same soil stratum is (rounded off to the nearest integer).

Hint:

Use the parabolic $T_v = (\pi/4)U^2$ relation for the 25% and 50% points to find the fixed $d^2/C_v$ constant, then switch to the $1.781-0.933\log_{10}(100-U)$ formula for the 90% point (since U exceeds 60%).

Answer 1

Correct Answer: 173

Step 1: Understanding the Question.
Consolidation time is linked to the degree of consolidation U through the time factor $T_v$, and $T_v$ is different for U up to 60% and for U above 60%. The time itself is $t = T_v \, d^2/C_v$, where $d^2/C_v$ is a constant fixed by the soil and the drainage path, and does not change with U. We first use the 25%-50% data to pin down this constant, then apply it at 90%.

Step 2: Key Formula.
For $U \leq 60\%$:
\[ T_v = \frac{\pi}{4}U^2 \quad (U \text{ as a decimal}) \]
For $U > 60\%$:
\[ T_v = 1.781 - 0.933\log_{10}(100-U) \quad (U \text{ in percent}) \]

Step 3: Use the 25%-50% data to find the constant.
\[ t_{25} = \frac{\pi}{4}(0.25)^2 \cdot \frac{d^2}{C_v}, \qquad t_{50} = \frac{\pi}{4}(0.50)^2 \cdot \frac{d^2}{C_v} \]
\[ t_{50} - t_{25} = \frac{\pi}{4}(0.25 - 0.0625)\frac{d^2}{C_v} = \frac{\pi}{4}(0.1875)\frac{d^2}{C_v} = 30 \]
\[ \frac{d^2}{C_v} = \frac{30}{0.7854 \times 0.1875} = \frac{30}{0.14726} = 203.72 \text{ days} \]

Step 4: Find $T_{v90}$.
\[ T_{v90} = 1.781 - 0.933\log_{10}(100-90) = 1.781 - 0.933\log_{10}(10) = 1.781 - 0.933(1) = 0.848 \]

Step 5: Compute the time for 90% consolidation.
\[ t_{90} = T_{v90}\cdot\frac{d^2}{C_v} = 0.848 \times 203.72 = 172.75 \text{ days} \]

Final Answer:
Rounded to the nearest integer, the time required for 90% consolidation is 173 days.
\[ \boxed{t_{90} = 173 \text{ days}} \]