Question

Question Three: Chose answer (3 Mark)
The time required for 25% consolidation of a 40mm thick clay layer, drained on both ends, in the laboratory is 4 min. How long, in days, will take for a 2-m thick clay of the same clay in the fiel under the same pressure increment to reach 25% consolidation? The soil mass in the field is sandwiched between two thick layers of gravel, as shown.

Answer 1

Answer:
The time required for consolidation in the laboratory can be found using the equation:
T = (T_50 * H^2) / (4 * h^2)
Where:
T = time required for consolidation in the laboratory (4 min)
T_50 = time required for 50% consolidation in the laboratory
H = thickness of clay layer in the field (2 m)
h = thickness of clay layer in the laboratory (40 mm)
Since T_50 is unknown, we can rearrange the equation to solve for it:
T_50 = (T * 4 * h^2) / H^2
T_50 = (4 min * 4 * (40 mm)^2) / (2 m)^2
T_50 = (4 min * 4 * 1600 mm^2) / 4 m^2
T_50 = 16 min
Now, we can use the equation for time of consolidation in the field:
T_field = (T_50 * H^2) / (4 * h^2)
T_field = (16 min * (2 m)^2) / (4 * 40 mm)^2
T_field = (16 min * 4 m^2) / 1600 mm^2
T_field = 64 min
Converting 64 minutes to days:
1 day = 1440 minutes
64 min * (1 day / 1440 min) = 0.0444 days
Therefore, it will take approximately 0.0444 days for a 2-m thick clay layer in the field to reach 25% consolidation.