Question

The time for 50% consolidation of a 25 mm thick clay specimen (drained at top and bottom) in the laboratory is 150 sec. How long (in days) will it take for a 3 m thick layer of the same clay in the field under the same pressure increment to reach 75% consolidation? There is an impermeable shale layer at the bottom of the clay layer in the field I. 2. Use Figure 1 for the following Compute the vertical effective stress (a'v) in kN/m2 at the center of the clay layer prior to the construction of the footing. [Note: To compute ơ, you will first have to use the weight-volume relationships to compute the unit weight of the clay.] a) b) Assuming the value that you computed in (a) is representative for the entire thickness of the clay layer, calculate the ultimate primary consolidation settlement (Se) in mm that will occur in the clay layer from placement of the footing (assume Ao-20 kPa throughout the clay layer). Use the empirical relation: C.-0009x(LL%-10) to estimate the compression index. c) Using Terzaghi's 1D consolidation theorem, calculate and plot the predicted settlement of the footing as a function of time (i.e., AH(t) vs. time), where AH(t) is in mm and time is in days. Estimate cv using Figure 9.15 in the textbook (cv vs. LL) assuming the sample is undisturbed. [Compute ΔΗ(t) at enough times so that you can reasonably construct the requested plot. Also, compute ΔΗ(t) for times long enough that AHinal is reached. )

Answer 1

Given data U = 50% consolidation t = 150 sec Bick H = 25mm d = H/2 = 25/2 = 12.5mm (because draned top and bottom) T_v = pi/u U^2 = pi/4 times 0.5^2 = 0.197. Therefore we know that T_v = c_v t/d^2 0.197 = c_v times 150/12.5^2 c_v = 0.197 times 2.5^2/150 c_v = 0.197 times 156.25/150 c_v = 0.197 times 150.25/150 c_v = 0.205 mm^2/sec. Now Bick H = 3m U = 75% consodilation d = H 3m = 3000mm. (because of bottom will be impable). U = 75% U > 0.6% T_v = 1.781-0.933/050 (100 - U%) = 1.781- 0.933/050(100-75) 0.476. T_v = c_v t/d^2 Rightarrow t = T_v d^2/c_v. = 0.476 times 3000^2/0.0205. = 2092957.3 sec = 242024 days.