Please use the information from question 2 to solve question 3.
Thanks
0 ITIT 2. (15 Points) An 8-m thick normally consolidated clay layer is doubly drained as shown below. (This means that a very pervious layer compared to the clay exists on top of and under the clay layer.) The embankment applied an average vertical stress increase of 100 kPa to the clay layer. The water table is located at the mid depth of the sand layer and is a hydrostatic condition. Assume the sand above the water table is dry. Pore water pressure, u (kPa) 20 40 60 80 100 120 140 160 180 200 0 Y=20 kN/m Sand Yst=21 kN/m² 4 Soft clay Ysat=22 kN/m eo = 0.6 OCR = 1 Cc= 0.25 C = 3X10-8 m²/s 12 • Dense sand Depth (m) 8 oo =24 kN/m 16
3. (15 Points) Repeat Problem 2, but for the case of single drainage. Assume that under the clay there is impervious shale instead of dense sand as shown below. Pore water pressure, u (kPa) 20 40 60 80 100 120 140 160 180 200 0 0 =20 kN/m Sand =21 kN/m IN 프 4 Soft clay 75=22 kN/m e = 0.6 Depth (m) 8 OCR = 1 Ce = 0.25 c = 3x10 m/s 12 Impervious shale 16 a) Fill the table below, and estimate the excess pore water pressure (Au) remaining in the clay after 5 year for the depths in the clay layer of 0, 2, 4, 6, and 8 m. Show a sample calculation for z=2 m. z/H UZ (%) Depth from ground surface (m) 4 6 8 10 12 z (m) in the clay layer 0 2 Au (kPa) (t= 5years) 4 6 8 b) On the graph provided above, plot the initial pore water pressure before the load was applied (t=0"), the initial pore water pressure after the load was applied (I=0), and the pore water pressure after 5 years (t=5 years). Use Yw=10 kN/m² for your calculation. c) Compute the primary (final or ultimate consolidation) settlement (sc) to 1 decimal place in [cm]. Primary settlement, sc (cm)
Please use the information from question 2 to solve question 3. Thanks
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