Question

7) (Total 20 pts) The figure shows a 20 m thick layer of normally consolidated clay (Y -18.6 kN/m3) that is one-dimensionally loaded by Δσ,-50kPa. The clay layer is below a 3 m thick layer of granular fill (y = 19.6 kN/m3), and a dense, compacted glacial till (impermeable) underlies the clay. The water table is located at the top of the clay layer. A 1-D consolidation test is performed on a 2.20 cm thick, doubly drained specimen from the middle of the clay layer. When the stress conditions from the field (including the Δσ.-50 kPa) are applied to this specimen, it takes 1 minute for 50% average consolidation to occur. a) From the lab test data, determine c, for the soil. (5 pts) b) Compute effective stress at 18 m depth below surface (15m into the clay layer) 4 years after the Δσν is applied to the clay layer. (10 pts) Compute the average degree of consolidation 4 years after Δσν application. (5 pts) c) 50 kPa 3 m Granular fill,ly -19.6 kN/m Clay, 18.6 kN/m3 20 m Dense glacial till

Answer 1

3m fill, Glay 2 2 a given Laboratory test dada, 50kPa H = 2-20 cm thick = 2:2x10 m double drainage time = 1 minute 50/- Consolidation 1 granular for u=50% 84 19.6kN/m Tv- 5 (50) = 0.196 for clouble drainage, d- y = 2,2x10 3 = 1.1 x10?m V4= 18.6 kw In Now Tv = Cut → a= Tyds 20m Cv = 0.196 X (1.4x30 -72 2:37 x 10 m 1 minute moin b) ou onu = 50 kPa + 19.6x3 + 15 X 10.6 Dense Glacial till = 387. 8 kPa U = 15x9.81 = 147.15 kPa (effective stess) = 387.0- 147.15 = 240.65 kPa c. after 4 years of application of ou Tu = out ; d = H = 20m of clay layer as there is single drainage cv is constant for a given soil. Cu = 2:37x10 mx 60 min x 24 kr mis mus day X 365 day = 0.124 m²/ year year Tv = 0.124x4 = 0.00124 < TV 50=0.196 202 TV (202 0.00124 = II (1U) 2 => %0 = 3.970). 4. Please rate if you Úke the answer