Question

soil

Q1) An oedometer test was performed on a 2 cm thick clay sample drained on both sides taken from the mid clay layer shown below, after 10 minutes 50% consolidation is reached . a) How long time in days would the same degree of consolidation(50%) be reached in the field ?. b) How much the Final settlement would be reached in the clay layer.? c) How much the preconsolidation pressure of a specimen. ? 3m 2m Sand surcharge Y = 17 kN/m sand Y = 18kN/m VW.T clay e. = 1, CC=0.2, Ysat = 20kN/m . A , O.C.R.= 1 7m Sand layer

Answer 1

Solution Griven, thickness of Clay Sample ( 42 ) = 22cm = 2x tom time to reach 50%. consolidation (t,) = 10 min As the 3 Οχιο - " 10 m) H, 2 clay is is drained on both sides, then length of drainage path () Relation between time consolidation and length of drainage path is of t = Tvde Су => ta del (9) let the time to reach 50-7. consolidation in field is to thickness of sample (Hg) = 1 m length of drainage path (de) the above relation, we get dan 1 - 3.5 m २ using 2 ta ti २ २ ta = ti ( 2 ) 슈 슈 taa 3.5 102 1225000 min ta : î ta 1225000 60 X 24 x 365 year

ta - 2.33 3 years Thus time to required 504. consolidation in field is 4.33 years (b) and (e) 3m is not Sand surcharge 7=11 kN/m3 am Sand 7 = 18 kN/m3 V 7m clay e=1 • A Ysat = 20 kN/m3 C -0.2 OCR 21 When Sand sw wicharge applied effective stress at point (A) is = (18x) + (20 X 3-5) –(3.5Yw) Assuming Yw = 9.91 KM/m3 r. - 36 + 70-(3.5%981) = 71,665 knyme 3 슈 수 Thus preconsolidated pressure of clay layer is 71.665 After applying sand surcharge, Change in effective Stress (15) = 3X17 = 51 KM/m2 settiement is given by C. Ho To+A5 AH - log ( 5th Ite

Не Но 1) eo - I сс = 0,2 2) ДңО.4x7 71-465 + 51 log 1+1 11.665 -> Дң О. 16 338 m) -) Дң : 163, 39 mm Hence, Final settlement is 163.38 mm. preconsolidated pressure of specimen is 71.665 kN/me