Question

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 eo = 1, Cc=0.2, Ysat = 20kN/m . A O.C.R.= 1 7m Sand layer

Answer 1

Solution :- Whole solution is shown in below images.

Part (a) :- First calculate the coefficient of consolidation (Cv) from the test data performed on odometer. After calculation of Cv, we can calculate the time required to reach the same degree of consolidation in field using the formula shown in images. The time required for the same degree of consolidation (50%) to be reached in the field is 850.7 days.

Part (b) :- First calculate the initial effective overburden pressure (i.e. when sand surcharge is not placed) at the centre of clay layer i.e. at point A. Now calculate the increased effective stress at centre of clay due to sand surcharge (∆σ'). The final settlement in the clay layer is now calculated using the formula shown in below images. The final settlement in the clay layer is obtained as 163.4 mm.

Part (c) :- As O.C.R. (over consolidation ratio) is given as equal to 1 (i.e. soil is normally consolidated), hence, the maximum value of stress that the soil has ever experienced in past i.e. preconsolidation pressure is equal to the initial effective overburden pressure (i.e. when sand surcharge is not placed) at the centre of the clay layer. The preconsolidation pressure of the specimen is obtained as 71.665 KN/m^2.

To find: - (a) Time in days that would be required for the same degree of consolidation (50%) to be reached in the field. (6) final settlement be reached in the layer. (c) Preconsolidation pressure (T) of a specimen. that would clay Solution :- 2cm . Part (a):- Given: Odcometer test alata :- Clay sample thickness , h lay sample is drained on both sides • 50% consolidation is reached after 10 minutes. field data:- Clay thickness, H = 7m lay layer is drained on both sides • Time (t) be calculated for the same degree of consolidation 50% to be reached in field) . For odeoneter test :- Tv Cut ha Here, Tu= Time factor Cu = coefficient of consolidation t = 10 minutes (for 50% consolidation) h = lay sample thickness = 2cm (اجا FOH 50-1. consolidation, Time factor, Tu Hoe 50 Tv - 71 4 Com = 0.1963

Substitute all respective values in eq (i), we get Ty=Cut for both side drainage 2² le Clay layer thickness - 0.1963 = Cu (10 min) -> = Cv 0.01963 cm² min -4 2. ») Cv. = 0.01963 x 10 m (60x24 ) day Cv = 2.8267 x 10 in /day 7 Ч 50 100 Now, for field :- for 50% consolidation, Tu= 0.1963 Cu = 2.8267x169 m/day Since, drainage is from both sides, lay layer thickness = H Hence, {Here zm} (1) (2.8267 x 1 2 Tu= Cut Here H= 7m 0.1963 = in day). t 2 7m 2 t = 0.1963 x (3.5)2 days (2.8267 x10-3) It = 850.7 days It is the required solution 2

Part (b):- Sand 13m surcharge VY=17KN/m3 X=18 KN/m3 2m Sand RW.T. e = 1 O.GR. = 1 Im lay Cc = 0.2 3.5m Centre of clay layer. A Ysat = 20 kN/m3 Sand layer To calculate; Final Settlement (OH) in the clay layer. Solution - Final settlement (oh) in the clay layer is calculated as :- Cc logio >(10) OH Н. To +00 To Iteo Here To of clay layer, Point A Initial effective over burden pressure at the centre (ie. when sand surcharge is not placed) so = Increase in effective stress at centre of clay layer due to placement of sand surcharge. e = initial void ratio at beginning of consolidation le = 1 (Given) (c = Compression Index = 0.2 H = Initial thickness of lay layer. = 7m at the centre) so To = (1 sand X 2m) + (That w) x 3.5m (of lay layen ) Jo - (18 x2) + (20 - 9.81).x3.5 71.665 KN/m² 13

AT - due to sand surcharge (4 x 3m) ar = 1783 kN/m² AF - 51 kN/m 2 Substituting, C++, C = 0.2, T. = 71.665 kW/m2 51 kN/m² and H= 7m in equation (iii), we get, Дr AH 0.2 10g10 (7m) 71.665 + 51 71.665 1+1 AH (0.7) logo 122.665 71.665 m =) AH = 0.1634 m AH = 163.4 mm It is the required solution. Part (c):- Given - ACR. = 1 (50 so, soil is normally cally) consolidated = 1 As O.C.R. hence, the maximum value of stress that the soil has ever experienced in past il preconsolidation pressure (ſe) is equal to the initial effective overburden pressure (i.l. when sand surcharge is not placed) at the centre of clay layer. Therefore, Preconsolidation pressure (Fe) - To Fe = ((sand) x 2m + (Csat - Yw)x 3.5m clay FEF - (18x2)+(20-9.81) x 3.5 71.665 kN/m2 It is the required solutish 4