An embankment (3 m high, = 19 kN/m*) has been built at the site shown below whic
ID: 1732575 • Letter: A
Question
An embankment (3 m high, = 19 kN/m*) has been built at the site shown below which includes a 6m layer of normally consolidated clay underlain by an imperable layer. Following construction of the embankment, the pore pressure at the mid-point of the clay layer (P) is monitored employing a piezometer At a given time t the piezometer measures a value of the pore pressure u equal to 63.6 kPa a) How long has it been since the embankment was built (i.e. what is t)? Express t in days. b) What is the effective stress at P at time t? c) What is the final settlement caused by the construction of the embankment? d) What is the thickness of the clay layer at time t? e) Sketch pore pressure and effective stress trend with depth of the clay layer at time t (no calculations required). EMBANKMENT Pore pressure sketch Effective stress sketch 3 m CLAY ninl P e 0.953 c 4x10-3 cm2/s 20 kN/m3 0.32 IMPERMEABLE ROCKExplanation / Answer
b) Embankment is 3 m high. Stress at bottom of embankment = 19 X 3 = 57 kPa.
Total stress at P = (20 X 3) + 57 = 60 + 57 = 117 Kpa
Effective stress = 117 - 63.6 = 53.4 kPa
c) H = 6 m, delta P = Change in stress = 57 KPA, Cc=0.32, e0 = 0.953 , Po = (20 X 3) - (10 X 3) = 30 KPA
Final settlement = (Cc/(1+eo)) * log((Po+deltaP)/(Po)) * H
= (0.32/ (1+0.953)) * log ((30+57)/30) * 6
= 0.163 * log (2.9) * 6 = 0.978 * 0.462 = 0.451 m
1. Tv = (Cv * t) / (Hd2 ) where Hd = Drainage path = one way drainage = 600 cm, Cv = 4 X 10-3 cm2/sec
Degree of consolidation = ( (57 - (63.6-30))/57) = 41.05 %
This is approximately 40% of shaded area. Average degree of consolidation is approximately 40%.
From graph Tv = 0.15 (approximate value)
t = Tv * H2 / Cv =0.15 * (600^2) / (4 * 10^-3) = 13500000 sec = 156 days
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