Question

A 10-m depth of sand overlies an 8-m layer of clay, below which is a further depth of sand. For the clay, my-0.78 m2/MN and c 4.2 m'/year. The water table is at surface level but is to be lowered permanently by 4 m, the initial lowering taking place over a period of 40 weeks. Calculate the final settlement due to consolidation of the clay (one sub-layer assumption), assuming no change in the weight of the sand, and the settlement two years after the start of lowering.

Answer 1

Given m_v = 0.78 m^2/mn t = 40 vector & C_v = 4.2 m^2/year a) final settlement delta H = m_v H delta_6 H = 8/2 = 4m = d(two way drainage) delta_6 = 4 times 9.81 = 39.24 kN/m^2 m_v = 0.78 times 10^-3 m^2/kN delta H = 0.78 times 10^-3 times 4 times 39.24 = 0.1224 m final settlement delta H = 122.43 mm b) Initial lowering time = 40 weeks = 280 days average time of lowering = 280/2 = 140 days times left for settlement = (2 times 365) - 140 = 590 days = 1.616 year T_v = C_v t/d^2 = 4.2 times 1.616/4^2 = 0.4243 T_v = -0.9332 log_10 (1 -v) - 0.0851 0.4243 = -0.9332 log_10 (1 - v) - 0.0851 u = 0.7154 settlement = u times delta H Implies 0.7154 times 122.43 = 87.6 mm settlement 2 years after start of lowering = 87.6 mm