Analytical Chem Soil HW Solutions 4

# Analytical Chem Soil HW Solutions 4 - hydrostatic below...

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4 The saturation ratio is given by the volume of water divided by the total void volume, = 0.2126litre ÷ (0.52litre - 0.297litre) = 0.9534 S r = 95.34% If the soil were fully saturated, the volume of water would be (0.52litre - 0.297litre) = 0.223litre. The mass of water would be 0.223kg, and the water content would be 0.223kg 0.7874kg w sat = 28.32% The overall mass of the 0.52litre sample would be 0.223kg + 0.7874kg = 1.0104kg, and its unit weight (1.0101kg × 9.81N/kg 10 -3 kN/N) (0.52 10 -3 m 3 ) ⇒γ sat = 19.06kN/m 3 If the soil were dry but had the same specific (and overall) volume, the mass would be equal to the mass of solids alone, and the unit weight would be (0.7874kg 9.81N/kg 10 -3 kN/N) (0.52 10 -3 m 3 ) dry = 14.86 kN/m 3 Q1.4 An office block with an adjacent underground car park is to be built at a site where a 6m-thick layer of saturated clay ( γ = 20 kN/m 3 ) is overlain by 4m of sands and gravels ( γ = 18 kN/m 3 ). The water table is at the top of the clay layer, and pore water pressures are
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Unformatted text preview: hydrostatic below this depth. The foundation for the office block will exert a uniform surcharge of 90 kPa at the surface of the sands and gravels. The foundation for the car park will exert a surcharge of 40 kPa at the surface of the clay, following removal by excavation of the sands and gravels. Calculate the initial and final vertical total stress, pore water pressure and vertical effective stress, at the mid-depth of the clay layer, (a) beneath the office block; and (b) beneath the car park. Take the unit weight of water as 9.81kN/m 3 . Q1.4 Solution Initially, the stress state is the same at both locations. The vertical total stress σ v = (4m 18kN/m 3 ) (for the sands and gravels) + (3m 20kN/m 3 ) (for the clay), giving v = 132 kPa The pore water pressure u = (3m 9.81kN/m 3 ) = 29.4 kPa The vertical effective stress ' v = v- u = (132kPa - 29.4kPa) = 102.6 kPa Finally,...
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