Analytical Chem Soil HW Solutions 56

Analytical Chem Soil HW Solutions 56 - against t is linear...

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56 t, min 0.5 0 0.707 1.000 1.414 2.000 2.828 4.000 5.657 8.000 Initial settlement of 0.02 mm is probably due to trapped air. Figure Q4.5: Settlement against time Assume that the apparent initial settlement (0.02 mm) is due to trapped air. Then ρ ult 0.22 – 0.02 mm = 0.2 mm Ultimate vertical strain ε v = 0.2 mm ÷ 20 mm = 0.01 (1.0%) One dimensional modulus E' 0 = Δσ ' v / v = 100 kPa/0.01 = 10 MPa The initial slope of the graph of
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Unformatted text preview: against t is linear with slope d /d( t) = ult . (4c v /3d 2 ); thus c v = 3d 2 /4t x From Figure Q4.5, t x = 5.95 min 0.5 t x = 35.4 minutes drainage path length d = 10 mm (half the nominal sample height) hence c v = 10 2 mm 2 /35.4 minute = 2.12 mm 2 /minute divide by 60 10-6 to convert mm 2 /minute to m 2 /second c v = 3.53 10-8 m 2 /s...
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This note was uploaded on 12/28/2011 for the course CHM 4302 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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