Analytical Chem Soil HW Solutions 54

# Analytical Chem Soil HW Solutions 54 - v /d 2 ) = 1 / t x...

This preview shows page 1. Sign up to view the full content.

54 Figure Q4.4: Settlement against time Assume that the apparent initial settlement (c 0.045 mm) is due to trapped air. Then ρ ult 0.27 – 0.045 mm = 0.225 mm Ultimate vertical strain ε v = 0.225mm ÷ 20 mm = 0.01125 (1.125%) One dimensional modulus E' 0 = Δσ ' v / v = 25 kPa/0.01125 = 2.22 MPa Initially, R = ( 4 / 3 T), that is / ult = ( 4 / 3 ). (c v /d 2 ). t The initial slope of the graph of against t has slope d /d( t) = ult . ( 4 / 3 ). (c v /d 2 ) = ult / t x (see Figure Q4.4 and main text Section 4.6); thus ( 4 / 3 ). (c
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: v /d 2 ) = 1 / t x or c v = 3d 2 /4t x From Figure Q4.4, t x = 7.7 min 0.5 ⇒ t x = 59.29 minutes drainage path length d = 10 mm (half the nominal sample height) hence c v = ¾ × 10 2 mm 2 /59.29 minute = 1.265 mm 2 /minute divide by 60 10-6 to convert mm 2 /minute to m 2 /second c v = 2.11 10-8 m 2 /s c v = k.E' / γ w so k = c v . w /E'...
View Full Document

## This note was uploaded on 12/28/2011 for the course CHM 4302 taught by Professor Stuartchalk during the Fall '11 term at UNF.

Ask a homework question - tutors are online