Analytical Chem Soil HW Solutions 79

Analytical Chem Soil HW Solutions 79 - compression line and...

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

View Full Document Right Arrow Icon
79 Figure Q5.7c: q against p' for sample B For test B at yield, q = 35 kPa (scaled from Figure Q5.7c); p' = 140 kPa; p = 251.7 kPa; u = p - p' = 111.7 kPa. At failure, q = 45 kPa; p' = 100 kPa; p = 255 kPa; u = 155 kPa (d) If sample B had been subjected to a drained shear test from a cell pressure of 140 kPa, the effective stress path would have been given by p' = CP + q/3 Failure would have occurred on reaching the line joining critical states, q = Μ . p' We can calculate the value of the critical state parameter using the data for the end point of test A, = q/p' at failure = 45 kPa/100 kPa = 0.45 Thus drained failure for sample B would have occurred at q/0.45 = 140 + q/3 q = 74.1 kPa; p' = 164.7 kPa On the graph of v against lnp', the line joining critical states is parallel to the isotropic normal
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: compression line and the slope is given by = - v/ ln p' = (3.501 2.101) (5.159 3.219) = 0.722 from the data for the isotropic compression of sample A. We know that the end point of test A lies on the critical state line on the graph of v against lnp' at p' = 100 kPa, and that for a drained test on sample B p' at the critical state would be 164.7 kPa. Hence the change in specific volume during a drained shear test on sample B would be v = 0.722 (ln p') = 0.722 (ln164.7 ln100) = 0.360 To find the actual volume of water expelled, we must multiply this by the volume of solids V s (because V t = V s . v) to give...
View Full Document

Ask a homework question - tutors are online