Analytical Chem Soil HW Solutions 81

Analytical Chem - 5.26 q/Mp ln(p/p = 0(main text Equation 5.37 Also v = 1.893 = Γ – κ – .ln p 0 κ.ln(p/p(Equation Q5.8a see main text

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81 This is a non-standard test with p = constant = 400 kPa, so the pore pressure u c at the critical state is given by u c = 400 - p' c u c = 183.2 kPa p = 1 / 3 ( σ a + 2 r ) = r + q/3 = constant, i.e. Δσ r = - Δ q/3 where sr is the cell pressure. Hence the cell pressure at failure = 400 kPa – 192.9 kPa/3 cell pressure at failure = 335.7 kPa Summary table: kPa CP q p' p u v Start of shear 400 0 400 400 0 1.893 End of shear 335.7 192.9 216.8 400 183.2 1.893 We need to calculate some intermediate points. This may be done as follows. Stress states between the start of the shear test and failure all lie on the current yield locus, which has associated with it a current value of p' 0 (which defines the isotropic pressure at the tip of the Cam clay yield locus – see main text Figure
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Unformatted text preview: 5.26). q/Mp' +ln(p'/p' ) = 0 (main text Equation 5.37) Also, v = 1.893 = Γ + λ – κ – λ .ln p' 0 + κ .ln(p'/p' ) (Equation Q5.8a: see main text Example 5.6) At the start of the shear test, p' 0 = 400 kPa. At the end of the test, knowing that p' c = 216.8 kPa and v = 1.892, we can use the given values of Γ , λ and κ together with Equation Q5.8a to calculate that p' 0,c = 589.2 kPa Hence we can choose some values of p' 0 between 400 kPa and 589 kPa and substitute them into Equation Q5.8a to calculate corresponding values of p'. We can then substitute the pairs of values (p' , p') into Equation 5.37 to calculate the corresponding value of q, as tabulated below:...
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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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