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Unformatted text preview: 5.26). q/Mp' +ln(p'/p' ) = 0 (main text Equation 5.37) Also, v = 1.893 = Î“ + Î» â€“ Îº â€“ Î» .ln p' 0 + Îº .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 Î“ , Î» and Îº 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.
 Fall '11
 StuartChalk
 Analytical Chemistry

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