Analytical Chem Soil HW Solutions 77

Analytical Chem Soil HW Solutions 77 - t/V s = 150796/43074...

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77 q is given by the ram load Q divided by the current sample area A In an undrained test the total volume V t is constant = A 0 .h 0 = A.h = A.h 0 .(1 – ε ax ) where ax is the axial strain Δ h/h 0 , hence A = A 0 /(1 – ax ) and q = Q.(1 – ax )/A 0 . In a drained test the total volume V t is equal to V t0 .(1 – vol ) and A = V t0 .(1 – vol )/ h 0 .(1 – ax ) = A 0 .(1 – vol )/(1 – ax ) hence q = Q/A 0 × (1 – ax )/(1 – vol ) (see main text Section 5.4.3). σ 2 = 3 = cell pressure CP (measured); u is the pore water pressure (measured); and 1 = cell pressure CP + q, so p' = CP – u + q/3 (b) V t = V s + V v ; e = V v /V s ; so V t = V s .(1 + e) = V s .v and V t = V s . v The volume of solids V s is constant and given by V s = m s / ρ s = m s /G s . w = 116.3 g ÷ (2.70 10 -3 g/mm 3 ) taking w = 1 g/mm 3 ; hence V s = 43074 mm 3 At a cell pressure of 25 kPa (q = 0), the sample has a total volume V t given by V t = ( π 40 2 mm 2 /4) 120 mm = 150796 mm 3 giving a specific volume v = V
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Unformatted text preview: t /V s = 150796/43074 = 3.501 Hence we can calculate the specific volume during the consolidation stage of the test as v = 3.501 – Δ V t /V s as in the table below: CP, kPa (= p') 25 50 75 100 150 174 u, kPa 0 0 0 0 0 0 V t , cm 3 0 22.4 34.47 43.08 56.01 60.31 ln p' (p' in kPa) 3.219 3.912 4.317 4.605 5.011 5.159 v 3.501 2.981 2.701 2.501 2.201 2.101 During the shear test, p' = CP – u + q/3 and the cell pressure CP = 274 kPa. Hence CP, kPa 274 274 274 274 274 274 u, kPa 100 104 114 132 162 189 q, kPa 0 10 20 30 40 45 p', kPa 174 173.3 166.7 152.0 125.3 100.0 The state paths followed in terms of v against lnp' and q against p' are plotted in Figures Q5.7 a and b....
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