Analytical Chem Soil HW Solutions 74

Analytical Chem Soil HW Solutions 74 - Vt = Vs + Vv; e =...

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Unformatted text preview: Vt = Vs + Vv; e = Vv/Vs; so Vt = Vs.(1 + e) = Vs.v and ΔVt = Vs.Δv The first sample at a cell pressure of 300 kPa (q = 0) has a total volume Vt given by ‘ Vt = (π × 382 mm2/4) × 80 mm = 90729 mm3 It is saturated, so e = w.Gs (main text Equation 1.10); w = 0.572 and Gs = 2.61 so e = 1.493 Hence the volume of solids Vs = Vt/(1 + e) = 90729 mm3 ÷ 2.493 = 36393.5 mm3 (constant) When the cell pressure was increased to 320 kPa, the change in total volume ΔVt was -618 mm3 giving a change in specific volume Δv of –618 mm ÷ 36393.5 mm = -0.017. The slope of the isotropic normal compression line on a graph of v against lnp' is -λ, where λ = -Δv/Δ(lnp'). Hence λ = 0.017 ÷ (ln320-ln300) = 0.017/0.0645 ⇒ λ = 0.26 The same change in specific volume occurs on swelling from p' = 320 kPa to p' = 229 kPa, which is along a swelling line on the graph of v against lnp' which has slope -κ, hence κ = 0.017 ÷ (ln320-ln229) = 0.017/0.3346 ⇒ λ = 0.05 The specific volume on the isotropic normal compression line at p' = 1 kPa is (Γ + λ – κ), hence (Γ + λ – κ) – λ.ln300 = 2.493 (Γ + 0.21) – (0.26 × 5.704) = 2.493 ⇒ Γ = 3.766 Assume that the first test ends on the critical state line at q = Μ.p', giving Μ = 136.5 ÷ 133.7 = 1.02 (c) The second sample behaves elastically (in the sense that p' = constant during undrained shear) until the stress path reaches that of the first sample. The second sample then yields, and its stress path follows that of the first sample (the undrained state boundary) to failure at the same critical state. This is indicated in Figure Q5.6b. 74 ...
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