VIEWS: 128 PAGES: 14 POSTED ON: 6/17/2012
in equilibrium under principal stresses: with a pore a V and Porosity Consider soil element: whose volume =water pressure: = n Pore Pressure Coefficients σ1 u0 σ3 V, n σ2 Pore Pressure Coefficient, B Then, the element is subjected to an increase in stress in all 3 directions of σ3 which produces an increase in pore pressure of u3 σ1 + σ3 u0 +u3 u0 σ3 + σ3 V, n σ2 + σ3 Pore Pressure Coefficient, B As a result, the effective stress in each This increase in effective stress reduces the direction skeleton by the pore volume of the soilincreases andσ3 -u3 space. σ1 + σ3 σ3 - u3 σ + σ3 -3u3σ3 u0 +u3 V, n σ3 - u3 σ2 + σ3 Pore Pressure Coefficient, B σ3 - u3 Soil Skeleton σ3 - u3 V, n σ3 - u3 ΔVss CsV Δσ 3 Δu3 where: Vss = the change in volume of the soil skeleton caused by an increase in the cell pressure, σ3 Cs = the compressibility of the soil skeleton under an isotropic effective stress increment; i.e., the fraction of volume reduction per kPa increase in cell pressure Pore Pressure Coefficient, B σ3 - u3 Pore Space (Voids) σ3 - u3 V, n ΔVPS CVVV Δu3 σ3 - u3 where: VPS = the volume reduction in pore space caused by a change in the pore pressure, u3 CV = the compressibility of the pore fluid; i.e., the fraction of volume reduction per kPa increase in pore pressure Since: VV nV Then: ΔVPS CVnVu3 Pore Pressure Coefficient, B σ3 - u3 Assuming: 1. the soil particles are incompressible 2. no drainage of the pore fluid σ3 - u3 Therefore, the reduction in soil V, n σ3 - u3 skeleton volume must equal the reduction in volume of pore space Therefore: CSV Δσ 3 Δu3 CVnVΔu3 1 or: Δu3 Δσ3 1 n CV C S Pore Pressure Coefficient, B σ3 - u3 The value: 1 σ3 - u3 CV 1 n C V, n S σ3 - u3 is called the pore pressure coefficient, B So, u3 = B σ3 If the void space is completely saturated, Cv = 0 and B = 1 When soils are partially is estimated > 0 and B < 1 In an undrained triaxial test, Bsaturated, Cv by increasing the cell pressure by σ3 and measuring the resulting change in pore This is illustrated on Figure 4.26 in the text. pressure, u3 so that: Δu3 B Δσ 3 Pore Pressure Coefficient, A What happens if the element is subjected to an increase in axial (major principal) stress of σ1 which produces an increase in pore pressure of u1 σ1 + σ1 u0 +u1 u0 σ3 - u1 V, n σ2 - u1 Pore Pressure Coefficients As change in effective stress also in each the This a result, the effective stress changes of the minor soil skeleton and the by -u1 volume of the directions increasespore space. σ1 + σ1 σ1 - u1 -σ3 -u1 u1 u0 +u1 V, n -u1 σ2 - u1 Pore Pressure Coefficients σ1 - u1 If we assume for a minute that soil is an elastic material, then the Volume change of the soil skeleton can be expressed from elastic theory: - u1 ΔVss 3 CsV Δσ 1 3Δu1 1 V, n - u3 As before, the change in volume of the pore space: ΔVPS CVnVu1 Again, if the soil particles are incompressible and no drainage of the pore fluid, then: 1 3 CSV Δσ 1 3Δu1 CVnVΔu1 Pore Pressure Coefficient, A σ1 - u1 or: Δu 1 1 Δσ1 1 BΔ 1 1 3 CV 3 - u1 1 n C V, n S - u3 Since soils are NOT elastic, this is rewritten as: u1 = AB σ1 or Δu1 A Δσ 1 where A AB For different values of saturated soil test, determined by A value is A for pressure during the to be determined by where A of a pore a fully σ1coefficient canbeu1 is measured, although the values at failure are of particular interest: experiment measuring the pore water pressure during the application of the deviatorΔustress in an undrained triaxial test A 1 , B 1 for saturation Δσ 1 Pore Pressure Coefficient, A Normally Consolidated σ1 - u1 Figure 4.28 in the text illustrates the variation of A with OCR (Overconsolidation Ratio). Lightly Over- - u1 Consolidated V, n Heavily Over- - u3 Consolidated In highly overconsolidated (normally consolidated clays), A For heavilycompressible soils clays, A may lie <between -0.5 & 0 For lightly overconsolidated clays, 0 A < 0.5 ranges between 0.5 and 1.0 maximum historical effective stress OCR current effective stress Pore Pressure Coefficient, B The third pore pressure coefficient is determined from the response, u to a combination of the effects of increasing both the cell pressure, σ3 and the axial stress (σ1 -σ3) or deviator stress. If we divide through by σ1 From the two previous effects: u = u3 + u1 u3 = Bσ3 { u3 + u1 = u = B[σ3+A(σ1-σ3)] Δu Δu B Δσ { Δσ3 Δσ1 u1 1= BA(σ1 Δσ3 Δσ -σ ) A 1 Δσ3 1 3 Δu or: B1 (1 A) 1 Δσ or: B Δσ1 1 Δσ 1 Pore Pressure Coefficient, B Δu The third pore pressure coefficient is not a B Δσ 1 constant but depends on σ3 and σ1 With no movement of water (undrained) and no change in water table level during subsequent consolidation, u = initial excess pore water pressure in fully saturated soils. Testing under Back Pressure When a sample of the calculation extracted pressure This process allowssaturated clay isof the pore from the coefficient, B. ground, it can swell thereby decreasing Sr as it breathes in air The pore pressure can be raised artificially (in sync with σ3) Values of B 0.95 are considered to represent to a datum value for excess pore water pressure and then the to consolidate sample can be allowedsaturation. back to the in situ conditions (saturation, pore water pressure).