Pore Pressure Coefficients

1. in equilibrium under principal stresses: with a pore water pressure: whose volume = V and Porosity = n Pore Pressure Coefficients σ1 Consider a soil element: σ3 u0 V, n σ2

2. Pore Pressure Coefficient, B σ1 σ1+ σ3 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 σ3 σ3+ σ3 u0 +u3 u0 V, n σ2 σ2+ σ3

3. This increase in effective stress reduces the volume of the soil skeleton and the pore space. Pore Pressure Coefficient, B σ1+ σ3 σ3 -u3 As a result, the effective stress in each direction increases by σ3 -u3 σ3 -u3 σ3+ σ3 u0 +u3 V, n σ3 -u3 σ2+ σ3

4. σ3 -u3 Pore Pressure Coefficient, B σ3 -u3 V, n σ3 -u3 where: Soil Skeleton 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

5. σ3 -u3 Pore Pressure Coefficient, B σ3 -u3 V, n σ3 -u3 where: VPS = the volume reduction in pore space caused by a change in the pore pressure, u3 Pore Space (Voids) CV = the compressibility of the pore fluid; i.e., the fraction of volume reduction per kPa increase in pore pressure Since: Then:

6. σ3 -u3 the soil particles are incompressible no drainage of the pore fluid Pore Pressure Coefficient, B σ3 -u3 Therefore, the reduction in soil skeleton volume must equal the reduction in volume of pore space V, n σ3 -u3 Assuming: Therefore: or:

7. σ3 -u3 Pore Pressure Coefficient, B σ3 -u3 V, n σ3 -u3 is called the pore pressure coefficient, B So, u3 = B σ3 Thevalue: If the void space is completely saturated, Cv = 0 and B = 1 In an undrained triaxial test, B is estimated by increasing the cell pressure by σ3 and measuring the resulting change in pore pressure, u3 so that: When soils are partially saturated, Cv > 0 and B < 1 This is illustrated on Figure 4.26 in the text.

8. Pore Pressure Coefficient, A σ1 σ1+ σ1 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 σ3 σ3- u1 u0 +u1 u0 V, n σ2 σ2- u1

9. This change in effective stress also changes the volume of the soil skeleton and the pore space. Pore Pressure Coefficients σ1+ σ1 σ1-u1 As a result, the effective stress in each of the minor directions increases by -u1 -u1 σ3-u1 u0 +u1 V, n -u1 σ2- u1

10. σ1-u1 Pore Pressure Coefficients -u1 V, n -u3 As before, the change in volume of the pore space: 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: Again, if the soil particles are incompressible and no drainage of the pore fluid, then:

11. σ1-u1 or: Pore Pressure Coefficient, A -u1 V, n -u3 Since soils are NOT elastic, this is rewritten as: u1 = AB σ1 or where A is a pore pressure coefficient to be determined by experiment A value of A for a fully saturated soil can be determined by measuring the pore water pressure during the application of the deviator stress in an undrained triaxial test For different values of σ1 during the test, u1 is measured, although the values at failure are of particular interest:

12. Normally Consolidated σ1-u1 Figure 4.28 in the text illustrates the variation of A with OCR (Overconsolidation Ratio). Lightly Over-Consolidated Pore Pressure Coefficient, A -u1 V, n Heavily Over-Consolidated -u3 In highly compressible soils (normally consolidated clays), A ranges between 0.5 and 1.0 For lightly overconsolidated clays, 0 < A < 0.5 For heavily overconsolidated clays, A may lie between -0.5 & 0

13. Pore Pressure Coefficient, B From the two previous effects: u = u3 + u1 If we divide through by σ1 u3 = Bσ3 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. { u3+ u1 = u = B[σ3+A(σ1-σ3)] { u1 = BA(σ1-σ3) or: or:

14. Pore Pressure Coefficient, B 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 The third pore pressure coefficient is not a constant but depends on σ3 and σ1 This process allows the calculation of the pore pressure coefficient, B. When a sample of saturated clay is extracted from the ground, it can swell thereby decreasing Sr as it breathes in air The pore pressure can be raised artificially (in sync with σ3) to a datum value for excess pore water pressure and then the sample can be allowed to consolidate back to the in situ conditions (saturation, pore water pressure). Values of B  0.95 are considered to represent saturation.