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Consider a soil element: whose volume = V and Porosity = n V, n in equilibrium under principal stresses: σ1σ1 σ3σ3 σ2σ2 with a pore water pressure: u0u0.

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Presentation on theme: "Consider a soil element: whose volume = V and Porosity = n V, n in equilibrium under principal stresses: σ1σ1 σ3σ3 σ2σ2 with a pore water pressure: u0u0."— Presentation transcript:

1 Consider a soil element: whose volume = V and Porosity = n V, n in equilibrium under principal stresses: σ1σ1 σ3σ3 σ2σ2 with a pore water pressure: u0u0

2 Then, the element is subjected to an increase in stress in all 3 directions of  σ 3 which produces an increase in pore pressure of  u 3 σ1σ1 σ3σ3 σ2σ2 u0u0 σ 1 +  σ 3 σ 2 +  σ 3 σ 3 +  σ 3 u 0 +  u 3

3 As a result, the effective stress in each direction increases by  σ 3 -  u 3 V, n σ 1 +  σ 3 σ 2 +  σ 3 σ 3 +  σ 3 u 0 +  u 3  σ 3 -  u 3 This increase in effective stress reduces the volume of the soil skeleton and the pore space.

4 Soil Skeleton V, n  σ 3 -  u 3 where:  V ss = the change in volume of the soil skeleton caused by an increase in the cell pressure,  σ 3 C s = 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 Pore Space (Voids) V, n  σ 3 -  u 3 where:  V PS = the volume reduction in pore space caused by a change in the pore pressure,  u 3 C V = the compressibility of the pore fluid; i.e., the fraction of volume reduction per kPa increase in pore pressure Since:Then:

6 Assuming: V, n  σ 3 -  u 3 1.the soil particles are incompressible 2.no drainage of the pore fluid Therefore, the reduction in soil skeleton volume must equal the reduction in volume of pore space or: Therefore:

7 The value : V, n  σ 3 -  u 3 is called the pore pressure coefficient, B If the void space is completely saturated, C v = 0 and B = 1 This is illustrated on Figure 4.26 in the text. So,  u 3 = B  σ 3 When soils are partially saturated, C v > 0 and B < 1In an undrained triaxial test, B is estimated by increasing the cell pressure by  σ 3 and measuring the resulting change in pore pressure,  u 3 so that:

8 What happens if the element is subjected to an increase in axial (major principal) stress of  σ 1 which produces an increase in pore pressure of  u 1 V, n σ1σ1 σ3σ3 σ2σ2 u0u0 σ 1 +  σ 1 σ 2 -  u 1 σ 3 -  u 1 u 0 +  u 1

9 As a result, the effective stress in each of the minor directions increases by -  u 1 V, n σ 1 +  σ 1 σ 2 -  u 1 σ3 -u1σ3 -u1 u 0 +  u 1 σ1 - u1σ1 - u1 -u1-u1 - u1- u1 This change in effective stress also changes the volume of the soil skeleton and the pore space.

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

11 Since soils are NOT elastic, this is rewritten as: where A is a pore pressure coefficient to be determined by experiment  u 1 = AB  σ 1 or 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  σ 1 during the test,  u 1 is measured, although the values at failure are of particular interest: - u1- u1 - u3- u3 σ1 - u1σ1 - u1 or:

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

13 The third pore pressure coefficient is determined from the response,  u to a combination of the effects of increasing both the cell pressure,  σ 3 and the axial stress (  σ 1 -  σ 3 ) or deviator stress. From the two previous effects:  u =  u 3 +  u 1  u = B[  σ 3 +A(  σ 1 -  σ 3 )] or: If we divide through by  σ 1  u 3 +  u 1 =  u 3 = B  σ 3  u 1 = BA(  σ 1 -  σ 3 ) { { or:

14 The third pore pressure coefficient is not a constant but depends on  σ 3 and  σ 1 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 saturated clay is extracted from the ground, it can swell thereby decreasing S r as it breathes in air The pore pressure can be raised artificially (in sync with  σ 3 ) 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). This process allows the calculation of the pore pressure coefficient, B. Values of B  0.95 are considered to represent saturation.


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