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Hello everyone. So, far we have gone through the Bishop's Effective Stress Principle, which is modified after Terzaghi's Effective Stress Principle for the Unsaturated Soils. Bishop introduced one more strength parameter called effective stress parameter χ which is a function of matrix suction. The estimation of χ function from the suction control direct shear test and suction control triaxial test is seen. there were some more developments that took place later on because the χ parameter estimation is very difficult especially for fine-grained soils where the estimation of χ over a wide suction range is very difficult. So, independently there was some more test that was conducted by other researchers, further to evaluate whether the independent stress state variables that were used by Bishop are valid or not. So, such tests are called null type tests, so which will discuss. So, the null type test these tests are to understand the validity of different stress state variables. So, it is based on the manipulation of a different combination of stress statevariables. Changes to whether the total volume is the degree of saturation are observed are used as the measure for understanding whether they are independent stress state variables or not. So, null tests are conducted by Fredlund Morgenstern in 1977, this is very popular work and the concept is used even to date. So, this is published in A.C. geotechnical and geoenvironmental engineering in 1977. So, in this work, they have conducted more than 19 null tests on compacted cerulean soil, cerulean clay by controlling different stress parameters. So, as in triaxial cell one can control the deviatoric stress, the all-round pressure, the pore air pressure, and pore water pressure, of course, we have coarse for here and then you have higher entry porous disk located at this location. So, therefore, ua within the soil mass and uw within the soil mass can be controlled independently. So, therefore, they have changed these parameters such a way that whether all the parameters are increased to the same extent or decrease to the same extent. So, here if you observe test number 23 from their work, this is the initial conditions of the soil, where the soil was at equilibrium, total stress was 420.7, and pore air pressure is 278.7, and pore water pressure is 109.6. You can assume some units a kilopascal. Now, the Δ σ is increased to + 71.4 by changing the water pressure to + 70.3, and changing the air pressure to + 70.7, they want values of these parameters to be the same. So, while doing this experiment slight deviation was there; however, nearly changes in the values are constant. So, due to these changes, the sample dimension the sample volume did not change, or the water content did not change during the testing. Similarly, there is another test 25 number, this initial condition and so the changes and stresses are the same in Δ σ Δ uw and Δ ua. Similarly, several other test data are shown here. Either you increase this you increase all of them + or you decrease all of them. So, in the same direction they have been applied, and when these changes are made the sample volume did not change; that means, the shear stresses are not initiated within the soil sample. So, this is because the σ - ua, that is effective stress that net normal stress we use; which = σ +, there is a small increment Δ σ in the test number 23, - ua. Here, ua is increased the small value Δ ua test number 23. So, as this Δ σ and Δ ua are maintained to be constant, they get canceled. So, essentially you get σ - ua. Similarly, if you take ua - uw the increase ua + Δ ua and uw + Δ uw. ThisΔ ua and Δ uw both are the same. This remains ua - uw. Due to an increase in the stresses as shown here. So, therefore, these 2 variables can be used as independent state variables for defining the shear in the soils. Because earlier Terzaghi's had shown that itis effective stress that defines the state of the soil for shearing. And here Bishop has proposed that σ - ua, the net normal stress, and ua - uw these 2should be used as stress state variables and they are independent to each other. And that is proven by these null test by Fredlund and Morgenstern. So, σ - ua and ua - uw when they are kept constant. There is no shear-induced in the soil sample. So, therefore, there is no change in the volume or degree of saturation. So, these are a confirmatory test for Bishop's effective stress equation for unsaturated soils. Further, their observation states that it is not just one combination of σ - ua and ua - uw to use as 2 independent state variables, stress state variables. They further say that even if we use a combination of this is one set σ - ua and ua - uw. Another combination σ - uwand ua - uw, and other third combination are σ 3 σ - ua and σ - uw. Any of these 3 combinations can be used for defining the stress state in soil. So, here the σ - ua or σ 3 - ua, and ua - uw are maintained to be 13.8 and 58.6 changing from the initial value like this 44.8, so 13.8 and 58.6 are maintained. And in the secondtest same values are maintained, but different values of σ 3 and ua are given. And similarly, different values of ua and uw are given, but the ua - uw remains constant. So, in this all 4 test the σ 3 - ua and ua - uw are kept constant, then the volume of the sample did not change. So, group one that is σ - ua net normal stress, and ua - uw this combination qualifies as independent stress state variables. Similarly, group 2 isanalyzed there is σ - uw and ua - uw. So, here the same combination is maintained for all different tests by varying σ 3 uw ua independently. So, then the σ 3 - uw and ua - uw these 2 combinations kept constant for all these 4 tests and we will now the volume remains constant. Similarly, in group 3 the σ - ua and σ - uw are maintained to be constant for all these 4 tests then the volume remains constant. So, therefore, the effective stress equation that is given by Bishop, that is σ dash = σ - ua + χ,( ua – uw). This is what we used as the effective stress principle, modified effective stress equation from Bishop's proposal. And here the effective stress also could be used as σ - uw + some χ times ua - uw. Or this could be a σ - ua + χ σ - uw. So, any of these 3 combinations can be used to formulate the effective stress principle, and it could be used to define the shear strength of the soil at any given stress state. So, therefore, this test reveals that Bishop's independence stress state variables indeed valid for the unsaturated soils. Moreover, any of these 3 combinations are group one group 2 group 3 could be used as independent stress state variables. Fredlund and Morgenstern after their stress state variables definitions and stress state variable confirmatory test by in null type test in suction control triaxial and suction control consolidation cells, they are come up with a new form of modified column failure criterion for unsaturated soils. Before going into that, let us understand how the stresscan be represented graphically so that the Fredlunds extended more coulomb theory can be very well understood.
So, here the stress representation for unsaturated soils can be understood by solving some problems. Here there is one particular problem there is taken; that is some example. So, one shear testing is conducted on clay soil using a triaxial setup. So, the soil sample
which is at equilibrium with an all-round pressure σ3 = 300 kilopascal. So, this is an allaround pressure, and deviatoric stress of 100 kilopascal is acting. So, this is σ 3 and that is Δ σ, and there is a pore air pressure on the sample is 200 kilopascal. And the pore water pressure is 100 kilopascal. This soil state can be represented graphically as we know the stress state variables for unsaturated soils we can choose net normal stress
there is σ - ua and ua - uw. So, as this is a triaxial test you will have σ1 and σ3. Therefore, we can draw the Mohr’s circle. So, this can be plotted as τ on the y-axis and σ - ua on the x-axis. But as we have suction also, matrix suction. So, that can be plotted on the third axis. So, this is the third axis, so z-axis for ua - uw. So, if that is represented on the y z-axis. Now the stress of the soil can be represented on ua - uw is 100 kilopascal. So, this is located at 100 kilopascal here. So, this is 100 kilopascal. The more circles stress circle
can be drawn on this axis, so which is at 100 kilopascal on the z-axis. So, here the σ 3 - ua and σ 1 - ua can be obtained. σ1 - ua that is net normal stress in the y-direction on the soil sample is 300 + 100, that is 400 - ua is to 200. So, therefore, this is 200 kilopascal σ 1 is 400, - ua is 200. So, therefore, this is 200 kilopascal, and σ 3 is 300 and ua is 200 therefore, this is 100 kilopascal. So, therefore, there is 100 kilopascal here, and there may be 200 kilopascal somewhere here. So, this is100 kilopascal and this is 200 kilopascal, σ 1 - ua is 200 kilopascal. So, maximum stress is at the centre, that is 50 kilopascal and maximum stress exists at 45 degrees from the
major principal stress and this is a maximum value of 50 kilopascal.
So, this is how the state of the stress can be represented graphically on a 3-dimensional plot. Similarly, let us take another example. So, there is a soil mass which is located on a slope. So, therefore, it has the 3 stresses acting major principal stress, and the net normal stress along with major principles is σ 1 - ua is 400 kilopascal. And intermediate stress is 100 kilopascal, and the minor principal stress is 50 kilopascal. So, if these stresses vary during the monsoon state. The stress state changed from one season to another season. This is season one and in season 2. The stress state changed to
σ - 1 u σ 1 - ua is 360 kilopascal, and σ 2 - ua is 90 kilopascal. And σ 3 - ua is 45 kilopascal. And in season one the matrix suction in the soil is 250 kilopascal. So, this is a relatively dry season. And in season 2 the matrix suction decrease to 50 kilopascal. So, how to represent the change in the stress state of soil? This can be again drawn on the 3 axis plot, σ - ua on the x-axis, and τ on the y-axis, and the third axis for ua - uw. So, here on the suction axis at 250 kilopascal somewhere here, you draw horizontal line.
So, this is σ - ua parallel to the σ - ua axis. So, here on this the σ 1 - ua is 400, say 400 and σ 2 - ua is 100, and σ 3 - ua is 50. So, you have one stress so this is how it is plotted. So, this is so this value is 50, and this value is this goes parallel to this axis. So, this is 100 and this is 400. Similarly, in another season 2, in the matrix suction decrease to 50. So, assume that this is 50 and horizontal if you draw and here this became 360. So, slightly decreased this value and here it is 90 and here it is 45. And this is how it can be plotted stress state of the soil. So, in the case of Bishop's effective stress principle, we have one test data. So, that is ua is 100 kilopascal and uw is 0 kilopascal. So, in axis translation let us understand how the Bishop's modified more coulomb equations can be represented graphically. So, let us assume one problem where ua is 100 kilopascal, and uw is kept at atmospheric, so this is
0 kilopascal. And in test one, σ 1 is 1,000 kilopascal and σ 3 is 400 kilopascal at failure; these are observed data. And in test 2 when the σ 3 is maintained 600 kilopascal, the soil field at σ 1 of 1600 kilopascal at failure. If χ is known = 0.5 because it depends on matrix suction. So, matrix suction is constant in these 2 tests. So, χwill be constant for these 2 tests is χis known there is 0.5, then how to represent this? Here in the same way you can represent on wax is sets the τ an x-axis is σ - ua. So now this is the third axis, now in this case because the equation is τ f = C` + σ - ua + ua - uw time χ times tanϕ. So, this is intercept + this whole thing, the one axis is σ - ua another axis is τ f, and other axis could be χ times ua - uw. So, this χ ua - uw. Then at one particular χ * ua - uw, ua - uw is suction is 100 kilopascal, χ is 0.5. So, this 50 kilopascal, χ times ua - uw that is 50 kilopascal. Then in one test, the soil failed at σ 3 400. And σ 1 is 1,000 kilopascal. So, this the test one 400 and 1,000. In test 2 when the σ 3 is 600 kilopascal, it failed at 1600 kilopascal somewhere here.So, this is test 2 and this is test 1. This is how it can be represented because we have when we have all of 3 axes defined one is τ f another one is σ - ua. And another one is ua - uw times χ as third axis then the stress state of the soil can be very well defined using graphically, by graphically.
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