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Module 1: Concept of Suction Stress

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Video 1
Hello everyone, let us discuss more details about Suction Stress characteristic curve. And how to determine the Suction Stress characteristic curve and further to evaluate the effective stress of unsaturated soil from suction controlled direct shear test data and suction control triaxial test data. The suction stress characteristic curve is a characteristic curve represents the variation of suction stress with whether matric suction or normalized volumetric water content. The suction stress consumes energy that is required for the change in energy of soil water due to several physical-chemical factors and capillary effects are consumed in suction stress, suction stress. So, the suction stress is the suction stress equation as we have seen is a σs` = it consists of stresses due to physical-chemical forces + the stresses due to capillary forces +. So, due to the effect of surface tension and this is a cohesion intercept which we use at asaturated state. So, this we started writing it as so, this one the effective stress can bewritten as σ` = σ - u a that is net normal stress - σs`. In further works of Lu et al, this suction stress is represented as σ s as suffix σ s in this manner. So, this can be written as σ - u a - σ s. So, accordingly, the σ s is the suction stress = - of u a - u w. If u a - u w is less than or = 0 which means that when the pore water pressureis positive; so, negative suction means it is a pore water pressure is positive. So, that is normal the consolidation test or any other test where the pore water pressures are more than 0. Any consolidation or shear strength tests where the pore water pressure is positive. So, in that particular case, the suction stress = the pore water pressure itself. So, therefore, when you substitute this 1 here σ s here so, this σ` becomes σ - u a -( u a - u w). So, therefore, this = σ - u w. So, this is the effective stress equation given by Terzaghi itself. So, this boils down to the Terzaghi’s effective stress equation. So, this work is from Lu et al in 2008 2010 sorry. So, in the previous work of Lu and Likos, they have modified or expanded the Terzaghi’s effective stress equation by incorporating physical-chemical forces. And if it is unsaturated soil then all the capillary forces are also included. So, however, in their further work, they again approximated their suction stress characteristic curve equation into a simplified manner where this boils down to Terzaghi effective stress equation for saturated soils. So, for unsaturated soils where u a - u w is greater than 0 then σ` is a function of u a - u w this we called suction stress characteristic curve, sorry σ s is a function of u a - u w. So, this is for saturated soils where the pore water pressure is positive. And this is for unsaturated soils where the pore water pressure is negative. So, therefore, the suction is above 0 you have suction in the soil. So, in the Bishop’s approach as we have said earlier Bishop in 1950 in 1950’s he has given the effective stress equation that is σ` =, σ` = σ - u a that is net normal stress + χ *( u a - u w). So, here he says that χ varies in the same manner as normalized volumetric water content. He often assumes the value of χ to be normalized volumetric water content or this could be the degree of saturation also. So, therefore, comparing this particular equation expression with this particular expression given by Lu et al σ s suction stress becomes simply - degree of saturation S times u a - u w. So, therefore, it is an extension of as they have using the degree of saturation here it is an extension of Bishop’s approach and expansion of Terzaghi’s equation for unsaturated soils. So, therefore, effective stress σ` = σ - u a - of - S e theyhave introduced to Lu et al introduced a term as c I will explain what it is times u a - u w. So, instead of degree of saturation if we use a c which is defined as the degree of saturation at any given water content - degree of saturation divided by 1 - Sr. So, this is defined in this particular manner c = S - Sr by 1 - Sr. So, this = σ - u a -. So, - if - so, this is the same as Bishop’s expression except that they have used a normalized degree of saturation here. So, for saturated soils S c = 1 for saturated soils S e equals 1 which boils down to Terzaghi’s 1-dimensional consolidation equation sorry Terzaghi’seffective stress equation. And when I say S equals 0 this is a dry condition which is σ` = σ - u a. So, that equals equivalent to the Bishop’s expression for dry soils. So, here the additional advantage they have given is that they assume that S e is a function of matric suction and which assumes a form of (Refer Time: 09:27) equation.So, the (Refer Time: 09:29) equation which is going to be written as 1 over 1 + alpha times u a - u w whole power n whole power m, but they assume m to be dependent on n. So, this is 1 - 1 by n. So, this is a form they assume to be existing between S e and u a - u w. So, this S e could be normalized volumetric water content also that is a big θ which is θ - θ s sorry θ r by θ r s - θ r which is similar to this.So, you can assume normalized volumetric water content also instead of Sc. So, they assumed this form to be existing then the σ s can be simplified as - u a - u a - u w - u a - u w divided by 1 + alpha u a - u w whole power n whole to the power of n - 1 by n. Theexpression for suction stress and as a continuous form can be defined as the suction stress characteristic curve. The dependency of suction stress on matrix action can be defined in a continuous manner using this expression. So, this is one expression and another expression is in terms of S e. So, S e here u a - u w can be written as S e S e power n by n - 1 and - - 1 whole power 1 by n 1 by alpha so, this is u a - u w. So, when you write u a - u w in terms of S e this isso, what you get and when you substitute this in the suction stress characteristic curve equation. So, we get - S e by alpha S e power n by 1 - n and - c is taken in here this becomes 1 - n - 1 whole power 1 by n where S e varies between 0 and 1. So, this is another expression in terms of S e. So, if soil-water characteristic curve data for any given soil is known under applied under any given stress state either applied mechanical stress or something. So, then the suction stress characteristic curve or the effective stress of the soil can be determined by simply substituting these parameters. Here alpha is the bubbling pressure or is related to, is related to bubbling pressure are Air Entry Value AEV and n a is the pore size distribution factor.So, essentially the SWCC of the soil can be directly utilized to determine the soil suction stress characteristic curve. So, from that, we can obtain the effective stress of the soil. So, this is very useful. So, then you can obtain the effective stress by simply summing upσ - u a - σ s anyway - a sign you have there. So, this is simply + u a - u w divided by 1 + alpha u a - u a - u w whole power n whole power n - 1 by n. So, here u a - u w is greater than or = 0. So, σ` is an effective stress equation. So, for several soils when the S c versus u a - u w is plotted so, this can be plotted as like this for go to 0. This is for clay soils, this is for silt, this may be for sand. The Suction StressCharacteristic Curve can be drawn u a - u w and S e for clay soils. So, this is how the SSCC varies for clay soils for silt soils. As we have seen yesterday the Suction Stress Characteristic Curve decreases and goes to 0 sorry this is σ s in negative kilo Pascal. Then for sandy soils, this goes somewhat like this. The suction stress characteristic of increases and nearly becomes constant and then goes to 0 because the capillary effects are dominant only at some particular value and beyond that is after the air entry value. After the air entry value, the capillary effects are predominant, and beyond that again as a degree of saturation goes to 0 approaches 0 thecapillary effects are absent and suction stress approaches 0 for sand soils and silts. Similarly, if this is plotted in terms of u a - u w and σ s; so, σ s sorry suction stress on the y-axis and u a - u w on the x-axis. So, suction stress increases and becomes nearly constant for clay soils and which increases and decreases beyond certain suction value for silt soils, for sandy soils this increases and decreases and goes to 0 at higher section values. So, this is how the suction stress characteristic curve varies for different soils these are the solutions this is the data obtained from closed-form solutions for different by assuming different values of alpha and n. Here the clay soil has a very high value of air entry value. So, alpha is accordingly adjusted and some n value is assumed. And similarly silt value the alpha value slightly got changed because the air entries slightly smaller and the sand air entry is smaller. So, that is how these three curves are generatedon SWCC using equation in using m and n dependent equation. So, once this is derived and their c versus σ s is directly obtained by substituting these values into this equation and for any given S c which has c changes from 0 to 1. So, knowing the alpha and then n parameters from the SWCC equation. For clay sand silt and sand these parameters are substituted and then σ s is plotted in this manner. Similarly, this third figure σ s versus u a - u w is obtained using this particular equation by again substituting alpha and n parameters into this equation, and u a - u w is varied over a wide suction range and then σ s is plotted in this manner. 
Video 2
So, they have proposed two approaches for estimating the suction stress characteristic curve. One is a semi-quantitative approach. So, Lu et al in 2010 they have come up with two approaches Semi-quantitative Approach. This approach is used when soil-watercharacteristic curve data is not known only the shear strength data from suction controlled direct shear test and suction control triaxial tests are known then this particular approach could be used to derive the effective stress variation within the soil mass the suction shows the variation with either suction or volumetric water connected. So, for the direct shear testing the Mohr-Coulomb failure criterion is τf equals to this is for direct shear τf equals to c` + σ - u a + - σ s tanϕ` because effective stress is σ - u a - σ s. So, this is the effective stress the total stress or net normal stress - the suction stress is the effective stress. So, the Mohr-Coulomb failure criterion is τf equals to c` + σ` tan phi and σ` is substituted here then this is the equation. So, therefore, the suction stress can be obtained by for any given matrix section. At any given matrix suction the suction stress is σ s = - τf - c` - σ - u a at failure tan ϕ` by tan ϕ`. So, this equation can be used for determining the suction stress characteristic curve if thesoil-water characteristic curve is not known. So, let us solve the example problem. When the suction control direct shear when the shear test data from suction control direct shear tests are obtained on some silty soil this is a generated synthetic data.So, the number of tests performed and the suction value control σ - u 8 failure that is net normal stress which is which can be controlled and τf which is measured. So, here the first number of tests is 9 to saturated two tests are conducted at the saturated condition. So, suction varied in this manner in these tests. So, the net normal stress varied in thefirst two tests which are saturated. So, different shear strength will be obtained. So, from that the strength parameters of saturated soils can be obtained for remaining tests the net normal stress is maintained constant only the matric suction is varied. So, this test data is used this test data were used earlier for demonstrating the Bishops approach for the prediction of χf calculation. So, the same test data is used here. So, here. So, for saturated tests and soil is saturated for the saturated condition. So, τf = C` +. So, this is simply σ, σ` or you can write σ - u a and suction stress is 0 u as tan ϕ`. So, therefore, solving these two equations 294 = C` + 300 tan ϕ` and 136 = C` + 120 tan ϕ`. Solving these two we get ϕ` = 41.3 degrees and C` = 30.4 kilo Pascal. So, knowing these two shear strength parameters so, the effective stress parameters; so, shear strength drain parameters ϕ` and C`. So, we can obtain the suction stress at any given matrix section value. So, the σ s = - here if you are given the suction stress characteristic suction stress values. So, this is a 0 here and 0 here and for the third data test three suction stress is - τf is 156 - C` is 30.4 - σ u a is 120 tan 41.3 divided by tan 41.3. So, this value comes out to be - 22.97 kilo Pascal. So, we can substitute the data for all the values - 22.97and for this is - 41.18 and this is for - 50.29 - 55.97 - 59.4 - 61.67 - 55.98. So, these are the suction stress values. And we can also compute for comparison χf because this equation is quite similar to the Bishops equation. So, the Bishopsequation if you compare that is σ - u a + χf u a - u w. If you compare these two equations σ s is nothing but - χf u a - u w. So, therefore, χf = - σ s by u a - u w. So, here u a - u w is 0 χf is nothing, but 1 this is a maximum value is 1. So, this is 1 and here. This is a - 20.97 divided by - 25 this is 0.918 and similarly if you get all the values 0.824 and 0.503 and 0.28 0.148 0.123 and 0.075. So, this is a χf data and. If you compare our earlier calculations where we estimated the bishops effective stress parameter using the same data and if you compare the χf data is the same. So, therefore, it is not a big modification we have done here using the suctions stress characteristic curve only therepresentation of effective stress is more effective here in this case. 
Video 3
Similarly, in the semi-quantitative approach for triaxial testing; if you have a data from suction control tri axial test then σ s can be represented as - σ 1 - u a at failure - σ 3 - u a at failure tan square 45 + ϕby 2 - 2 C` and 45 + ϕ` by 2 divided by 2 tan 45 + ϕ` by twotimes tan phi -. So, this is the expression that can be used to determine the suction stress if the major principal stress at failure and - principal stress at failure are the net normal stress in the vertical direction and net normal stress in the horizontal direction or radially it directions are known at failure. And we can determine the suction stress. So, let us solve another example of this. So, the test data, here also we utilize 9 test data. So, this is a matrix suction in kilo Pascal. This is varied in this manner 0 10 25 50 100 200 400 500 750. So, as the matrix action is only the saturated state at the saturated state only one test is conducted. So, therefore, this must be for sand for sandy soil and σ 1 - u a at failure this is 180 these are observed data. So, σ three is maintained constant. So, we can obtain the σ s we can calculate the χf and we can also estimate the ϕb. So, that we can compare Bishop’s approach and Fredlund’s approach we with SSCC. So, now, using the first test data either to obtain the anglophone intersection either you can use this equation. So, here anyways c is 0 because it is sand you can directly use this expression or you can use sin phi - is σ 1 - u a at failure - σ 3 - u at the failure by σ 1 - u at failure + σ 3 - u a at failure. Because when you are plotting tau versus σ and if this is a failure envelope and this is more circulant this is a failure envelope and this is the angleof internal friction ϕ` and if you take sine. So, this is a σ 1 - this quantity is σ 1 - u a - σ 3- u a whole divided by 2 when this is major principle stress at failure and this is minor principle stress. And this quantity = σ 1 - u a + σ 3 - u a by 2. So, if you take sine then this psi divided by this one so, you get this expression. So, if you use this expression 180 - 50 by 180 - 50 by 200 - 50 sorry 180 + 50 from the first expression. So, you get ϕ` which equals 34.42 degrees. Once ϕ` is obtained you can calculate σ s suction stress = - σ - u a is σ 1 - u a is 200, 200 - σ 3 - u a is 50 tan square 45 + 34.42 by 2. And anyways to see this is the whole thing is canceled for sand and divided by 2 and 45 + 34.42 by 2 times and 34.42.