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Video 1
Hello everyone, we will discuss how the experimental data can be analyzed, the experiment data of time from a modified shear test we discuss and we also discuss how to analyze this data? let us discuss what type of data be obtained from Suction Controlled Shear Test. And, also how to analyze this data to interpret the effective stress concept for Unsaturated Soils will look into. So, as I explained in the previous lectures, the unsaturated strength parameters can be analyzed by considering two independent trusted
variables, such as net normal stress and matrix suction. And, Bishop is a person who introduced these two independent trusted parameters and lumped into one single equation for defining the effective stress equation.
So, using the modified more column concept by introducing this effective stress principle into the equation one can obtain the effective stress parameter χ can be estimated and χ is a function of weather volumetric water content or matric suction. So, this functional form can be obtained from the experimental data. So, let us understand what kind of
experimental data we obtained from suction controlled shear tests such as suction control direct shear and suction controlled triaxial test. And, from the data how to interpret this effective stress parameter we will understand. It is seen that many researchers when they have conducted tests on the suction controlled shear test. Especially suction controlled direct shear test setup, they observed that the strengthened soil varied in this particular manner. So, when they plotted, generally in our direct shear test, we obtain shear stress versus shear strain we written in terms of γ. And, the stress versus strain relationships for NC soils look somewhat like this, and, for the OC soils, it appears somewhat like this. And, this is called strain hardening behavior and
this is strain-softening behavior because after this the shear stress decreases so, it is called strain-softening behavior. So, now what kind of results you get this is the normal direction for saturated soils and this is for unsaturated soils. So, here you can plot τ for one particular σ normal stress. So, here when you plot τ versus γ are often this is plotted as stress ratio also. The stress ratio is defined as τ by the normal stress. Here this is the net normal stress σ - ua because this test is conducted under one particular net normal stress. So, when you control the particular one net normal stress you get τ versus γ, where is shear stress versus shear strain, you can also keep changing these values. So, that you will get a stress ratio versus this one it is observed that for nearly saturated soils this is the behavior of our NC soil. So, this is for suction ua - uw = 0 so, this is for saturated soils.So, as you increase the suction it is observed that it increases in this particular manner.
So, this is another particular suction ua - uw = 10 kilo Pascal. And, similarly when you increase 50 kilo Pascal, interestingly if you increase the suction value beyond this, it is observed that it behaves in this particular manner.
So, let me redraw this with different colour with different marker colour. So, this is the behavior at ua - uw = say 200 kilo Pascal. And, this is increased further this is how it changes? And, the peak stresses the ultimate stress increases. So, this is an interesting trend you get when the suction control shear tests are conducted on unsaturated soils when the stress ratio kept on increasing. So, this could vary in the range of say 0.5 this one, and this can go up to say 2.5 also stress ratio value and this can around say 10 percent strain. So, an interesting trend is that as the suction increases the shear stress and ultimate shear stress increases. And, this exactly behaves like strain hardening behavior, but beyond certain suction, value says more than 50 kilo Pascal. It exhibited strain-softening behavior it has a peak that has peak behavior peak stress and after that, it reached the ultimate. So, this ultimate stress now again keeps increasing with the increase in the matric
suction. This is a trend this is shown by the researcher’s Lu and Likos 2005 using a modified direct shear test. And, further if you generally you know these tests are conducted between τ and γ, that is shear stress and shear strain and different normal stresses. So, this is at one particular normal stress and when you conduct with different normal stress. So, this is with σ2 where σ2 is not principal stress by the way this is one particular stress one particular normal stress, and this value is more than σ1. So, then as the normal stress increases the ultimate shear stress increases. So, in this
particular case under the given normal stress as the matric suction increases the shear stress increases ultimate shear stress increases. Then, generally what will do is we in direct shear stress for saturated soils we plot τ versus σ. So, under different normal stresses when you conduct the test you get different ultimate shear stress values, then when you plot. So, you get a trend like this. So, this is the angle of internal friction ϕ, if you have intercept then you get a cohesion term. But, if it is pure sand then you will not have cohesion intercept and you will havethe only angle of internal suction value. So, what kind of a trend we get in this particular case. So, here when you plot τ on the y-axis and you plot net normal stress that is σ - ua on the
x-axis, then for s = 0 you get the same trend. So, the angle is ϕ`so, this is for s = 0 kilo Pascal. As s increases, you see that the critical shear plain critical straight line also increases. So, this is s = say 200 kilo Pascal and this is for 400 and this is for 850 kilo Pascal this s is nothing, but ua - uw sorry, this is matric suction ua - uw. So, this is the trend you get and the angle nearly reminds the same, the angle does not significantly change, the angle of internal suction value appears to be nearly constant. Does not significantly change for different matric suction values that relationship between τ versus σ - ua. So, the relationship between shear stress versus net normal stress for different matric suction values the angle of internal suction value is nearly the same. So, this is another observation and often we plot the dilatancy effect.

Video 2
So, in our shear stress test direct shear test we have 2 halves. So, this is soil and this issoil again and you have say porous plate etcetera the top and bottom. And this is connected to the rigid base then when you apply load. So, this will be the reaction and this starts moving and you connect to a dial gauge. And, you can note down what is the strain this is the shear stress that is applied, and shear strain can be determined from thisgauge. And, also apart from this, there is one normal stress applied, that is σ applied and
there is one another dial gauge is kept for recording the strain in the vertical direction. So, here you get shear strain and here you get vertical strains. So, when you have this kind of setup often we measure the dilatancy effect by plotting with shear strain, what is the normal strain we get or the vertical displacement you get, that can be observed and of positive is compression and negative is a dilation or swell or increase in the volume. So, when the volume decreases that is positive and when volume increases that is dilatancy negative. And, we often observe that the NC soils behave in this manner and OC soils are densely compacted soil exhibit this kind of behavior. So, this is for N C or normally consolidated clay or the loosely compacted sand and this is a densely compacted sand so, these soils exhibit dilation behavior. We have also seen this is because when the particles are in the loosest possible state then assume this particular configuration and when you shear. So, during shearing operation, they eventually rolled down to denser packing. So, because of which the volume decreases. So, because of this denser packing, they exhibit strain hardening behavior. And, as well as there is a volume decrease, that takes place during measuring, this is with loosely compacted sands. In the case of densely
compacted sands initially soil itself in a denser packing. So, this will be configuration, and then when you shear the soil particles have to roll up on these particles. So, this is one particle and this is another particle so, this particle has to roll up and to the other particle. And, because of which the stress increases and it reaches peak value and after that the stress drops. And, strain-softening is exhibited because of this particular behavior. And, because of this the volume of the sample increases initially you know initially small settlement that takes place after that the dilation is increase in the volume that takes place when the soil is sheared. So, this is observed in saturated soils. Now, if a test similar test conducted in unsaturated soils in the suction control test, then the behavior is shown here. So, the dilatancy is plotted on the y-axis, and on the x-axis, either normal strain or sorry shear strain γ or horizontal displacement is plotted on the xaxis. So, the dilatancy is represented so, here there is an interesting behavior for s = 0 the suction is 0 ua - uw = 0, it exhibited this behavior.And, when suction is slightly increased this is ua - uw = 0 kilo Pascal. And, the suction is slightly increased, this is the trend that exhibited this is 10 kilo Pascal. And, interestingly if you keep on increasing the dilation increases so, this is the behavior it shows.
So, this is maybe at suction value ua - uw = say 400 kilo Pascal. So, as I mentioned in the previous slide also that as the suction increases the soil exhibits strain-softening behavior. So, which is similar to that the volume of the sample increases as shear strain increases so, as horizontal displacement increases the vertical displacement observed on the thing is dilatancy dilation is observed in the soils. So, here also even though sample whatever the condition initial it is initiated when the soil is in a looser state or something it exhibited similar behavior, but when it has some suction that some water is presented in the soil something like this. And here when you
when we are shearing so, the particles have to overcome the tensile stresses are tension within the force. And, it has to overcome that because of which initially it shows an increase in the shear stresses. So, it reaches a peak value and once these surface tension bonds are broken, then it exhibits a strain-softening behavior therefore, overcome the surface tension forces in the soil force. So, it exhibits dilation behavior similar to densely packed soil so, this is a very interesting behavior that we understand from such stress. So, now, the question is how to analyze such data, to interpret the results in the field. So, the laboratory test helps us in establishing constitutive relationships such as stress versus strain behavior. And, once this behavior is very well known, then we can utilize these relationships directly to understand how the failure takes place within the soil in the field directly.
So, here we need to establish the unknown parameters, apart from cohesion. And, angle of internal friction these two is material constants and another parameter that is available that is known as the effective stress parameter. If these parameters are determined in the laboratory for a given soil under different stress state conditions, then the behavior of the soil in the field can be predicted very well.

Video 3
So, earlier have shown that the Bishop has given effective stress principle;, modified the effective stress principle for soils, unsaturated soils, which can be written as the effective stress principle given by Bishop. So, this is modified for unsaturated soils given by Bishop in 1959, which is σ`= σ - ua + χ( ua – uw). And, when you substitute in the more column equation are simply the column principle, the modified theory is τf at failure = τ = C dash + σ - ua + χ ua - uw * tan ϕ. Here the interesting thing is that the determination of χ is very complex, but at failure, the χ can be determined for non-value of the stress state. So, at failure, if we know these trusted variables, we can determine the C dash ϕ`and ψf. So, therefore, that is possible then ψf can be written as τf - C dash - σ - ua ϕ tan ϕ`divided by (ua – uw)f tan ϕ`sSo, by knowing the strength shear at failure. So, the ψf can be obtained and for different values of matrix suction can be obtained, because ψf depends on either volumetric water content or matrix section. So, ψf function in terms of
matrix action or θ can be determined from experimental measurements. So, let us see how the experiments are conducted. So, from this particular data it is evident that, when we control the σ - ua independently and ua - uw see when we control the σ - ua and ua - uw independently. And, when the soil fails at that particular state if these two parameters are known these two stress state variables are known. So, then by conducting series of tests by varying these trusted variables we can obtain ψf value.So, let us see how we do it? So, the modified direct shear or suction controlled sheared box text, which is similar to the normal shear direct shear operators, where you have two
halves, where the soil is available and you have a porous at the top
and bottom. Here soil is available and here also soil will be available. So, this is a shear plane along which the shearing takes place when load is applied and, this is a course for a stone and this is higher inter disk. And, here this whole thing kept in a airtight chamber and shear stress are applied and this one box is confined. So, therefore, the other box can be sheared and similarly you have dial gauges kept on bottom and top and there is a provision to apply σ - ua net normal stress. And, here the air can be control inside so, air pressure can be controlled and this is connected to water chamber. So, u w can be controlled through this and u a can be controlled through this. So, uw can be kept at atmospheric pressure ua can be controlled to any values so, that ua - uw inside the soil can be controlled. And under this suction control environment the shear test are conducted by varying the normal stress are the net normal stress and ua - uw within the soil mass. Set failure when
the soil fails this ua - u w at failure is know and σ - u at failure is also known. By knowing these two things we can obtain the ψf parameter let us see how I estimate. Let us solve simple example so, here the data of some synthetic data is generated for understanding how we estimate the ψf parameter these are at failure and this is also in kilo Pascal. So, 1st set of test is a test number so, 1st test is conducted by maintaining a suction value of 0. So, this is a data of sandy soil on suction control direct shear. So, when u a - u w is controlled to 0; that means, fully saturated soil, this is typically your normal direct shear test, but here you have a dryness control. So, therefore, so pore water pressure can be better control. So, the σ - ua is kept 200 kilo Pascal so, then when the σ - ua is kept 200 kilo Pascal, then the τf the shear stress at
failure is observed to be 150 kilo Pascal so, this the 1st test. And, when you conduct 2nd test by maintaining a suction value of 20 kilo Pascal and σ - ua is maintained to be 120 kilo Pascal and τf is observed to be 105. So, the 1st test is the saturated test, because, in our Bishops expression and ua - uw = 0, it boils down to expression, then again this is sandy soil. So, the dash is not present. So, only ϕ`is present and ϕ`can be now determined using the 1st test. And, subsequent tests would help in understanding how ψf changes with change in the suction value, knowing the ϕ`which is nearly constant that is an observation I had shown in the beginning that the ϕ`nearly is constant.
So, the increase in τ dash with an increase in ua - uw and the changes in τ verses σ - ua for different suction values is similar and in fact, there the critical state lines are parallel to each other. So, therefore, ϕ`could be nearly the same so, therefore, we are issuing that ϕ`remains the same and then, when suction is increased how ψf changes can be obtained. So, therefore, the tester also controlled tester also conducted in the same manner. One
test is conducted saturated and other tester conducted at unsaturated conditions. And, 50 kilo Pascal and similarly 200 500 and 800, nearly 6 tests are conducted; 6 tests are conducted under 50 kilo this is kept constant after the first test. So, because suction values varied the τf started increasing, 172 240, and slightly decrease here onwards 210 205. So, now, this is a measure data from suction control tests and this is this need to be estimated this one. The ψf value for the first test, because it is the saturated soil the σ dash = Bishop’s effective suspenseful σ - ua + ψf ua - uw in this if ψf becomes 0, this is completely dry test dry soil and ψf = 1; that means, these to get canceled and this is σ - uw this is effective suspenseful for completely saturated
soils.So, therefore, ψf needs to be 1 and this particular condition ua - uw = 0 should not be substituted in the previous equation. The previous equation is ψf = τf - C dash – (σ – ua)f tan ϕ`by ua - uw failure tan ϕ`. So, here if you substitute ua - uw = 0, then ψf becomes infinity it does not have any meaning. So, here we know that this effective suspenseful itself is that σ - uw so if you substitute directly and anyways ψf should be 1. So, we do not need to substitute here there is no ua - uw that in it. So, ψf maximum value that can take is 1 so, the maximum value is 1 ψf infinity means in fact, the maximum value is 1, that also can be (Refer Time: 35:25), but the ψf value should not be obtained from this equation by substituting ua - uw = 0.
Otherwise, if you interpret that the maximum value of ψf is 1. Therefore, ψf = infinity means ψf = 1. So, other values can be obtained by first by obtaining the ϕ`value, this is the material constant see if the ϕ`is obtained then other values of ψf with ua - uw can be obtained. So, from the first equation itself, you can estimate that τf = σ - uw. So, the effective stress this applied stress is nothing, but effective stress at a particular point. So, therefore, ϕ`= τf by σ - uw the applied stress σ - ua is nothing, but σ - uw know because of total stress, becomes effective stress when the pore water pressure completely dispute. So, therefore, ϕ`= tan inverse of 150 by 200 so, this = 36.9 degrees.
So, if this quantity is known this is the material constant, which this value is known, then other values can be obtained by varying the ua - uw that is it. So, for one particular data that is a 2nd test. This is τf is 105 - C dash is not there. So, this is a sand and σ - ua is 120 tan 36.9 degrees divided by ua - uw is 2 at failure tan 36.9 degrees so, this = 0.9923. So, this ψf value is 0.9923, when the ua - uw = 20. And, this value is 0.824 for ua - uw = 50 and 200, this value is 0.593 and at 500 this value is 0.169 at 800 this value is 0.098. So, this is how the ψf can be obtained from the modified direct shear test operate test data.

Video 4
Let us see how these values could vary? So, this is the variation of swcc this is normalized volumetric water content versus suction in kilo Pascal tell the trend is like this so, if it is plotted on a normal scale. So, then the variation of ψf is the normalized volumetric water content say varies from 0 to 1 and ψf varies from 0 to 1. So, variation is linear it could be linear like this. So, there is a problem with a pen so, this does not appear to be linear, but this is linear right. So, so there is a this is how the ψf varies with volumetric water content. If it is with suction as suction increases ψf becomes smaller and smaller unit goes to 0 0 to so, say 800 kilo Pascal or something and this is 0 to 1. When the suction is 0 this value = 1 and when suction is very high this value goes to 0. And, the experimental observation shows that this need not be linear also. In some cases, it may be non-linear and it could be varying like this or it could be varying like this etcetera. So, there are several empirical equations proposed to estimate ψf value. So, one such an empirical equation is which are given in Lu and Likos textbook are derived by several people like Khalili etcetera. One of such empirical relationship is ψf = ψm by ψe power - 0.55, when ua - u w are ψm is the matric suction value is more than ψe. And this = 1 when ψm is less than or = ψe so, what it ψe? Ψe is a suction corresponds to air entry for drawing data or air expression pressure, for pressure for wetting data so, what does it mean? When, if you have a soil characteristic curve data, if you plot on a semi-log x-axis log scale, then you will have a relationship somewhat like this. So, this is for drawing data so, this is for wetting. So, this particular point, wherever this deviates from 1, that is an air entry value this is ψe air entry and here this particular point is ψe2 so, this is for drawing this is for wetting. So, because we have seen that there are histories that exist for the soil-water characteristic curve, the drawing data exhibits a different trend compared to the wetting data. Generally, this is the initial drawing and this is the main wetting, but if you do main drying data, then it shows very good hysteresis data like this. So, this point and this point again merge this is the main drawing. So, essentially initially when normalized volumetric water content data is 1, then ψ value is 1 and if the data the normalized volumetric water content θ value is less than 1, big θ is less than 1, then ψf value takes this particular form. So, corresponding matric suction will be substituted and this is a bubbling pressure, in case of drying soil and this is air exposition pressure for the wetting soil when you
substitute power - 0.55 should give ψf. This is an empirical relationship based on several observed measurements from the suction control test.Similarly, you have another available relationship, that is a very simple second empirical relationship that is ψf = θ by θ s power k, or simply ψf = normalized volumetric water content, that is θ - θ r, by θ s - θ r. Here θ by θ s power k, when the k values change the nonlinearity between volumetric water content and ψf value change so, for example, I will plot. So, this is ψf and this is θ big θ. For k = 1 this exhibits a linear relationship for other k values for example, k = 2 it exhibits this behavior, for k = 3 non-linearity increases, this is how it varies? So, however, anyways we have very limited data from the suction control shear test, because there are several limitations associated with this test one major limitation is that the maximum suction range, would be in the range of 15 kilo 100 Pascal. If you are using, if you are controlling the suction the shear test by utilizing access
translation technique or it would be around 2000 1000 to 2000 kilo Pascal. If we are controlling the suction by controlling the suction osmotically, in an osmotic control test also the semipermeable membrane would start degrading after nearly 10 days this is research outcome by some researchers. So, because of these limitations, the equilibration time required for the soil to establish for the soil to come to one particular suction value takes time. So, within this equilibration time if the semipermeable membrane degrades, then it allows the peg solution to go into the soil. So, because of such a situation, there is a limitation on the available data, especially for fine-grain soils most of the data is available is for either sands or silt soils.
And, that to the measured shear test data in a small range of suction values. A very large range of metric suction values only recently this is attempted by controlling the suction within the shear test by (Refer Time: 48:13) vapor equilibrium techniques. And, combining all these data’s and considering the initial compaction etcetera, the test would be more complex and the analysis is very tough. So, these are the major some of the major limitations we have with test data. Apart from that as a matrix suction value increases within the soil mass the shear strength increases, but, this does not increase linearly or this cannot be directly proportional proportionality to suction value. There is because we have seen in our basic soil mechanics that thestrength behavior of sands in a completely saturated state and completely dry state is the
same. So, the factor of safety for infinite slope behavior of sands, when the soil is completely saturated or when the soil is completely dry the factor of safety remains the same, because the angle of internal friction is ϕ`only. However, due to the presence of small water film within the soil mass, there is a surface that is available which increases the strength of the soil. So, this increase in the strength that is with the increase in the matrix suction the increase in the strength of the soil is not proportional directly proportional to
the entire range of metric suction values. So, even though matrix as a matrix suction increases initially the shear strength of the soil increases, but beyond certain suction values the shear strength should drop and it comes to again back to the same value as it has initially at a fully dry state. So, such
concepts could be developed if more and more experimental data are available. So, based on the measurements we can build the effect we can develop models we can develop better constitutive models for understanding the behavior of unsaturated soils. Thank you.