Notes d'étude
Study Reminders
Support
Text Version

Extended Mohr-Coulomb Criterion - Non Linearity

We will email you at these times to remind you to study.
• Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Sunday

Video 1
Hello everyone. let us discuss some more details about the Extended MohrCoulomb Criterion given by Fredlund et al. So, here some issues related to the extended Mohr-Coulomb criterion will be discussed. This is the Extended Mohr-Coulomb Criterion given by et al. Here the τf the shear strength is written as c` + σ - ua at failure times tan ϕ`+ ua - uw at failure times tan ϕb. So, when we represent this equation graphically, τf on the y-axis and σ - ua on the x-axis.At any given suction value, you expect a linear relationship like this as angle ϕ`and this intercept is c1`, this is not c`. If the matric suction for this particular data is 0 then this is nothing, but c`. The angle of internal friction does not vary if you tested with another matrix suction, then you get another relationship with different intercept, c2` or something and this is also essentially ϕ`.So, this is with 1 ua - uw and this is with another ua - uw. So, you get another interceptc2`. So, the same equation, this equation can be represented in this manner; so, in two dimensional on τf and σ - ua at f. On the other hand, if I represent τf with respect to ua - uw with matric suction at a given σ - ua net normal stress, then I expect the relationship to be a linear and which is something like this. And the angle is ϕb at ua - uw this 0, this is nothing, but c`. This is what so, far we have discussed, and we said that, it has a great advantage in representing shear strength profile, shear strength envelop graphically and which can be understood clearly. However, there are some issues in representing this whole equation. In later, some researchers found that the ϕb is no longer constant, ϕb varies with matric suction. So, Gan et al in 1988, observed that the shear stress τf versus matric suction data varies is in this particular manner. So, interesting in that the ϕb is highly non-linear initially takes ϕb, and as the suction increases the ϕb value decreases. This ϕb at this particular pointsay b is less than at this particular point. Similarly, it observed that it nearly approaches to 0 even. And some more research in 1989 by Escario et al. So, this work is published in a conference in a twelfth international conference; Conference on International Conference on Soil Mechanics and Foundation; Foundation Engineering in 1989. The title of the work is strength and the formation of partlysaturated soils. So, in this work, we showed that when they found the shear strength of the soil with different matric suctions at a given σ - ua, the σ - ua is kept constant which is 120 kilo Pascal. For two different soils, one is red silty clay and another one is Madrid Clay, Gray Clay. So, the strength profile varied in this manner. So, this is for red silty soil, this is for Madrid Gray Clay. So, initially, it has a certain ϕb then as the suction increases the ϕb starts decreasing. Similarly, Madrid Gray Clay also exhibited the same behavior. Interestingly, ϕb becomes 0 at one particular point and even started decreasing; itbecomes negative ϕb at very high matric suction values. So, ϕb is no longer constant. Then in an earlier discussion using Bishop’s effective stress principle, when we are discussing that the shear strength can be represented as c` + (σ – ua) f tan ϕ` +χ (ua – uw) f tan ϕ`.So, this is from the Bishop and Fredlund one τf = c` + σ - ua tan ϕ` + sorry, here this should be χf (ua - uw ) tan ϕ`. So, this is ua - uw and tanϕb. If you compare these two equations, the χf ua - uw is just replaced with tanϕb and because of this the representation of the shear strength envelope is very easier. And we could see that theshear strength envelope varies with an angle ϕb when it is a represented with tau, τf versus ua - uw. And when it is represented with tau versus σ - ua, the angle of internal friction is ϕ`. So, only the representation change; however, the this essentially the Χf tan ϕ` when it got replaced with tanϕb, the Fredlund and Bishop relationship, the difference between these two relationships is simply understood by replacing Xf tan ϕ` with tanϕb. So, this is the earlier data we used for triaxial test data, to analyze the triaxial test data using Bishop’s effective stress and principle. So, where the Xf estimated for different matric suction values. So, as a matric suctionvaried from 0 to 75 kilopascal the X varied from 1 to 0.064 So, in this particular case, I get ϕb from Xf. If, I estimate ϕb from X f knowing the angle of internal friction for this particular soil; so, the angle of internal friction 34.41 which is constant which is does notvary then ϕb is nothing, but tan inverse of X f tan ϕ`, so, we write this. So, here X f is 1 then ϕb = ϕ` so, that is 34.41. So, this is the same as ϕ` and in this particular case this is 31.011 and it decreases 24.49, 15.7, 9.51, 5.56, 3.36, 3.15, and 2.51. So, the strength parameter ϕb varied from ϕ` to very small value such as 2.51. So, if we can draw this, τf on the y-axis and ua - uw on the x-axis, ϕb as τf increases this valueincreases and which becomes nearly constant or 0. So, this is how the experimental observations are also. So, therefore, the ϕb by replacing Xf tan ϕ` with tanϕb, the equation slightly got changed, but this is the same representation as the effective stress principle given by Bishop. So, not much improvement except that graphically this can be represented very well.
Video 2
So, now it is very clear that ϕb is no longer constant. So, therefore, using the Fredlund’s approach, we get the strength parameters like c` ϕ` and the ϕb which is again a functional form which is dependent on the matric suction ua - uw. So, ϕb is no longer constant, thefunction is non-linear. How to handle this nonlinearity in ϕb? Fredlund suggested that three ways, it can be resolved. Perhaps, the τf versus ua - uw, when you have this data up to air entry value. So, this ua - uw at air entry ϕb can be approximated as ϕ`. So, beyondthat this non-linear portion can be ignored and which can be replaced with a straight line. So, this can be some ϕb then using two linear portions, this is one linear zone, the linear zone from here to here and this is another linear zone this from here to here.So, this is how it can be approximated, and then ϕ` and ϕb both can be used in this manner; that is one approach in the first approach considering two linear zones. Second, he, suggested a conservative solution where you can consider from ua - uw = 0, you can consider the entire profile varies with an angle ϕb. So, this is a highly conservativesolution. So, considering ϕb throughout this is a constant approach.So, generally the observations from experiments; so, it is understood that the ϕb value nearly equals ϕ` in the beginning. As a matric suction increases the ϕb starts decreasing and it approaches 0 and even it goes to negative is very high suction values. So, therefore, Fredlund suggests that the ϕb can be replaced with ϕ` up to the air entry matric suction. And beyond that, a linear approximation can be used for the estimation of the representation of the strength envelope. The second approach he considers ϕb us constant, when you consider ϕb us constant it is less than ϕ`, so, this is a conservative approach. And the third one, he suggests considering piecewise linear. So, the entire non-linear curve can be approximated as linear by considering several segments we need to segment the profile can be considered as linear and this can be used in the design. So, a piecewise linear approach, the third one is a piecewise linear approach. Let us try to understand with a simple problem, how to address this particular problem using a piecewise linear approach? Direct Shear test data is considered for demonstrating how, how to estimate the strength, strength parameters using extended MC criterion where the nonlinearity in ϕb is considered? So, Direct Shear test data, here several tests are conducted. So, this is the test number 1, 2 like that and σ 1, σ - ua. So, this is net normal stress atfailure and these are all 1 kilopascal. And the other one is τf, this is at kilopascal. So, τf is shear stress at failure and the matric suction ua - uw at failure. So, this is also in kilopascal. So, the first data whether the net normal stress of 100 kilopascal applied and ua - uw is 10, then the measured shear stress at failure is 55 Kilopascal. Similarly, under the same condition of the same matric suction when the σ - ua is increased to 300 so, the τf value is increased to 150. Another set of data is considered, this is varied 100 and 300 only in all the test, only the matric suction is increased. Then the observed shear strengths varied as 74 and 170. And similarly, the 5th and 6th stress data is also shown. Here, also the σ - ua is varied 100 and 300 and matric suction is 100, then my τf is 98 and 193. And another set with a matric suction as 300, here this is varied 100 and 300 so, then this is 178 and 273, these are the observations and last set 11 and 12.So, this is 400 kilopascal and then this is 100 and 300. So, that τf values are 196 and 290. So, this is a data observed from suction control direct shear test data; however, this is synthetic data which is generated using certain strength parameters for demonstration. So, here considering the first two sets of data; so, this is one set of data with matric suction 10 and this is with 50, and with 100, this is with 200, this is with 300 and this is with 400. Now, considering the first set of data there is data set a then τf = c1` + σ - ua at failure times tan ϕ`. This is an expression, if you expand this c1` this is c1` + ua - uw times tanϕb. So, here anyways the ϕb is not known. Now we express this as one intercept unknown intercept which is also not known, and ϕ` is also not known; that is how we solve other state variables are known τf and σ - ua f are known. So, when you substitute form data a. So, this is a 300 = c1` + 150 tan ϕ` and this is 100 = c1` + 55 tan ϕ`. When we solve this, you get ϕ` which is tan inverse of 95 by 200, which is 25.4 degrees and c1` = 7.55 kilopascal.As this test is at ua - uw of 10 Kilopascal. So, the c1` = c`, this is one set. And similarly, using data b that is at ua - uw = 50 Kilopascal. So, this is 170 = c2`, this intercept is different because at higher matric suction value + 300 sorry, here there is a mistake inwriting τf is 150 and σ - ua is 300. And here this is 55 and this is 100 and here this is tan ϕ` and here this is 74 is strength shear strength and c2` + 100 is the net normal stress and tan ϕ`. Again this is solved ϕ`; we obtain as 25.64 which is nearly the same as earlier ϕ` and c2` = 26 Kilopascal from the second set. So, these are the one strength parameter and one intercept, we got. Similarly, using data c, we get 193 = c3` + 300 tan ϕ` and this is 98 = c3`, 100 tan phi. So, ϕ` is 25.4 and c3` is 50.5 kilopascal. And data d ϕ` is tan inverse of 234 - 140 by 200 which = 25.2 and c 4 dash = 92.83.  Similarly, the data e when we take and we consider ϕ` is 25.4 and c5`is 130.55kilopascal. Data f ϕ` is 25.2, c6` is 149 Kilopascal. So, from this you, we get an angle of internal friction which is nearly 25.4 degrees even after averaging it and which is nearly the same. So, the intercept values have changed from 7.55 Kilopascal to 149 Kilopascal.So, if this data is given for ua - uw at f and the intercept values. When it is 10 Kilopascal, this is 7.55 and this is 50 then this is 26, if this is 100 then this is 50.5. And if this is 200, this is 92.83 and this is 300 this is 130.55, and when this is 6 when 6 point that is 100,400 kilopascal suction this is 149. So, the intercept if you plot τf versus ua - uw so, thefirst intercept at 10 kilopascal small value of suction is 7.55 and then at 50, this is 26, 50.592, 130, and 149.So, essentially this curve, the envelop if you see this is how it varies. So, this is exactly the experimental data also showed that it is highly non-linear. So, initially, at the first point, this angle = ϕ`. As a suction increases the ϕ` ϕb decreases ϕb is less than ϕ` and nearly approaches to 0, right. So. So, this ϕb also can be estimated by considering two data points. Here, we consider p square is linear. So, we take two data points, this is 1, this is 10 kilopascal and this is 50 suction and you have a 100 kilopascal. So, we take different zones, in each zone, we consider that the variation of τf with ua - uw is linear. So, when this assumption is valid then we can consider two sets like this and we consider c` or c1` = c` + ua - uw times tanϕb. So, using these expressions we can estimate what is ϕb. So, when we estimate c1` is 7.55 for 10 kilopascal. So, that is c` +, so, this is 10 tanϕb. So, here two unknowns we have c` and ϕb. And from the second point, this intercept is 26, this is c` + this is 50 tanϕb. So, when we solve this, we get ϕb = 25, 24.7 degrees which = ϕ`. And when you substitute c` comes out to be 3 kilopascal a very small value, this is from 1 and 2. So, using 3 and 4, we can also use the second data set and third data set as well. So, we use this one and this one and we can use this one and this one as well. So, when we use 2 and 3, we get ϕb as tan inverse of 50.5 - 26 by 50 which = 26, just slightly higher, and c` comes out to be 1.6, this has decreased. The c` which is the cohesion intercept which should be constant got decreased. And if you take data 3 and 4 ϕb value is 22.94. So, this data we took.And this is lesser than ϕ` and c` is 8.17 kilopascal. And if we take 5 and 6, ϕb comes out to be 10.45 decreased very much, much less than ϕ`. And c` is very high 75.3 kilopascal. The interesting observation is that c` is a cohesion intercept at ua - uw = 0. So, thiscohesion intercept should be constant for any given soil; however, this is also changing along with ϕb that is, because this is non-linear. So, if I redraw this curve τf versus ua - uw, this is the non-linear behavior, we have observed. So, initially, the angle is ϕ` only and as it starts decreasing this angle is decreasing this ϕb is less than ϕ`.So, the intercept value starts increasing. So, this is the c, this is the c` and now the new the intercept is some other intercept, this is maybe c1` you can put or c 1 double dash. So, this intercept starts increasing because a ϕb is getting decreased as ϕb is getting decreased. At one particular point, the ϕb when becomes 0 at that particular point youhave the cohesion intercept very high value. So, this value is as high as 75.3 kilopascal. So, this c` is just 3 kilopascal and the c1` where ϕb is very small, where ϕb here is maybe 10.45 degrees, when this becomes 0 then this further increases the cohesion interceptbecomes very high. So, because this curve is highly non-linear when the cohesion intercept is not constant which also started varying with ϕb. So, as this ϕb is highly non-linear and we are approximating linear in a given range of matric suction values and estimating using the Extended MC Criterion by piecewise linear approach, the estimated values may be erroneous, if this is highly non-linear.
Video 3