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Extended Mohr-Coulomb Criterion – Triaxial Test Data

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Video 1
Hello everyone, we have been discussing the interpretation of suction controlled direct shear test data using the Extended M-C Criterion. So, let us discuss how to interpret the suction controlled triaxial test data using the extended MC criterion in this lecture. So, in the triaxial test setup, which is a suction controlled, here we independently control ua uw. So, therefore, you can control the suction and you can also control the all-round pressure and the deviatoric stress. So, then you apply the deviatoric stress so that you get the shear failure and you get the deviatoric stress at failure and pore water pressure etcetera. Then you can find the strength of the soil at different suction values using extended MC criterion so, let us solve one simple problem. So, here we have some synthetic data that is a data representation of extended triaxial test suction controlled triaxial test. So, you have series of tests conducted say test I, test 1consists of u a - u w, so where the suction is maintained to be 0 kilo pascal. So, essentially the test is conducted at a fully saturated state. And the all-round pressure, sothere is a σ2 - u a, which is nothing but which =σ 3 - u a, because this is radial applied in triaxial setup, so which =10 kilo pascal. And then the failure of the soil took place, when the major principal stress value of 60 kilo pascal. Similarly, another test is conducted in test II. Here the test is again conducted at saturated state u a - u w =0 kilo pascal and the similarly the σ 2 and σ 3 or σ 2 - u a and σ 3 - u a both are increased. So, then the failure took place when σ - 1 a 1 σ - ua of 100 kilo pascal. So, this is one set of data conducted at saturation state, so that we can get up you can obtain C and ϕ values. Another set of test data conducted say test III, where u a - u w is maintained to be 60 kilopascal and σ 2 - u a and σ 3 - u a both are at 10 kilo pascal. So, this is similar to thefirst test, except that the suction value is a non-zero or it has some suction value, the test is conducted at an unsaturated state. So, we expect that when the suction is increased, even though all-round pressure is the same, you expect to fail at a higher load, so thiswould fail at 100 kilo pascal. Similarly, another test is conducted test 4, where the suction is maintained to be constant 60 kilo pascal, but the all-round pressure is changed to the same value as test 2, so that is 30 kilo pascal. So, as the all-round pressure is increased by keeping the suction value constant, then definitely the σ1 - ua increases, so this value becomes 142 kilo pascal. So, it failed at higher major principle stress. So, higher deviatoric stress fails so, this is another set of data. So, from the first set of data, one can obtain the C value and ϕ value. And from this data, again you get C1` here C` and you get ϕ b. Now, here again, you get ϕ` so, using these two test data, you can get ϕ b. So, this is how we get, let us understand how we can solve this problem. This also can be represented in a matrix form, the test results one can be represented as σ 1 - u a, when does σ 2 - u a, σ 3 - ua this is 0, this is 0 =60, 10, 10. So, it can be represented in this manner also, this is test 1, either way, we can represent the data in either way. So, now the extended MC criterion given by Fredlund et al that is a σ 1 - u a =σ 3 - u a times tan square 45 + ϕ by 2 + 2 C` tan 45 + ϕ` by 2. Here, this is C1`; this becomes C`,when matric suction is 0, so, this is the expression we use for saturated soils also. So,when this is when you have suction in the soils, when this is defined for unsaturated soils, this is C1`, where C1` =C` + u a - u w times tan ϕ b fine. Now, first test data we can substitute the values of σ 1 - u a that becomes 60 kilo pascal and σ 3 - u a is 10 kilo pascal, then tan square 45 + ϕ` by 2 + 2 c. Here, anyways u a - u w is 0, so I substitute the value of 0, then it C1` is nothing but C`. So, this is 2 C` tan 45 + ϕ` by 2. From the 2nd test data, again we can write this one as. Solving these two equations, weget if you subtract the 1st equation from the 2nd equation, then we get ϕ value. Φ` value is nothing but 2 times tan inverse square root of 40 by 20 - 45, then this =14 19.47. So, ϕ` is 19.47, if we substitute this value in this, we get C` =14.14 kilo pascal, of course, this can be substituted into the other equation also. Either way, you get the C` the cohesion intercept, which is 14.14. Similarly, we can use the other test data test 3 data when we use, so we get 100 =10 tan square 45 + ϕ by 2 + 2 C1`, here you have an intercept because the suction value is nonzero tan 45 + ϕ` by 2. Similarly, the 4th data here is 142, which =30 tan square 45 + ϕ` by 2 + 2 C1`, these are the same C1` because, at the same suction value, the tests are conducted + ϕ` by 2.When this is solved, you get this expression so, when you solve for ϕ`, ϕ` is 20.7 degrees. And when you substitute either of these equations, then you get C1`, which =27.8 kilo pascal. So, if you observe, if you notice the ϕ` value ϕ` value from the first two test data, you got 19.47 so, from the last two test data, you got 20.7.So, when we take average value, this is 19.47 + 20.7 divided by 2, this gives a value of around 19.8 degrees and C` anyways we have 14.14. So, we got two of the strength parameters is estimated. We can estimate ϕ b dash by using those two expressions, wherethe C1` =C` + u a - u w tan ϕ b. If we utilize this expression, C1` is known that is a 27.8, C` is also known that is 14.14 + u a - u w, when the intercept is C1` is 60 kilo pascal that is into tan ϕ b. So, when wesolve this, we get ϕ b, which is nothing but 12.8 kilo pascal. So, ϕ b is less than ϕ` in this case. So, these three are the strength parameters of the extended MC criterion given by Fredlund et al., so, this is how we interpret the triaxial test data. And then you can write the expression that is a τf=(σ - u a)f tan ϕ` is 19.8 + u a - u w F tan ϕ b is 12.8 degrees + C`, C` is 14.14. So, this is the modified MC criterion for this particular soil. So, therefore, at any given stress state σ - u a and u a - u w, the strength of the soil shear strength of the soil can be estimated using this particular expression. So, the stress state of the soil is defined, then we understand the behavior. If the stress state changes with time due to some monsoon season, during monsoon season generally the water infiltrates into the ground or infiltrates into the soil slopes, then the suction values decrease. When these suction values decrease, how the strength changes can be evaluated using this particular expression. 
Video 2
So, the same thing can be solved graphically similar to what we did for direct shear test data, let us examine, how we do it. So, here in the extended MC criterion, we have all the test data given so, this is using test I and II data sets. This is at one particular u a - u w that =u a - u w is 0 kilo pascal. So, the three-dimensional diagram, there is a three-dimensional surface failure surface can be now approximated τfand this is σ - u a and the other axis is u a - u w. So, this is a three dimensional thing, we use to plot earlier. Now, because we are plotting at one particular u a - u w, so you can now consider τfhereand σ - u a here and this is at one particular u a - u w that is u a - uw in this particular case is 0 kilo pascal. So, as is 0 kilo pascal, you can directly plot it here itself, this is 0 now. So, directly you can plot it here, so that is what we have done, the τfor the shear stress on the y-axis and net normal stress σ - u a on the x-axis. So, when this is plotted, now we have two test datasets. So, two test data sets, one is 1 σ 1 - u a σ - u a is 60 kilo pascal, this is 60 kilo pascal and σ 2 - u a and σ 3 - u a both are 10 kilo pascal, so this is 10. So, this is one more circle and similarly, the other test data that is a σ 1 - u a is 100 kilo pascal and σ 2 - u a are σ 3 - u a is 30 kilo pascal, so this issecond more circle. So, you draw a line joining to joining these 2 circles so, this is the line you get and the angle of this line is ϕ` angle of internal friction. The intercept here is C` itself because u a - u w is 0 here. So, the C1` =C` that is a 14.14 or 14 kilo pascals so, this graphically we can directly get this data. Second, using 3rd and the 4th test data that is test data III and IV we can obtain shear stress versus net normal stress by recognizing the values of σ 1 - u a and σ 3 - u a. σ 3 - ua value is sorry, for this particular data that is test data III, this is a test data 3 test III. Test III data the σ 3 - u a data is 10 kilo pascal and this value is 100 kilopascal, σ 1 - u a is 100 kilo pascal. And similarly, the test IV, the σ 3 - u a is 30 and σ 1 - u a is 140 142kilopascal so, this is 140 and this is 142. So, now again we join a line to these two circles so, this is a failure plane. So, this failure plane once it is drawn, the angle is this is the angle of internal friction that is 20.5 degrees, and the intercept is C1`, this is 27 kilo pascal. So, now you got C1` and ϕ`. Now, you can take the average value of ϕ`, earlier you got a 19.47 or 4, now 20.5. So, we take the average and we get the average value nearly 20 degrees we get. So, this can be represented in this particular manner. The first test data is this test I and this is test II and this is test III and this is test IV. So, this test I and test II both are conducted at u a - u w of 0. So, this is u a - u w axis, the z-axis so, when u a - u w is 0 on this axis, this particular test data is conducted. So, τf versus σ - u a at u a - u w =0 so, this is the data so, this is the failure plane that is angle of internal friction is 19.5. Similarly, at u a - u w value of 60 kilo pascal, these tests are conducted. So, this is represented in this manner and this line is the failure plane line that is 20.5 degrees and which is nearly parallel to this line because the values are very close right. Now, as we have the values of C1` = C` from the first two test data, which is which arethat are conducted at 0 kilo pascal u a - u w. Though so the matric suction is 0 correspondingly, the values of the value of this one are 14 kilo pascal. So, corresponding to matric suction 0, the C 1 value is C1` =C 1 that is 14 kilo pascal. And corresponding to matric suction 60, the C 2 dash value is 27 kilo pascal, so another intercept. So, now if you join these two lines, you get ϕ b that is 12.4 degrees. The intercept, in fact, is anyways it is starting at matric suction 0, so you got C`. So, therefore, C` is estimated, which is 14 kilo pascal. And ϕ` is estimated, which is 20 degrees and ϕ b is estimated, which is 12.4 degrees. This data, which is obtained, graphically is very close to the analytical technique. This data is very close to the data obtained from the analytical technique. Now, knowing these three parameters estimated, now you can draw the failure surface. So, the extended more MC criterion gives a surface so, this is the surface, the pinkcoloured one is the surface failure surface. So, here earlier data which we have drawn from the test 1 and test 2 and test 3 and test 4 so, this is estimated to be 19.5, which is ϕ`. And this 20.5 is also ϕ`, you can average it and give the values of ϕ` has 20, so you can,take ϕ` as 20 in both the cases, but here it is as it is drawn. So, then here from here to here, here the intercept is 27.5 kilo pascal from this point at 60 kilo pascal u a - u w. So, this is a 27.5 kilo pascal intercept and here the intercept is 14 kilo pascal. So, a line joining these two with respect to the axis, which is parallel to this one matric suction line in between these two the line, which joins the two intercepts. And another line, which is parallel to the matric suction line, the angle made between these two lines is your ϕ b, so that ϕ b is 12.4 degrees. So, for this surface, this angle or this angle is 20 degrees that is ϕ` and this angle with respect to u a - the line parallel to u a - u w is ϕ b. So, once ϕ b is known and ϕ` is known, we can draw the failure surface. So, this is the failure surface for the extended MC criterion using triaxial test data.
Video 3
Let us solve another problem, so, here the data is summarized in this manner. So, test data and u a - u w, so which is controlled and σ 3 - u a, which is maintained and σ 1 - u a, which is observed. So, now test 1, u a - u w is maintained to be 40, and the σ 3 - u a ismaintained to be 30, and this σ 1 - u a is observed to be 100. For test II, this is again 40 u a - u w is 40 and σ 3 - u a is 80, so at higher all-round pressure the test is conducted by maintaining the same matric suction, therefore the σ 1 - u a should now increase, this value is 200.Then now, the matric suction is increased to 120 kilo pascal, and the σ 3 - u a is now 30, then σ - u a should be higher than 100, which is observed to be 180. And the IV test with the same matric suction, but increasing the all-round pressure 280, the same value as thesecond test now they should be higher than 200, because the matric suction is higher than the second case, so this is 280 so, this these are the observations. Here there is one interesting aspect that is a reason why I am discussing this particular example so; here notest is conducted at a fully saturated state. In the case if of Bishop’s effective stress, Bishop’s extended MC criterion, Bishop’s modified Mohr-Coulomb criterion to estimate c ϕ values and ψf, one needs to conduct the tests at a fully saturated state, at least two tests should be conducted. If you have cohesion intercept also, then a minimum of two tests need to be conducted, so that you can estimate cohesion and an angle of internal friction. And when these two values areestimated, other tests can be conducted at higher matric suction values, so that you can obtain ψf value at the corresponding matric suction value. However, in this case, even though you have cohesion intercept, you do not need to conduct the tests at a fully saturated state without knowing the test data at a fully saturated state, you can obtain the entire Mohr-Coulomb failure, you can obtain the failure surface. Let us see how we have obtained? So, briefly, I discussed this using these two test data sets, you can obtain C1` and ϕ`. And using these two test data sets, you can obtain C 2 dash and ϕ`. So, using these two test data, you should be able to get C` and ϕ b, let us see how we get. So, from the first two sets of data, we get when we substitute in the MC criterion, there is σ 1 - u a =σ 3 - u a tan square 45 + ϕ by 2 + 2 C1` tan 45 +. This is at one particular u a - u w, so, here when we substitute σ 1 - u a from the 2nd test data, this is 200 =80 tan square 45 + ϕ by 2 + 2 C1` tan 45 + ϕ` by 2. From the 1st test, this is 100 40 tan square 45 + ϕ` by 2 + 2 C 2 dash. After all, this is conducted at one particular suction, which is not the same as the suction value, which is used in the first two test data Ist test sorry, here the this is C1` because this is conducted at the same suction. So, now when you simplify this simply 100 and this is 40 tan square 45 + ϕ` by 2. Thus, if you solve for ϕ`, you get ϕ` =19.47 degrees, which is the same value as we obtained in the previous case. So, you can obtain the C1` C1` =14.15 kilo pascal. Just in the previous cases, you can directly substitute ϕ` will ϕ` value in one of these expressions, you can get C1`, which is 14.15 kilo pascal. Similarly, this is 1 and this 2, when you use sorry this is 2 and 1 so, this is 2, and this 1. And similarly, when you use 4, 280 =80 tan square 45 + ϕ by 2 + 2 C 2 dash, because these two are conducted at u a - u w of 40 kilo pascal. And this one is 180 30 sorry this is 30, so this is 50, so that is why this is half, and you get the same value ϕ`. So, here this is 30 tan square 45 + ϕ` by 2 + 2 C` tan 45 + ϕ` by 2. These two are u a - u w of 120 kilo pascal; this is 4th and this 3rd data. So, when we solve this, we get the same ϕ`, which is 19.47 degrees, and C 2 dash is 42.43 kilo pascal. So, from the expression, that C1` =C` + u a - u w times tan ϕ b. Using this expression, you can substitute C1` that is 14.15 =C` + u a - u w is 40 tan ϕ b. And in the secondexpression, C 2 dash is 42.43 =C` + 120 tan ϕ b. When you solve this, you get ϕ b =19.5 kilo pascal, which is the same as your ϕ, ϕ is nearly the same and the C is 0 kilo pascal.So, when the intercept this is intercepted is 0 that means, this is sand, the test data is for sand. So, you got ϕ`, which is the same as 19.45 or 5 kilo sorry degrees, sorry this is degrees. So, you have ϕ`, ϕ b dash, and C`, all these thee strength parameters could beestimated just by utilizing four test data sets. So, these test data sets are essentially we obtained by maintaining one particular suction value, two test data sets could be obtained by changing the all-round pressure and another set of data can be obtained bymaintaining different matric suction but changing the all-round pressure. So, if we have such data, we can obtain all the strength parameters estimated. 
Video 4
Quickly let us see how graphically this could be done. So, here as we have done earlier, the net normal stress that is σ - u a and τfare plotted, this is at one particular u a - u w that is u a - u w is 40 kilo pascal. So, at 40 kilo pascal, when we use the test data 1, this is for test data 1 and this is test data 2, then we obtained the angle of internal friction that is 19.5 degrees. And the intercept, C1` this is not =C` because this test is conducted at one particular matric suction value, so this is 14 kilo pascal. Then the other two test data were used, where this is test data III and this is test IV. So, the angle of internal friction is the same that is 19.5 degrees. This is C 2 dash which is not 14; here we could see that this is 42, so these are around 42.5 so, this is 42.5 kilo pascal. So, the angle of internal friction is the angle of internal friction value is the same for both the cases, but the intercept value changes from the first set of data and the second set of data because of matric suction value change changed. Now, here this is 14, and this is not 27, this is C1`, and this is C 2 dash, this is 42.5 kilo pascal and here 19.5 and this is also 19.5. Sometimes, the measurement errors would be there while plotting. And here this is a test data I, II, III, and IV so, this is if you observe,which is on an axis of this is the matric suction axis. So, this is at a value of 40 kilo pascal so, this is 40 and this is 80, 120, 160. So, hereagain at 120, so that is another matric suction value, where the tests were conducted at 120, so this is drawn so, this is how now, we locate the test data of all the four test data on the three-axis plot. Now, once we have the ϕ`, ϕ` in the cohesion intercepts, the first intercept value 14.14 kilo pascal or 14 kilo pascal, which is obtained at a matric suction value of 40. And another intercept is obtained that is 42.5 at 120 kilo pascal. So, when these two test data are plotted on τfversus matric suction plot so, when you draw the straight line joining these two points, the line passes through the origin. So, the cohesion intercepts C` =0, which is at matric suction of 0 so, this we obtain for graphically. Now, the angle of internal the angle, which indicates ϕ b is 19.5 degrees so, this data is how we get ϕ b estimated. Now, we have all the data ϕ`, ϕ b, and C`. So, now this data this line is with ϕ b sorry this line is with ϕ`, this line with horizontal makes ϕ` 19.5. And this line joining these two planes is ϕ b that is also 19.5, for this particular case, in this particular case, the ϕ` =ϕ b. So, here this line joins from the origin and at a 40 kilo pascal of matric suction, the intercept is 14. And at 120 kilo pascal matric suction, the intercept is 42.5. And this line with respect to the horizontal line, with respect to the matric suction line, parallel to this. This angle is again ϕ b, in this case, ϕ b =ϕ`. Therefore, this is a 19.5 and this line is parallel to this line these two are parallel. And again, this and these two are parallel and this is ϕ` and again this is ϕ`. So, this represents a failure envelope, which is a surface so, this particular surface can be drawn due to the model given by Fredlund et al. So, in summary, if we compare the modified M-C Criterion, which is given by Bishop, and extended Mohr-Coulomb criterion, which is given by Fredlund et al, there is a slight difference. Here the strength parameters are C`, ϕ` and ψf, which is a functional form,which is a function of u - u w. In Fredlund et al, the strength parameters are C`, ϕ`, and ϕ b. So, these are the strength parameters in Bishop and these are the strength parameters in Fredlund et al.The tests the triaxial tests or direct shear test, we need to conduct these tests to estimate these strength parameters, we require suction control direct shear test data or suction control triaxial test data in a specified manner. For example, in the case of Bishop’s model, we acquire the test data at u a - u w =0 that is at a fully saturated state. At a fully saturated state, we require 2 sets of data minimum because, from these 2 tests of test data, we can obtain C` and ϕ`. If you have sand, then anyways the cohesion intercept does not exist at a saturated state, so one test data would be sufficient. And then you need to conduct a series of tests. Test III, IV like that series of test data to obtain the values of ψf. Ψf, in this case, is 1, but other cases ψf need to be estimated ψf can be estimated using the equation expressions for different suction values. When youmaintain different suction values in different tests, you can obtain ψf value. Ψf value changes from one at a saturated state to nearly 0 in a nearly dry state.In Fredlund et al case, we do not require series of test data, we just require four test data, two test data should be conducted at one particular u a - u w, two tests need to beconducted test 1 and test 2. So, in these two tests, you can vary all-round pressure σ 3 - u a, if it is a triaxial test, or you can vary the net normal stress if it is a direct shear test. And conduct another two tests, test III and IV again by varying σ 3 - u a or net normal stress in direct shear and conduct these two tests. So, from this test, you obtain the cohesion intercept to one particular cohesion interceptat u a - u w and the angle of internal friction. And from these two tests, you get another cohesion intercept and angle of internal friction. So, by solving these two expressions for cohesion intercepts, we get C`, ϕ` and ϕ b by solving these two expressions for cohesionintercept, we get C` and ϕ b. Anyways we already got the ϕ`, so we get all the strength parameters. We do not require to conductive in the test set fully saturated state in the case of the Fredlund et al model that is extended MC criteria. So, this way, we can obtain thestrength parameters using a modified MC criterion and extended MC criterion by conducting a different set of tests under suction control in suction control triaxial or suction control direct shear test. Thank you.