 So, for these type of samples normally we do Venshia test. So, somewhere in between 2D and 3D falls the category of a Venshia test, those of you have done engineering mechanics properly will realize that it is a good example of the bearings which you talked about. So, what a Venn is like these type of soils which are very sensitive I will bring them but I cannot take out the sample from the sampler and what I will be doing is I will be keeping them in a casing or a sampler itself and then I will lower down a vane, please see the videos which I have asked you to look into. So, this is sort of a vane, this could be a laboratory vane or this could be a, this is the shearing which you are applying at the torque. So, this could be a laboratory vane or it could be a field vane. Most of the bridges in offshore environment are designed by conducting these steps by mounting the setup on a boat or a ship. So, a good example would be Bandarali ceiling or MTHL which is going on right now, it is a huge project in bombacity, thousands of Venshia tests should be done because there is no point in taking out the sample, samples are very soft. If I draw the free body diagram of this type of a system, you know, this is how it looks like, this is the blade of the vane, AA, BB, radius is R, length is L, sometimes they use H also to define this, the movement is going to occur about this axis by applying a torque. So, draw the free body diagram, this you must have done in your engineering mechanics course. This end is resting in the sample, this could also rest inside the sample or it could be outside the sample depends upon the boundary conditions. So, suppose if I assume a situation where the entire vane is contained in the sample itself and then I am rotating it, imagine you have a highly viscous fluid like honey and take a spoon and try to rotate it or the way you make coffee, alright, so you have to stir it or you have to twist it. So, this is a type of a test which is being done over here. I hope you will realize that this is how the cohesion will get mobilized. So this is the vertical phase, this is the cohesion which is getting mobilized, do not mistake this as this CV, so this will be, I will write as vertical cohesion. Units of cohesions are normally KPA or kilo Newton per meter square, soft sensitive material, so we do not use phi over here. What about this point? This is the axis of rotation, so I will assume a pressure distribution or the cohesion mobilization in this way, at the center point the cohesion will be 0 and this is the distribution of cohesion. Horizontal plane on which the cohesion is getting mobilized, so I will write this as horizontal cohesion. I cannot use the term CR because CR we have used for radial consolidation also, coefficient of radial consolidation. Now rest is simple, you have the symmetry about this point and this is how the cohesion is getting mobilized, I hope you can solve this problem very easily. You can obtain this term, the total torque required would be 2 pi R into L or H multiplied by CV, this is the force which is acting multiplied by R, so this is the component of the torque which is coming due to the mobilization of cohesion in the vertical plane and plus the horizontal plane, so there are 2 tangles. Now to solve this problem the way you did in your end bearing cases, you have studied end bearing in a mechanics course, you must have, so this is sort of end bearing, if I take a pointed chalk and if I rotate it, the cohesion mobilization at the tip is 0, so this becomes an end bearing, most of the piles, structural piles not the biological piles, so most of the piles are end bearing systems clear, so you go deep inside the ground and tamp them in the hard rock and the entire bridge can be located on the top of it. So when you are designing this type of, analyzing this type of system, what I have to do is just to clearly remind you, if this is the CHT horizontal cohesion and this is at R, I can go by analogous triangles and this is your C value, C at R and you can derive relationship between CHT upon R equal to C upon R1 or whatever alright, ideally CHT should have been equal to CT ideally, but for taking into account the heterogeneity and anisotropy of the system of the soils, we have split it in two parts. So the way you will write this term is now pi R square into C value, so C value will be C1 let us say and this is acting at R into dr, so this is your R1 into dr, this is dr thin element and multiplied by the lever arm. So lever arm at this point would be R1 correct, substitute this term and integrate it over to obtain the value, so you can substitute here C1, integrate it and then get the terms that will give you the total value of the torque. What is the advantage of doing this type of test? You are realizing there are two unknowns, CVT and this will be C1, C1 is the function of CHT. So CVT and CHT are unknowns, fine, surface area, oh you are right, so this will be I had written is this correct now, so this is a surface area R1 into dr at which C1 is acting multiplied by 2R by 3, this is the point of application of the total C value, C1, vertical part is the surface area of the fin or the vein, I have mobilized C value, so this is 2 pi R into H is the curved surface on which CV is acting multiplied by R is the moment, from this point if I take moment of this point this will be this term, yeah 2 pi R into dr correct, so this becomes your annular serving 2 pi R into dr on which C1 is acting and this is the moment. Now what you realize here is that the principal unknowns are CVT and CVHT, two unknowns, now the question is how we will solve this, so these type of tests are very trivial in the sense that at the same location you have to do two times the test, so that you get tau 1 and tau 2, so if you have two equations minimum then only you can get the value of CVT and CVR, so if you perform this test two times you have two equations, you have CVT value, CVR value you can solve this. Now one of the beauty of the systems is that if I am dealing with let us say heterogeneous system, suppose there could be a situation where there is a leryfication of the soil and I know that this is soil 2 and this is soil 1, so problem becomes more complicated now, so now you have CVT1, CVT2 and this remains your CHT and now you require three equations to obtain the heterogeneity of the system, so depending upon the situation you can perform these tests. Now there is one parameter which is to be defined here which is known as sensitivity parameter, now this is what is the ratio of the remolded sample of the soil, remolded shear strength of the soil depended by undrained shear strength or undisturbed shear strength, this is another classification scheme which is used to deal with the sensitive soils, very soft and sensitive soils, S parameter is used and typically the value of S is you know and the description, so need not to remember this when you become a designer you will get all these things in the course but just to get a feel of it when you have S equal to 1 insensitive, 1 to 4 this is low sensitive, incidentally these are the OC soils and we will describe them in details, 4 to 8 are sensitive, 8 to 16 are extra sensitive and greater than 16 vehicles sink not only in the sands, there is a possibility where this may happen in the soft sensitive clays also and there we describe the quick sand condition and remember all this description is for very soft and sensitive clays which are mostly obtained in marine environment and I am sure nowadays most of the infrastructure development is happening where in the offshore regions because good land has already been used by your grand-grandparents and you are now left with to do with the engineering in such type of deposits which are extremely difficult to understand sample train, so from this point onwards engineering with soft clays starts, this will be yeah correct, yes please take care of all this, I can give you the concepts yeah you are right, so now this is correct, so 2 r1 by 3 yeah you are applying the integration from this point to this point, so yeah the way I have defined this is I am taking only this surface correct and I am varying C from this point to this point because there is a discontinuity at this point.