 Hello, this is Dr. Mahesh Kallanshati, Associate Professor, Department of Civil Engineering, Valtran Institute of Technology, Swalapur. In this session, we will discuss about the evaluation of soil strength parameters by Unconfined Comparation Test. The learning outcome will be, at the end of this session, students will be able to evaluate soil strength parameters that is cohesion by performing Unconfined Comparation Test. As we are aware of that, there exist two approaches of finding the strength parameters of the soil, C and phi. The first approach is laboratory test and another one is the field test. In the laboratory test, we have direct box shear test, triaxial shear test, unconfined compression test and vane shear test. Whereas in the field, we can perform a vane shear test, pressure meter test, static cone penetrometer test and standard penetration test. Out of these tests, now today, we will discuss about the Unconfined Comparation Test. Now, the Unconfined Comparation Test is a special case of triaxial compression test, in which the confining pressure is zero. The cylindrical specimen of the soil is subjected to major principal stress sigma 1 only till the specimen fails. And this particular experiment is performed on only saturated cohesive soil. So, in the figure, you can see a typical arrangement of the loading. So, it is basically a loading frame wherein we have an arrangement of testing this soil sample. Basically, a cylindrical soil sample is kept and then the compression load is applied with the help of the loading frame. And we keep on applying this load till the failure takes place. And simultaneously, we measure the compression taking place in the soil sample. So, this is what is an explanation regarding the determination of Cu that is cohesion, which is a undrained one, because this experiment necessarily is a undrained test. So, in this case, you can see a soil sample subjected to major principal stress sigma 1 only. And we apply this sigma 1 till the failure takes place. And the failure will take place along one particular plane, which is inclined at an angle of alpha to the major principal plane. Since we do not have any confining pressure, so sigma 3 is 0. And whatever sigma 1 we obtain from the experiment, that is the vertical stress that we can use and we can draw the Mohr's circle. So, in this particular diagram, you can see a Mohr's circle is drawn, where sigma 3 is 0. So, at the origin and sigma 1 is present here. And the Mohr's circle is plotted. Then we plot the failure envelope. Now, the failure envelope of this soil will be horizontal because the soil is purely cohesive. So, it does not have any inclination. Only cohesion is present. Therefore, necessarily the failure envelope will be horizontal. So, a horizontal tangent line is drawn here, which represents the failure envelope. From this Mohr's circle, now we can derive the equation for cohesion. So, since the cohesion is nothing but the intercept of the failure envelope to the vertical axis. So, from this figure, this cohesion is equal to the radius of the Mohr's circle. And radius of the Mohr's circle is nothing but it is half of sigma 1 or half of q u, which is presented here. So, from the Mohr's circle, it is very clearly we can define or we can derive the equation for cohesion as half of unconfined compressive strength. So, this sigma 1 is also called as unconfined compressive strength. Now, this sigma u, sorry, q u that is a unconfined compressive strength is determined as a ratio of failure load to the final area. And when we perform the test, you find such kind of crack is developed. So, it is a typical failure shown in this picture. A cylindrical sample is subjected to failure. And usually this failure plane will be 45 degree because the inclination of the failure plane, we know it is 45 plus 5 by 2. But since in this case, the phi is 0, so the failure plane will have an inclination of 45 degree with the major principle plane. So, that is been proved here through this Mohr's circle. When we perform the experiment, actually there are some criterias which are to be followed. The failure criteria is considered as the failure load at which the soil sample fails or a 20 percent strain whichever occurs first. Now, determination of final area because the stress is to be determined as the load divided by the final area because in the test what happens at the time of failure, the volume or the area of the soil sample is usually higher than the initial area because of the bulging. So, we must calculate the final area at the time of failure. Now, for determining the final area, we use this ideology because our test is a undrained test. So, our assumption is that the test is undrained, it means that no volume change of the sample is taking place. Therefore, we can write a equation initial volume is equal to final volume. So, the initial volume of the soil sample will be initial area into initial length and the final volume of the sample will be final area into final length. So, again we will go for some simplification. Now, the final length of the sample is nothing, but it is the initial length minus the change in length. So, therefore, now we can write the equation A2 equal to now rearranging the terms. So, we can write the equation for final area that is called A2 is V1 upon L1 minus change in length and this V1 is nothing, but again A1 into L1 and let us divide the numerator and denominator by L1. So, we get the equation for A2 like this and in other words we can write this as A2 is equal to A1 divided by 1 minus change in length upon original length. So, this change in length upon original length is nothing, but it is a strain. So, A1 upon 1 minus strain gives you the final area. So, using this equation we find the final area. Now, the equation for cohesion can be also derived from the principal stress relationship equation. So, we know that whenever the soil mass is about to fail there exists one relationship between sigma 1 and sigma 3 and this relationship is given by this equation sigma 1 is equal to sigma 3 tan square alpha plus 2 c tan alpha. So, this 45 plus 5 by 2 is also called as alpha it is a inclination of failure plane. Now, since in our case no confining pressure is there therefore, sigma 3 is 0 and phi also is 0. So, therefore, this equation will reduce to sigma 1 is equal to 2 c because sigma 3 is 0 means what the first equation or the first term becomes 0 and in the second term since phi is 0 it is tan 45. So, tan 45 is 1. So, we get a equation sigma 1 is equal to 2 c and therefore, c is equal to sigma 1 by 2 and sigma 1 is also referred as unconfined compressive strength. So, by using the principal stress relationship also we can derive the equation for cohesion. Let us take a review questions. Look at these two MCQ questions. So, read it carefully. The first question an unconfined compression test is good for four options are given. Second question a clay soil is having unconfined compressive strength of 40 kilopascal its cohesion is four options are given. So, think over it take a pause and resume it once you get the answer. So, these are the answers the first question unconfined compression test is good for saturated cohesive soil. So, we discussed in the earlier slides that this particular test is exclusively done on cohesive soils and the second question the clay soil is having unconfined compressive strength of 40 kPa its cohesion is 20 because we know that cohesion is half of unconfined compressive strength just now we derive the equation for this. Let us take an example in an unconfined compression test a sample of clay 100 mm long and 50 mm in diameter fails under a load of 150 Newton at 10 percent strength calculate the shearing resistance taking into account the effect of change in cross section of the sample. So, in this example the test is performed on a clay sample and the dimension of the cylindrical specimen is given it is 100 mm long specimen and 50 mm diameter and the load corresponding to failure is 150 Newton and the at the time of failure 10 percent strain is taken place it means that we should consider the failure load as a failure criteria because the failure is not crossed the 20 percent strain and you have to find the shearing resistance taking into account the effect of change in cross sectional area. So, we need to find the change in cross sectional area and we have to get the cohesion. Now, we know that the final area that is at the time of failure is a 1 upon 1 minus change in length upon l. So, the initial area is pi by 4 into d square because the diameter is 50 mm. So, we got the value as 21.8 sorry 2187.7 mm square and unconfined compression strength is determined as 68.75 and then we using the equation cohesion is half of unconfined compression strength we got the value as 34.38 kilo Newton per meter square. These are the references which are used for the presentation. Thank you.