 Hello, this is Dr. Mahesh Kalyanshati, Associate Professor, Department of Civil Engineering, Vulture Institute of Technology, Solaapur. In this session, we will discuss about the evaluation of shear strength of soil by laboratory vane shear test. The learning outcome will be, at the end of this session, students will be able to evaluate shear strength of soil by performing vane shear test in the laboratory. As we are aware that there are many approaches of finding the strength parameters of the soil or even the shear strength of the soil, some of the methods in the laboratory are, we can perform the direct box shear test or triaxial shear test or unconfined compression test or vane shear test. Whereas on the field, we have the methods like vane shear test, pressure meter test, static cone penetrometer test and standard penetration test. Now, out of these, today we will discuss about the vane shear test. Now, the vane shear test can be performed in the laboratory as well as on the field. So, today we will focus on the laboratory vane shear test. This is a typical setup of the laboratory vane shear test. So, as you can see in this picture, in the laboratory, we get such kind of small instrument where basically we have a rod and there is a blades connected to this rod and this particular veins are inserted into the soil specimen which are filled in this mold and then this particular rod is rotated till the failure takes place. So, such kind of setup we have in the laboratory. Also on the field, we can apply the same logic and we have the apparatus where you see a rod, torque rod and then at the bottom there are the blades and these blades are rotated with the help of the torque machine here. So, the vane shear test is a quick test either it is used in the laboratory or in the field. It is used usually for a very soft soil because in the hard soil, there is a difficulty in penetrating these blades and also there will be a difficulty in applying the torque to this rod. So, usually on a soft soil, this test is performed and it is an undrained test basically. So, we do not have any drainage kind of concept here, it is a destructive kind of test. Vane shear tester consists of four thin steel plates called veins which are welded orthogonality to steel rods. So, the another picture you can see here this particular rod at the bottom the four blades are connected orthogonal to each other. A torque measuring arrangement such as calibrated torsion spring is attached to the rod which is rotated by warm gear and warm wheel arrangement. Now after pushing the veins gently into the soil, the torque rod is rotated. So, first of all we need to fill the soil sample or we have to prepare the soil sample in this mold and then this particular torque rod is to be penetrated into the soil sample gently and torque is applied to the rod with a uniform speed of 1 degrees per minute. So, it is very slow, very small rate is used. Rotation of veins shears the soil along the cylindrical surface. So, when the rotation takes place then it shears the soil along the cylindrical surface as shown here, such kind of fill your plane is developed. Rotation of spring in degrees is indicated by a pointer moving on a graduated dial attached to the warm wheel shaft here. The torque T is then calculated by multiplying the dial reading with a spring constant. So, we have a spring constant. So, knowing the dial reading and using the spring constant we can find out the actual torque applied on the sample. A typical laboratory vein is 20 mm high and 12 mm in diameter. So, we have a different dimensions for the different setup. So, for the laboratory setup we have the vein height 20 mm, so this is what is the height of the vein it is usually of 20 mm and the diameter is almost 12 mm. So, this diameter is of 12 mm and the thickness of the blade is about 0.5 to 1 mm. Whereas, the field shear vein is from 10 to 20 cm in height. So, for field we have a bigger dimension of these blades it is 10 to 20 cm in height and from 5 to 10 cm in diameter with a thickness of about 2.5 mm. So, before we proceed let us have a review questions. So, three questions I have posted just read it. First one is the vein shear test is performed in four options. The second question is which strength parameters are determined in vein shear test again four options and the third question is in vein shear test torque rod is rotated at a uniform speed of four options. So read it carefully give the answers and then resume the video. These are the answers the vein shear test is performed in both laboratory and field. So, this is the only test which we can perform in the laboratory as well as in the field. The second question which strength parameters are determined in vein shear test now this is an experiment where we do not get the strength parameters we get directly the shear strength of the soil. Therefore, cohesion is not possible to find friction angle also is not possible to find right. So, therefore the first option is not correct second option is not correct even third option is also not applicable. So, none of the these because we are not able to find C and phi in the vein shear test. So, basically we find the shear strength. The third question in vein shear test the torque rod is rotated at a uniform speed of one degrees per minute this is the correct answer. Now let us go for the determination of shear strength. Now with reference to this particular vein of some specific sizes we will derive the equation. Let us say the tau f is the unit strength of the soil and h is the height of the vein d is the diameter of the vein and t is the torque required to cause the failure. And we can derive the equation for the torque for two different cases. First one let us assume that the top end of the vein is embedded in the soil. So, the top end of the vein is also embedded in the soil. So, it means that the top and the bottom ends are taking part in the shearing of the soil. So, we have two surfaces one is at the top another is at the bottom. So, these two surfaces are resisting the force or resisting the torque. And if this is the case then the total torque is given by this equation pi d square into tau f in bracket h by 2 plus d by 6 where we know the meaning of these terms. So, small d is the diameter of the vein and tau f is the shear strength h is the height of the blade and d is the diameter. So, from this equation or it is again clear that we can find the value of shear strength tau f we can find if I know the torque. So, the torque required to cause the failure can be determined in the laboratory by performing the test and based on whether the only one part is taking part in the shearing or both the surface is taking part in the shearing we can use the equation. Then if suppose only bottom part is taking part in the shearing it means top surface is above the soil mass only bottom surface is taking part in the shearing in that case the equation for torque is given here. So, the difference is only instead of d by 6 you have to use d by 12 here. So, again here knowing the t value we can find out the shear strength of the soil. So, here look at this equations we can get directly the shear strength. So, we are unable to find any strength parameters. Let us take an example a vein 10 centimeter long and 8 centimeter in diameter was passed into a soft clay at bottom of the borehole torque was applied and gradually increased to 45 Newton meter when failure takes place. Subsequently the vein rotated rapidly so as to completely remove the soil. The remolded soil was sheared at a torque of 18 Newton meter calculate the cohesion. So, it means they have given two conditions one is natural condition and there is a remolded condition in the natural condition we require the torque of 45 Newton meter and in the laboratory we require the torque of 18 Newton meter and in both the cases you are supposed to find what is the cohesion or shear strength in fact. So, the first case if I look at natural state. So, the torque is known to us the h is given 10 centimeter and the diameter is known to us 8 centimeter. So, using this formula which we discussed just now we get the value as tau f that is shear strength as 3.54 Newton per centimeter square and this particular shear strength is nothing but it is a cohesion because there is no resistance offered or the no strength developed because of the friction angle therefore the shear strength itself is the cohesion that is 35.4 whereas in the second case if I consider the remolded case the torque required was 1800 Newton centimeter and again the another equation we use here for that and using that equation we get the shear strength as 14.1. So, the cohesion also is 14.1. So, we can find the cohesion value or the shear strength value is different for different conditions first for the natural condition if you test the strength is 35.4 and in a remolded state if you test the strength is 14.1 kilo Newton per meter square. These are the references used for this presentation. Thank you.