 Now, what I am drawing all the time is a triaxial element, why triaxial? You have the ground surface, you take any point over here and take out the sample, truly speaking the sample is getting exerted by the stressors in three directions, clear? So this is what is known as a triaxial state, three stressors acting on any element in the material, triaxial, very realistic situation but highly complicated to deal with because if you remember first of all we are dealing with a very complicated material like soils where we assume them as isotropic homogeneous, clear? We assumed isotropic homogeneous, you might be violating these situations, clear? So material is very complicated and then the complication comes in the form of the three dimensional analysis. So what I said is let us make life comfortable, let us try to understand 2D situations, we are dealing with the 2D situations by assuming that the effect of the stress which is perpendicular to the plane is ignored, these type of problems are known as plane strain problems, alright? So these are the plane strain problems, it is in a plane, alright? The reverse situation would be a plane stress problem, so right now we are going to deal with only plane strain problems. So if I take out this element and if I show here as sigma 1 and sigma 3, the statement of the problem is find out the state of stress at this plane which is inclined at an angle alpha of let us say 35 degree, sigma 1 is 52 kPa, sigma 3 is 12 kPa, find out the state of stress at this point which is nothing but sigma and tau, I hope you can solve this problem. Take this as an assignment, go back to your hostel, spend 20 minutes, master this, if you do not master this I can assure you, you will not follow anything next time onwards in the class. So try to solve this, the first thing you have to do is complete the Mohr's circle, you can do it very easily, sigma 1 is known, sigma 3 is known, radius is known, center is known, everything is known. Once this is known, 52, 0 is acting here, 12, 0 is acting over here, pole is known, clear? Second part of the problem, what it says is find out the state of stress at 35 degree angle, I have 2 ways to solve this problem, the state of stress is known, sigma 1, sigma 3, this point is 35 degree, what should I be doing now, where this plane is going to lie, can I project this plane on the Mohr's circle, do that and obtain the sigma and tau value. So the first stage is going to be MC, the second is going to be finding out sigma and tau, there are 2 ways of doing this now, I want you to do all these things and hence I am skipping steps, I can obtain sigma and tau by 2 methods, one is the graphical, this is what we have been doing, taking the help of the Mohr's circle, second is analytical, so this is analytical solution, sigma 1 is known, sigma 3 is known, alpha is known, sigma tau is known, clear? So if you do the analytical method which is nothing but your equations, try to match the answers and see whether you get the same answer or not, the answer you should be getting is, state of stress would be 39 kPa and 18.6 kPa, where is your guess, where the pole is, pole we have already identified, 35 degree line if I draw, this is the place where sigma and tau is acting, fine? So by definition, the state of stress has given me pole and pole is giving me a plane which is cutting the Mohr's circle where the state of stress is to be identified, this is part clear? Now most of the time in geomechanics, the problems are not going to be so simple, what is going to happen? Like in this case, if I consider this as the embankment, railway embankment and this happens to be the critical plane, you remember the definition of critical plane? At this plane, tau is getting or becoming higher than tau critical. So the failure has to happen, at this point the element looks like this, at this time this point the element looks like this, at this point the element looks like this, truly speaking what is happening is the axis is getting rotated, so we call this as the rotation of axis and what is going to happen if I am analyzing this type of situation from point 1 to 2 to 3 to n, how results are going to change? So second class of the problem which we normally talk about is, this is the first category of the problems. The second class of the problem would be if there is a complete rotation of the plane, that means this element rather than sitting horizontal, it might be sitting on a plane which is beta, rest of the things remain same, I might be having sigma 1, sigma 3 and again the statement of the problem is find out the state of stress at a plane which is let us say alpha equal to 35 degree, this is what is known as axis rotation. This plane 1, 1 has been rotated by an angle beta and the question is whether tau sigma or sigma tau is going to be same or changed when I do the same analysis for 35 degree angle. Do you think something substantial is going to change or everything remains same? I can still start with sigma 1, sigma 0, sigma 3 is 0, where these stress are going to act on the planes which are inclined to the horizontal plane now, try to solve this problem. Now let us take a situation where we call it as a generalized state of stress. So this is the C type of problem, the state of stress is defined as this is the element of the soil, you have a compressive stress acting over here, there is a tensile stress acting over here, this is a compressive stress, this is a tensile stress, so tensile stress is always negative, compressive stresses are always positive. The state of stress is you have shear stress acting at 2 MPa, the nomenclature is if I consider a point on this stress vector and if the direction of shear stress is clockwise, alright, if this is clockwise, we consider this as positive. If I consider a point over here and if I see the direction of the shear stress is anticlockwise, this becomes negative, alright. Solve the same problem and find out the state of stress, evaluate sigma and tau that is the state of stress at alpha equal to 30 degree, be evaluate sigma 1 sigma 3 when alpha equal to 30 degree, so this will come out to be take a graph paper and analyze this, this will be 6.4 minus 4.4, determine the orientation of major and minor principal stresses, in this case the orientation of the major and minor principal stresses was 0 because this happens to be the major principal stress and this happens to be the minor principal stress. The sign convention is the angles are always assumed to be positive when they are clockwise. Clockwise angles on the Mohr circle are positive, so you will be getting this as 11 degree and 101. I hope you will realize that the 2 planes are always inclined at 90 degree mutually, so 11 plus 90 would be 101, they are perpendicular planes, clear, sigma 1 sigma 3. The t problem is find the maximum shear stress, just to give you an idea the maximum shear stress acts at the point where you have this point and this point, clear, so these are the points of tau maximum, one is positive, another one is negative, the way we have shown over here, this is positive, this is negative, all right. Suppose if I ask you a question on which plane the maximum shear stress is acting, not difficult, once you have drawn the Mohr circle you get the sigma 1 sigma 3, identify the pole, you know the point at which the maximum shear is acting, this point when connected to the pole is going to give you a plane on which the state of stress is acting which is the maximum stress and its orientation would be this divided by 2 because we have been talking about 2 alpha, fine, so try to solve this. And this comes out to be 2 alpha equal to 90 degree, so alpha is 45 degree and the state of stress would be plus minus 5.4 MPa, this another class of problem which is more interesting, more practical but slightly difficult to assimilate, slightly trivial situation. So the triviality comes from the point that suppose on a tau sigma plane if I plot 2 points which are sitting very close like this that is a P1 and P2 point, all right, not very difficult but slightly trivial because you have to do something special to create a Mohr circle here what property you will be utilizing and you guess use the principle of quarts and perpendicular to the chord passing through the center, so try to solve this type of problem and the problem statement is 2 planes A and B are separated by an unknown angle, let us say this angle is theta on plane A the state of stress is sigma A equal to 10 kPa and this is 2 kPa, kPa is kilopascal, fine, 100 kilopascal is 1 kg per centimeter square, plane A lies 15 degree from the horizontal, the state of stress at point B sigma B is equal to 9 kPa and tau B equal to minus 3 kPa, the answers for this problem would be sigma 1 sigma 3 comes out to be 10.65 and 3.3 kPa and the value of theta comes out to be 46 degree, try to solve this problem, so what we have done in today's lecture is I have created a situation like this and try to explain to you what is the significance of state of stress in the material and then we have tried to answer this question by applying the concepts of Mohr's circle, alright, because as an engineer as a technologist most of the time you come across the question if this is a practical problem and if I have to construct something or provide some utility how I should be going ahead.