 See, another situation which normally we come across while designing the retention schemes is when the backfills are sloping. So far, you must have realized that we have not talked about the backfills with inclination because we did not want to defy the Rankine's theory alright. So a Rankine wall is a smooth vertical wall and the backfill is always horizontal and which is retaining the frictional material at the backfill material. So one of the deviations we did from the Rankine's earth pressure theory is that we have introduced the concept of a C5 soil keeping a vertical wall and horizontal backfill. Now we are going to deviate further from this situation where I would say find out the earth pressures for sloping backfills. A very complicated situation would be you have a sloping retaining wall and it is retaining a backfill at an inclination of I and this angle we can define this as beta. The slip surface is going to be now like this a complicated case. So if you do the analytical solutions it will be difficult for you to obtain the earth pressure coefficients k and kp for the situation and hence the pa and pp. Better would be either we go for Coulomb's method which I will be talking about later then we can create a Coulomb's block and mechanistically we can analyze this, this we will do subsequently. Another way would be go for the graphical schemes which I said in one of the lectures that they have become outdated nobody uses them due to the advent of very precise numerical codes, finite element methods and boundary element methods which people are using for solving these type of problems. There are several softwares which can give you these analysis very quickly. So let us slightly you know easy make this problem easy by assuming that still this is a ranking wall and in this ranking wall which is vertical the backfill happens to be inclined ok. So this is the initial case when the backfill was horizontal. If I would have taken an element of the soil mass very close to the wall this is how it would have looked like fractional element ok. And we would have been having sigma v, sigma h due to the condition when the backfill is horizontal. But now what has happened? The backfill has got inclined with an angle of i. So geometrical compatibility is lost and the new geometrical compatible system would look like this, element would look like this is this ok. Your sigma v is acting perpendicular to the plane not in the vertical direction this is sigma h which could be Pa or Pp all right. We like to analyze this situation and by using simple concept of axis of rotation of the principle stresses which you had studied long back. You remember we did several cases for finding out the Mohr circle. If this is the element of the soil sigma v is known, sigma h is known, this is the inclination plane theta and then you were supposed to find out the state of stress here ok at a point O which is passing through a plane inclined at an angle of theta. So you are supposed to find out sigma tau as a function of sigma v, sigma h, theta alright and we did a reverse problem also. What we did is we created a situation where the element itself was inclined at an angle of theta. So that means if this is the element of the soil which is sitting at an angle of beta and then being acted upon by sigma v, sigma h and then you are supposed to find out it could be anything it could be with respect to horizontal or I can also define the state of stress which is parallel to the surface inclination beta. So it could be anything. Now what we can do is we can use the concept of Mohr circle very easily to solve this type of problems. And you guess how? So draw the Mohr circle ok, this is the Mohr circle for frictional material. So this becomes the failure envelope, phi angle, what is the state of sigma 1, sigma 3, this is sigma 3, this is sigma 1 that means this is sigma z sorry sigma v and this is sigma h and sigma v is if this is the depth of a point from the horizontal backfill at a depth of z. So this was equal to gamma into z alright. Now what has happened? Because of rotation with eye angle the state of stress which you have shown as sigma h sigma v has got rotated. So this is the new state of stress which is going to act. The sigma v has gone from here to here and sigma h has gone from here to here that is the only difference. And this inclination is i. So what I need to find out? I need to find out let us say k, a coefficient of earth pressure under active earth pressure condition. Can I use the geometry to solve this problem alright, perpendicular from the point of tangency at the center, draw a perpendicular from this center to this chord, this is perpendicular. We can connect the chord by the radius, this is okay. We normally define this as O, C and we can take a and b as the two points here. So this is a and this is b, we can have this as b and e. The most interesting thing here to follow is when i was 0 what was the value of k, a, k value was sigma h upon sigma v correct. So that means sigma 3 divided by sigma 1 was the k value, is this alright? Now this is equal to O, a divided by O, b in the new system. Is this fine or not? Have you understood this thing? Rest is all geometry. So the concept is axis rotation can still be depicted on the Mohr circle as a rotation of the planes. The value of k will be O, a divided by O, b. And this point is clear, rest is all simple to follow. What is O, a equal to? O, a is equal to now you can write in terms of O, d, O, d minus d, a divided by O, b, O, d plus d, a and d, a is equal to d, b. So this can also be written as d, O, d minus d, b and d, b correct. From the strangle e, c what can be obtained? This is phi, O, c is common. So e, c is equal to O, c into sin of phi, is this fine? From strangle O, d, c we can obtain d, c equal to O, c sin of i. And remember here we have used the equality ad equal to d, b. e, c is nothing but the radius of the circle. So now can you substitute for O, d? What is O, d value? O, d can also be written as O, c cos of i because this angle is i, O, d equal to O, c cos of i fine and d, b equal to which is equal to ad. So how will you find out d, b? We can use another property of strangle d, b, c which is d, c square equal to d, c square plus d, b square. So d, b is nothing but d, c square minus d, c square under root of and that will be equal to d, b. So if you solve this expression what is that you are going to get? And if you substitute it over here you will be getting kA will be equal to, can you try this cos i minus under root of cos square i minus cos square phi divided by positive summation of these terms, is it not? So here this will be plus and rest of them will be same. So this is the value of k which we have obtained. Now let us check it quickly. If i is 0 what is going to happen? This will be 1 minus cos square phi over 1 minus cos square phi this becomes cos of 0 is 1, 1 minus sin phi over 1 plus. So that means whatever we have derived is alright. This is a good example of how axis of rotation can be utilized to compute the earth pressures which are acting on the walls. So we have discussed a lot of cases for finding out the earth pressure which are acting on the retaining systems. And now we are going to use these concepts to solve the problems which are happening in real life. Any questions?